course phy 201 yPy饨̲Student Name:
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19:53:46 `q001. The accepted value of the acceleration of gravity is approximately 980 cm/s^2 or 9.8 m/s^2. This will be the acceleration, accurate at most places within 1 cm/s^2, of any object which falls freely, that is without the interference of any other force, near the surface of the Earth. If you were to step off of a table and were to fall 1 meter without hitting anything, you would very nearly approximate a freely falling object. How fast would you be traveling when you reached the ground?
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RESPONSE --> 2ad= vf^2 where v0 = 0 2* 9.8m^2/s^2= vf^2 19.6 m/s^2= vf^2 vf= 4.43 m/s 4.43m/s/2= vave= 2.22 m/s vave * t = d 1 m/2.22m/s= t= .45 seconds So, I would land .45 seconds after I jumped, traveling 4.43 m/s ( or 443 cm/s)
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19:55:11 You would have an initial vertical velocity of 0, and would accelerate at 9.8 m/s^2 in the same direction as your 1 meter vertical displacement. You would also have a slight horizontal velocity (you don't step off of a table without moving a bit in the horizontal direction, and you would very likely maintain a small horizontal velocity as you fell), but this would have no effect on your vertical motion. So your vertical velocity is a uniform acceleration with v0 = 0, `ds = 1 meter and a = 9.8 m/s^2. The equation vf^2 = v0^2 + 2 a `ds contains the three known variables and can therefore be used to find the desired final velocity. We obtain vf = +- `sqrt( v0^2 + 2 a `ds) = +- `sqrt ( 0^2 + 2 * 9.8 m/s^2 * 1 m)= +- `sqrt ( 19.6 m^2 / s^2) = +- 4.4 m/s, approx. Since the acceleration and displacement were in the direction chosen as positive, we conclude that the final velocity will be in the same direction and we choose the solution vf = +4.4 m/s.
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RESPONSE --> ok
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20:06:48 `q002. If you jump vertically upward, leaving the ground with a vertical velocity of 3 m/s, how high will you be at the highest point of your jump? Note that as soon as you leave the ground, you are under the influence of only the gravitational force. All the forces that you exerted with your legs and other parts of your body to attain the 3 m/s velocity have done their work and are no longer acting on you. All you have to show for it is that 3 m/s velocity. So as soon as you leave the ground, you begin experiencing an acceleration of 9.8 m/s^2 in the downward direction. Now again, how high will you be at the highest point of your jump?
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RESPONSE --> You are asking at what point my initial velocity of 3m/s is cancelled out by my acceleration of -9.8 m/s^2. at the highest point in my jump, my velocity is 0. so my change in velocity is vf- v0= - 3m/s acceleration (-9.8 m/s^2)= dv/t -9.8m/s^2 = -3m/s/t t= -3m/s/-9.8m/m/s^2= .306 seconds. so I would be jumping up for .306 seconds at an average velocity of (3m/s + 0)/2 or 1.5 m/s. I would jump up 1.5 m/s * .306 seconds = .459 m or 45.9 cm above the ground.
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20:08:52 From the instant the leave the ground until the instant you reach your highest point, you have an acceleration of 9.8 m/s^2 in the downward direction. Since you are jumping upward, and since we can take our choice of whether upward or downward is the positive direction, we choose the upward direction as positive. You might have chosen the downward direction, and we will see in a moment how you should have proceeded after doing so. For now, using the upward direction as positive, we see that you have an initial velocity of v0 = + 3 m/s and an acceleration of a = -9.8 m/s^2. In order to use any of the equations of motion, each of which involves four variables, you should have the values of three variables. So far you only have two, v0 and a. {}What other variable might you know? If you think about it, you will notice that when objects tossed in the air reach their highest point they stop for an instant before falling back down. That is precisely what will happen to you. At the highest point your velocity will be 0. Since the highest point is the last point we are considering, we see that for your motion from the ground to the highest point, vf = 0. Therefore we are modeling a uniform acceleration situation with v0 = +3 m/s, a = -9.8 m/s^2 and vf = 0. We wish to find the displacement `ds. Unfortunately none of the equations of uniformly accelerated motion contain the four variables v0, a, vf and `ds. This situation can be easily reasoned out from an understanding of the basic quantities. We can find the change in velocity to be -3 meters/second; since the acceleration is equal to the change in velocity divided by the time interval we quickly determine that the time interval is equal to the change in velocity divided by the acceleration, which is `dt = -3 m/s / (-9.8 m/s^2) = .3 sec, approx.; then we multiply the .3 second time interval by the 1.5 m/s average velocity to obtain `ds = .45 meters. However if we wish to use the equations, we can begin with the equation vf = v0 + a `dt and solve to find `dt = (vf - v0) / a = (0 - 3 m/s) / (-9.8 m/s^2) = .3 sec. We can then use the equation `ds = (vf + v0) / 2 * `dt = (3 m/s + 0 m/s) / 2 * .3 sec = .45 m. This solution closely parallels and is completely equivalent to the direct reasoning process, and shows that and initial velocity of 3 meters/second should carry a jumper to a vertical height of .45 meters, approximately 18 inches. This is a fairly average vertical jump. If the negative direction had been chosen as positive then we would have a = +9.8 m/s^2, v0 = -3 m/s^2 (v0 is be in the direction opposite the acceleration so if acceleration is positive then initial velocity is negative) and again vf = 0 m/s (0 m/s is the same whether going up or down). The steps of the solution will be the same and the same result will be obtained, except that `ds will be -.45 m--a negative displacement, but where the positive direction is down. That is we move .45 m in the direction opposite to positive, meaning we move .45 meters upward.
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RESPONSE --> ok
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21:09:26 `q003. If you roll a ball along a horizontal table so that it rolls off the edge of the table at a velocity of 3 m/s, the ball will continue traveling in the horizontal direction without changing its velocity appreciably, and at the same time will fall to the floor in the same time as it would had it been simply dropped from the edge of the table. If the vertical distance from the edge of the table to the floor is .9 meters, then how far will the ball travel in the horizontal direction as it falls?
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RESPONSE --> so we need to know how long the ball is in the air. vertical distance is .9 m initial vertical velocity is 0 so, .9m = 1/2 (9.8m/s^2)t^2 1.8 m= 9.8m/s^2(t^2) .18m^2/s^2= t^2 .43 seconds in the air. now horizontal velocity * time= horizontal distance. 3 m/s * .43 seconds = 1.29 meters
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21:11:29 A ball dropped from rest at a height of .9 meters will fall to the ground with a uniform vertical acceleration of 9.8 m/s^2 downward. Selecting the downward direction as positive we have `ds = .9 meters, a = 9.8 m/s^2 and v0 = 0. Using the equation `ds = v0 `dt + .5 a `dt^2 we see that v0 = 0 simplifies the equation to `ds = .5 a `dt^2, so `dt = `sqrt( 2 `ds / a) = `sqrt(2 * .9 m / (9.8 m/s^2) ) = .42 sec, approx.. Since the ball rolls off the edge of the table with only a horizontal velocity, its initial vertical velocity is still zero and it still falls to the floor in .42 seconds. Since its horizontal velocity remains at 3 m/s, it travels through a displacement of 3 m/s * .42 sec = 1.26 meters in this time.
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RESPONSE --> we had a difference in significant figures.....
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}xƝYΐk` Student Name: ambler assignment #008
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z{jצҵEɜԪwӘ Student Name: assignment #009
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22:48:41 `q001. Note that there are 10 questions in this set. .You have done the Introductory Force Experiment in which you used rubber bands and bags of water, and you understand that, at least in the vicinity of the Earth's surface, gravity exerts downward forces. You have also seen that forces can be measured in units called Newtons. However you were not given the meaning and definition of the Newton as a unit of force. You are also probably aware that mass is often measured in kilograms. Here we are going to develop in terms of an experiment the meaning of the Newton as a force unit. Suppose that a cart contains 25 equal masses. The cart is equal in mass to the combined total of the 25 masses, as indicated by balancing them at equal distances from a fulcrum. The cart is placed on a slight downward incline and a weight hanger is attached to the cart by a light string and suspended over a low-friction pulley at the end of the ramp. The incline is adjusted until the cart, when given a slight push in the direction of the hanging weight, is observed to move with unchanging, or constant, velocity (and therefore zero acceleration). The masses are then moved one at a time from the cart to the hanger, so that the system can be accelerated first by the action of gravity on one of the masses, then by the action of gravity onto of the masses, etc.. The time required for the system to accelerate from rest through a chosen displacement is observed with one, two, three, four, five, ... ten of the masses. The acceleration of the system is then determined from these data, and the acceleration is graphed vs. the proportion of the total mass of the system which is suspended over the pulley. It is noted that if the entire mass of the system, including the cart, is placed on the weight hanger, there will be no mass left on the incline and the entire weight will fall freely under the acceleration of gravity. Suppose the data points obtained for the 5 of the first 10 trials were (.04, 48 cm/s^2), (.08, 85 cm/s^2), (.12, 125 cm/s^2), (.16, 171 cm/s^2), (.20, 190 cm/s^2). Sketch these points on an accurate graph of acceleration vs. proportion of weight suspended and determine the slope and y-intercept of the line. What is your slope and what is the y intercept? What is the equation of the line?
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RESPONSE --> I did not get perfectly linear data (although it came very close). My best fit line had a y intercept of 15 cm/s^2. the slope was (190-48)/(2.0-.04)= (142/.16) 887.5 therefore the line is y= 887.5 mp + 15 cm/s^2 to check, using Mass Proportion .12 y = 887.5 * (.12) + 15 cm/s^2 This gives me a y value of 121.5 cm/s^2 this is in fact a point on my best fit line.
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23:02:52 Since there are 25 equal masses and the mass of the cart is equivalent to another 25 of these masses, each mass is 1/50 = .02 of the total mass of the system. Thus the first 10 data points should might have been something like (.02, 21 cm/s^2), (.04, 48 cm/s^2), (.06, 55 cm/s^2), (.08, 85 cm/s^2), (.10, 101 cm/s^2), (.12, 125 cm/s^2), (.14, 141 cm/s^2), (.16, 171 cm/s^2), (.18, 183 cm/s^2), (.20, 190 cm/s^2). The data given in the problem would correspond to alternate data points. The slope of the best-fit line is 925 x + 12.8, indicating a slope of 925 and a y intercept of 12.8. The 967.5 is in units of rise / run, or for this graph cm/s^2. If you calculated the slope based on the points (.04, 48 cm/s^2) and (.20, 190 cm/s^2) you would have obtained 151 cm/s^2 / (.16) = 950 cm/s^2. Whether this is close to the best-fit value or not, this is not an appropriate calculation because it uses only the first and last data points, ignoring all data points between. The idea here is that you should sketch a line that fits the data as well as possible, then use the slope of this line, not the slope between data points.
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RESPONSE --> 190 - 142= 48 ..... my best fit line included my first data point, last data point and my (.08, 85 cm^2) data point... using two middle points, I get (155-85)/(.16- .08)= 70/.08= 875 for my slope...... I still do not get a slope > 900......
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23:07:49 `q002. Do the points seem to be randomly scattered around the straight line or does there seem to be some nonlinearity in your results?
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RESPONSE --> it actually seems to increase in a decreasing manner at the end.
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23:16:16 The slope of your line should probably be somewhere between 900 cm/s^2 and 950 cm/s^2. The points should be pretty much randomly scattered about the best possible straight line. Careful experiments of this nature have shown that the acceleration of a system of this nature is to a very high degree of precision directly proportional to the proportion of the weight which is suspended.
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RESPONSE --> ok, I can see the randomness, and by redrawing a best fit line I get a slope of 916.7 cm/s^2
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23:20:40 `q003. If the acceleration of the system is indeed proportional to the net force on the system, then your straight line should come close to the origin of your coordinate system. Is this the case?
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RESPONSE --> My y intercept is still at about 11 cm/s^2..... what are you considering close? compared to 190 cm/s^2 it is :>.
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23:21:25 If the acceleration of the system is proportional to the net force, then the y coordinate of the straight line representing the system will be a constant multiple of the x coordinate--that is, you can always find the y coordinate by multiplying the x coordinate by a certain number, and this 'certain number' is the same for all x coordinates. The since the x coordinate is zero, the y coordinate will be 0 times this number, or 0. Your graph might not actually pass through the origin, because data inevitably contains experimental errors. However, if experimental errors are not too great the line should pass very close to the origin. In the case of this experiment the y-intercept was 12.8. On the scale of the data used here this is reasonably small, and given the random fluctuations of the data points above and below the straight-line fit the amount of deviation is consistent with a situation in which precise measurements would reveal a straight line through the origin.
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RESPONSE --> like I said, how do you define close!
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15:55:52 `q003. What is it that causes the system to accelerate more when a greater proportion of the mass is suspended?
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RESPONSE --> the suspended mass is accelerating due to gravity, while the mass that is not suspended is acting as an opposing force.
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15:56:57 The gravitational forces exerted on the system are exerted on the suspended masses and on the cart and the masses remaining in it. The supporting force exerted by the ramp counters the force of gravity on the cart and the masses remaining in it, and this part of the gravitational force therefore does not affect the acceleration of the system. However there is no force to counter the pull of gravity on the suspended masses, and this part of the gravitational force is therefore the net force acting on the mass of the entire cart-and-mass system. The force exerted by gravity on the suspended masses is proportional to the number of suspended masses--e.g, if there are twice as many masses there is twice the force. Thus it is the greater gravitational force on the suspended masses that causes the greater acceleration.
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RESPONSE --> i think i understand....
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23:07:15 `q004. This results of this sort of experiment, done with good precision, support the contention that for a given mass the acceleration of a system is indeed proportional to the net force exerted on the system. Other experiments can be done using rubber bands, springs, fans and other nongravitational sources of force, to further confirm this result. In another sort of experiment, we can see how much force is required on different masses to obtain a certain fixed acceleration. In such experiments we find for example that if the mass is doubled, it requires twice the force to achieve the same acceleration, and that in general the force required to achieve a given acceleration is proportional to the amount of mass being accelerated. In a certain experiment using the same cart and masses as before, plus several additional identical carts, a single cart is accelerated by a single suspended mass and found to accelerate at 18 cm/s^2. Then a second cart is placed on top of the first and the two carts are accelerated by two suspended masses, achieving an acceleration of 20 cm / s^2. Then a third cart is placed on top of the first to and the three carts are accelerated by three suspended masses, achieving and acceleration of 19 cm/s^2. A fourth cart and a fourth suspended mass are added and an acceleration of 18 cm/s^2 is obtained. Adding a fifth cart in the fifth suspended mass an acceleration of 19 cm/s^2 is obtained. All these accelerations are rounded to the nearest cm/s^2, and all measurements are subject to small but significant errors in measurement. How well do these results indicate that to achieve a given acceleration the amount of force necessary is in fact proportional to the amount of mass being accelerated?
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RESPONSE --> assuming identical carts and masses m1= 18 cm/s^2 2m1 20 cm/s^2 3m1 = 19 cm/s^2 4m1 = 18 cm/s^2 5m1= 19 cm/s^2 the average acceleration is 18.8 cm/s^2 or 19 cm/s/s. for each unit x being accelerated a mass of unit y must be added to keep acceleration consistant. this would indicate that force must be proportional to mass for consistant acceleration values.
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23:07:41 The accelerations obtained are all about the same, with only about a 10% variation between the lowest and the highest. Given some errors in the observation process, it is certainly plausible that these variations are the result of such observation errors; however we would have to have more information about the nature of the observation process and the degree of error to be expected before drawing firm conclusions. If we do accept the conclusion that, within experimental error, these accelerations are the same then the fact that the second through the fifth systems had 2, 3, 4, and 5 times the mass of the first with 2, 3, 4, and 5 times the suspended mass and therefore with 2, 3, 4, and 5 times the net force does indeed indicate that the force needed to achieve this given acceleration is proportional to the mass of the system.
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RESPONSE --> ok
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23:18:05 `q005. Now we note again that the force of gravity acts on the entire mass of the system when an entire system is simply released into free fall, and that this force results in an acceleration of 9.8 m/s^2. If we want our force unit to have the property that 1 force unit acting on 1 mass unit results in an acceleration of 1 m/s^2, then how many force units does gravity exert on one mass unit?
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RESPONSE --> If 1 force unit accelerates 1 mass unit by 1 m/s^2, and gravity accelerates 1 mass unit by 9.8 m/s^2, then f*m=a m= 1 a= 9.8m/s^2 so gravity is exerting 9.8 force units on the mass.
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23:18:17 Since gravity gives 1 mass unit an acceleration of 9.8 m/s^2, which is 9.8 times the 1 m/s^2 acceleration that would be experienced from 1 force unit, gravity must exerted force equal to 9.8 force units on one mass unit.
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RESPONSE --> yes!!
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23:18:49 `q006. If we call the force unit that accelerates 1 mass unit at 1 m/s^2 the Newton, then how many Newtons of force does gravity exert on one mass unit?
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RESPONSE --> 9.8 newtons of force is exerted by gravity on 1 mass unit
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23:18:57 Since gravity accelerates 1 mass unit at 9.8 m/s^2, which is 9.8 times the acceleration produced by a 1 Newton force, gravity must exert a force of 9.8 Newtons on a mass unit.
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RESPONSE --> yes
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23:20:04 `q007. The mass unit used here is the kilogram. How many Newtons of force does gravity exert on a 1 kg mass?
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RESPONSE --> again, mass * acceleration= force. acceleration is 9.8 m/s^2 * 1 kg mass= 9.8 newtons force
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23:20:12 Gravity exerts a force of 9.8 Newtons on a mass unit and the kg is the mass unit, so gravity must exert a force of 9.8 Newtons on a mass of 1 kg.
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RESPONSE --> yes
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23:23:19 `q008. How much force would gravity exert on a mass of 8 kg?
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RESPONSE --> 8 kg * 9.8 m/s^2= 78.4 kg*m/s^2 (newtons)
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23:23:38 Gravity exerts 8 times the force on 8 kg as on 1 kg. The force exerted by gravity on a 1 kg mass is 9.8 Newtons. So gravity exerts a force of 8 * 9.8 Newtons on a mass of 8 kg.
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RESPONSE --> ok
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23:25:34 `q009. How much force would be required to accelerate a mass of 5 kg at 4 m/s^2?
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RESPONSE --> 5kg * 4 m/s^2= force 20 newtons
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23:26:16 Compared to the 1 Newton force which accelerates 1 kg at 1 m/s^2, 2e have here 5 times the mass and 4 times the acceleration so we have 5 * 4 = 20 times the force, or 20 Newtons. We can formalize this by saying that in order to give a mass m an acceleration a we must exert a force F = m * a, with the understanding that when m is in kg and a in m/s^2, F must be in Newtons. In this case the calculation would be F = m * a = 5 kg * 4 m/s^2 = 20 kg m/s^2 = 20 Newtons. The unit calculation shows us that the unit kg * m/s^2 is identified with the force unit Newtons.
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RESPONSE --> I noted earlier that a newton is kg*m/s^2 ......
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23:27:27 `q010. How much force would be required to accelerate the 1200 kg automobile at a rate of 2 m/s^2?
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RESPONSE --> 1200 kg * 2m/s^2 (mass * acceleration...) = x newtons 1400 kg * m/s^2 = force 1400 newtons
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23:28:03 This force would be F = m * a = 1200 kg * 2 m/s^2 = 2400 kg * m/s^2 = 2400 Newtons.
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RESPONSE --> I wrote down the right equation, I just didn't do it!!
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bUIĎqX Student Name: assignment #011
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13:37:06 `q001. A cart on a level, frictionless surface contains ten masses, each of mass 2 kg. The cart itself has a mass of 30 kg. A light weight hanger is attached to the cart by a light but strong rope and suspended over the light frictionless pulley at the end of the level surface. Ignoring the weights of the rope, hanger and pulley, what will be the acceleration of the cart if one of the 2 kg masses is transferred from the cart to the hanger?
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RESPONSE --> okay, the force going forward is 2kg*9.8m/s^2, or 19.6 N (that is on the mass suspended). the original mass of the cart is 30 kg + 10*2 kg or 50 kg, therefore, when we move one 2 kg mass it becomes 48 kg. since force = m* a, then f/m = acceleration in this case 19.6 n/50 kg (altogether) = .392 m/s^2
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13:38:09 At the surface of the Earth gravity, which is observed here near the surface to accelerate freely falling objects at 9.8 m/s^2, must exert a force of 2 kg * 9.8 m/s^2 = 19.6 Newtons on the hanging 2 kg mass. This force will tend to accelerate the system consisting of the cart, rope, hanger and suspended mass, in the 'forward' direction--the direction in which the various components of the system must accelerate if the hanging mass is to accelerate in the downward direction. This is the only force accelerating the system in this direction. All other forces, including the force of gravity pulling the cart and the remaining masses downward and the normal force exerted by the level surface to prevent gravity from accelerating the cart downward, are in a direction perpendicular to the motion of these components of the system; furthermore these forces are balanced so that they add up to 0. The mass of the system is that of the 30 kg cart plus that of the ten 2 kg masses, a total of 50 kg. The net force of 19.6 Newtons exerted on a 50 kg mass therefore results in acceleration a = Fnet / m = 19.6 Newtons / (50 kg) = 19.6 kg m/s^2 / (50 kg) = .392 m/s^2.
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RESPONSE --> yes!!! that took a few minutes to figure out even using your stuff AND the book.
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13:51:05 `q002. Two 1-kg masses are suspended over a pulley, one on either side. A 100-gram mass is added to the 1-kg mass on one side of the pulley. How much force does gravity exert on each side of the pulley, and what is the net force acting on the entire 2.1 kg system?
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RESPONSE --> think of 1 kg mass without addition as side a the 1 kg mass with additional .1 kg is side b gravity is 9.8 m/s^2 and Force is measured in Newtons which is kg*m/s^2 side a has a net force due to gravity of 9.8m/s^2 * 1 kg, or 9.8 N side b has a net force due to gravity of 9.8 m/s^2 * 1.1 kg or 10.78 N If we think of this as an atwood machine, then the net force is simply fnet = fnetb-fneta or 1.02 newtons (- newtons if we think of down as negative.......)
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13:52:18 The 1-kg mass experiences a force of F = 1 kg * 9.8 m/s^2 = 9.8 Newtons. The other side has a total mass of 1 kg + 100 grams = 1 kg + .1 kg = 1.1 kg, so it experiences a force of F = 1.1 kg * 9.8 Newtons = 10.78 Newtons. {}{}Both of these forces are downward, so it might seem that the net force on the system would be 9.8 Newtons + 10.78 Newtons = 20.58 Newtons. However this doesn't seem quite right, because when one mass is pulled down the other is pulled up so in some sense the forces are opposing. It also doesn't make sense because if we had a 2.1 kg system with a net force of 20.58 Newtons its acceleration would be 9.8 m/s^2, and we know very well that two nearly equal masses suspended over a pulley won't both accelerate downward at the acceleration of gravity. So in this case we take note of the fact that the to forces are indeed opposing, with one tending to pull the system in one direction and the other in the opposite direction. {}{}We also see that we have to abandon the notion that the appropriate directions for motion of the system are 'up' and 'down'. We instead take the positive direction to be the direction in which the system moves when the 1.1 kg mass descends. {}{}We now see that the net force in the positive direction is 10.78 Newtons and that a force of 9.8 Newtons acts in the negative direction, so that the net force on the system is 10.78 Newtons - 9.8 Newtons =.98 Newtons. {}{}The net force of .98 Newtons on a system whose total mass is 2.1 kg results in an acceleration of .98 Newtons / (2.1 kg) = .45 m/s^2, approx.. Thus the system accelerates in the direction of the 1.1 kg mass at .45 m/s^2.
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RESPONSE --> again, that adding and subtracting thing......
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14:02:43 `q003. If in the previous question there is friction in the pulley, as there must be in any real-world pulley, the system in the previous problem will not accelerate at the rate calculated there. Suppose that the pulley exerts a retarding frictional force on the system which is equal in magnitude to 1% of the weight of the system. In this case what will be the acceleration of the system, assuming that it is moving in the positive direction (as defined in the previous exercise)?
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RESPONSE --> so the retarding force is 1% of the entire weight of the system, which is .01 * 2.1kg or .021N Now the force on the system is 10.78 N - .021N - 9.8N= .959 N .959N/2.1kg = .46 m/s^2
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14:06:27 We first determine the force exerted a friction. The weight of the system is the force exerted by gravity on the mass of the system. The system has mass 2.1 kg, so the weight of the system must be 2.1 kg * 9.8 m/s^2 = 20.58 Newtons. 1% of this weight is .21 Newtons, rounded off to two significant figures. This force will be exerted in the direction opposite to that of the motion of the system; since the system is assume to be moving in the positive direction the force exerted by friction will be frictional force = -.21 Newtons. The net force exerted by the system will in this case be 10.78 Newtons - 9.8 Newtons - .21 Newtons = .77 Newtons, in contrast with the .98 Newton net force of the original exercise. The acceleration of the system will be .77 Newtons / (2.1 kg) = .35 m/s^2, approx..
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RESPONSE --> I see, I did not read the question correctly-- when you said force was equal in magnitude to 1% of weight, I assumed you meant the percentage WAS the force.
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14:08:22 `q004. Why was it necessary in the previous version of the exercise to specify that the system was moving in the direction of the 1.1 kg mass. Doesn't the system have to move in that direction?
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RESPONSE --> The system does not have to move in the direction of the heavier mass if other forces are being applied to the system........
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14:08:45 If the system is released from rest, since acceleration is in the direction of the 1.1 kg mass its velocity will certainly always be positive. However, the system doesn't have to be released from rest. We could give a push in the negative direction before releasing it, in which case it would continue moving in the negative direction until the positive acceleration brought it to rest for an instant, after which it would begin moving faster and faster in the positive direction.
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RESPONSE --> ok
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22:27:59 `q005. If the system of the preceding series of exercises is initially moving in the negative direction, then including friction in the calculation what is its acceleration?
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RESPONSE --> The mass and acceleration due to gravity is still the same, as is the force applied by friction. It will still be .35 m/s^2 it's just that at first it will be negative acceleration....
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22:36:57 Its acceleration will be due to the net force. This net force will include the 10.78 Newton force in the positive direction and the 9.8 Newton force in the negative direction. It will also include a frictional force of .21 Newtons in the direction opposed to motion. Since motion is in the negative direction, the frictional force will therefore be in the positive direction. The net force will thus be Fnet = 10.78 Newtons - 9.8 Newtons + .21 Newtons = +1.19 Newtons, in contrast to the +.98 Newtons obtained when friction was neglected and the +.77 Newtons obtained when the system was moving in the positive direction. , The acceleration of the system is therefore , 1.19 N / (2.1 kg) = .57 m/s^2.
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RESPONSE --> oh, ok..... I didn't take the directions of force into account-----
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23:27:56 `q006. If friction is neglected, what will be the result of adding 100 grams to a similar system which originally consists of two 10-kg masses, rather than the two 1-kg masses in the previous examples?
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RESPONSE --> Again, this is an atwood machine where net force is determined by subtracting force on mass a from force on mass b. the force being applied by the acceleration due to gravity on the mass. f(a)= 9.8 m/s^2 * 10.1 kg= 98.98 n f(b) = 9.8m/s^2 * 10 = 98 n fnet = (9.8 m/s^2* 10.1 kg) - (9.8 m/s^2 * 10 kg) 98.98 N - 98 N = .98 n since the mass of the system is 20.1kg, then Acc = .98 N/20.1 kg= .049 m/s^2 Obviously this is a slower acceleration than in the first system. This makes since, as a percentage, the mass added to the second side is much smaller than that of the first system. 9.8 N will be the net force on the system. since the total mass is 20.1 kg
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23:28:21 In this case the masses will be 10.1 kg and 10 kg. The force on the 10.1 kg mass will be 10.1 kg * 9.8 m/s^2 = 98.98 Newtons and the force on the 10 kg mass will be 10 kg * 9.8 m/s^2 = 98 Newtons. The net force will therefore be .98 Newtons, as it was in the previous example where friction was neglected. We note that this.98 Newtons is the result of the additional 100 gram mass, which is the same in both examples. The mass in this case will be 20.1 kg, so that the acceleration of the system is a = .98 Newtons / 20.1 kg = .048 m/s^2, approx.. The same .98 Newton net force acting on the significantly greater mass results in a significantly smaller acceleration.
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RESPONSE --> yes.
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23:42:32 `q007. If friction is not neglected, what will be the result for the system with the two 10-kg masses with .1 kg added to one side? Note that by following what has gone before you could, with no error and through no fault of your own, possibly get an absurd result here, which will be repeated in the explanation then resolved at the end of the explanation.
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RESPONSE --> friction is an opposing force to the downward acceleration of the system (well, down on one side, up on the other......) If friction is once again 1 % of system weight, then 20.1 kg * 9.8 m/s^2= 196.98 N * 1% = 1.97 N force applied due to friction. So the final net force would be 98.98 N - 1.97 N - 98 N= - .99 N This would seemingly end in the lighter side accelerating downward...... however, if we remember that friction opposes ANY movement, then in this case we simply conclude that the force on the system is not enough to overcome the opposing force of friction and that the system does not accelerate due to gravity in this instance.
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00:00:16 If friction is still equal to 1% of the total weight of the system, which in this case is 20.1 kg * 9.8 m/s^2 = 197 Newtons, then the frictional force will be .01 * 197 Newtons = 1.97 Newtons. This frictional force will oppose the motion of the system. For the moment assume the motion of the system to be in the positive direction. This will result in a frictional force of -1.97 Newtons. The net force on the system is therefore 98.98 Newtons - 98 Newtons - 1.97 Newtons = -.99 Newtons. This net force is in the negative direction, opposite to the direction of the net gravitational force. If the system is moving this is perfectly all right--the frictional force being greater in magnitude than the net gravitational force, the system can slow down. Suppose the system is released from rest. Then we might expect that as a result of the greater weight on the positive side it will begin accelerating in the positive direction. However, if it moves at all the frictional force would result in a -.99 Newton net force, which would accelerate it in the negative direction and very quickly cause motion in that direction. Of course friction can't do this--its force is always exerted in a direction opposite to that of motion--so friction merely exerts just enough force to keep the object from moving at all. Friction acts as though it is quite willing to exert any force up to 1.97 Newtons to oppose motion, and up to this limit the frictional force can be used to keep motion from beginning. In fact, the force that friction can exert to keep motion from beginning is usually greater than the force it exerts to oppose motion once it is started.
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RESPONSE --> exactly
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00:14:31 `q008. A cart on an incline is subject to the force of gravity. Depending on the incline, some of the force of gravity is balanced by the incline. On a horizontal surface, the force of gravity is completely balanced by the upward force exerted by the incline. If the incline has a nonzero slope, the gravitational force (i.e., the weight of the object) can be thought of as having two components, one parallel to the incline and one perpendicular to the incline. The incline exerts a force perpendicular to itself, and thereby balances the weight component perpendicular to the incline. The weight component parallel to the incline is not balanced, and tends to accelerate the object down the incline. Frictional forces tend to resist this parallel component of the weight and reduce or eliminate the acceleration. A complete analysis of these forces is best done using the techniques of vectors, which will be encountered later in the course. For now you can safely assume that for small slopes (less than .1) the component of the gravitational force parallel to the incline is very close to the product of the slope and the weight of the object. [If you remember your trigonometry you might note that the exact value of the parallel weight component is the product of the weight and the sine of the angle of the incline, that for small angles the sine of the angle is equal to the tangent of the angle, and that the tangent of the angle of the incline is the slope. The product of slope and weight is therefore a good approximation for small angles or small slopes.] What therefore would be the component of the gravitational force acting parallel to an incline with slope .07 on a cart of mass 3 kg?
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RESPONSE --> It has been a while since I took trig, but this does make sense..... 3 kg * 9.8 m/s^2= 29.4 kg m/s^2 or 29.4 N, so the gravitational component of the force acting on the cart would be .07 (slope) * 29.4 N (weight) = 2.06
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00:16:30 The gravitational force on a 3 kg object is its weight and is equal to 3 kg * 9.8 m/s^2 = 29.4 Newtons. The weight component parallel to the incline is found approximately as (parallel weight component) = slope * weight = .07 * 29.4 Newtons = 2.1 Newtons (approx.).. STUDENT COMMENT: It's hard to think of using the acceleration of 9.8 m/s^2 in a situation where the object is not free falling. Is weight known as a force measured in Newtons? Once again I'm not used to using mass and weight differently. If I set a 3 kg object on a scale, it looks to me like the weight is 3 kg.INSTRUCTOR RESPONSE: kg is commonly used as if it is a unit of force, but it's not. Mass indicates resistance to acceleration, as in F = m a. {}{}eight is the force exerted by gravity. {}{}The weight of a given object changes as you move away from Earth and as you move into the proximity of other planets, stars, galaxies, etc.. As long as the object remains intact its mass remains the same, meaning it will require the same net force to give it a specified acceleration wherever it is.{}{}An object in free fall is subjected to the force of gravity and accelerates at 9.8 m/s^2. This tells us how much force gravity exerts on a given mass: F = mass * accel = mass * 9.8 m/s^2. This is the weight of the object. **
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RESPONSE --> ok. actually, my problem is units--- what units were we measuring slope in? cm/cm? cm/m? how would I note that as a product with newtons?
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00:20:26 `q009. What will be the acceleration of the cart in the previous example, assuming that it is free to accelerate down the incline and that frictional forces are negligible?
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RESPONSE --> Since the mass is 3 kg and the force acting on the cart (parrallel to the slope) is 2.1 kg/m/s^2 (I used your answer), then acceleration is 2.1 kg/m/s^2/ 3kg (f/m=a) the acceleration would be about .7 m/s^2
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00:20:39 The weight component perpendicular to the incline is balanced by the perpendicular force exerted by the incline. The only remaining force is the parallel component of the weight, which is therefore the net force. The acceleration will therefore be a = F / m = 2.1 Newtons / (3 kg) = .7 m/s^2.
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RESPONSE --> ok
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00:27:08 `q010. What would be the acceleration of the cart in the previous example if friction exerted a force equal to 2% of the weight of the cart, assuming that the cart is moving down the incline? [Note that friction is in fact a percent of the perpendicular force exerted by the incline; however for small slopes the perpendicular force is very close to the total weight of the object].
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RESPONSE --> 3 kg = mass weight = 9.8 m/s^2 * 3= 29.4 N friction force = 2% * 29.4= .59N 29.4 n * .07 = 2.06 Newtons parallel to cart , perpendicular force cancels out. 2.06 newons down slope - .59 n friction = 1.47 N down slope. acceleration = f/m 1.47 kg/m/s^2/ 3kg= .49 m/s^2
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00:29:06 The weight of the cart was found to be 29.4 Newtons. The frictional force will therefore be .02 * 29.4 Newtons = .59 Newtons approx.. This frictional force will oppose the motion of the cart, which is down the incline. If the downward direction along the incline is taken as positive, the frictional force will be negative and the 2.1 Newton parallel component of the weight will be positive. The net force on the object will therefore be net force = 2.1 Newtons - .59 Newtons = 1.5 Newtons (approx.). This will result in an acceleration of a = Fnet / m = 1.5 Newtons / 3 kg = .5 m/s^2.
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RESPONSE --> Ok, I did not round up to .5
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01:02:01 `q011. Given the conditions of the previous question, what would be the acceleration of the cart if it was moving up the incline?
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RESPONSE --> these are the questions that get to me.....! acceleration is still due to net force. in this case gravitational forces are negative (away from the current motion) since the gravitational forces are still exerted DOWN the incline at 2.1 newtons . this time, however, friction is ALSO a negative in opposition to upward movement. -2.1 + -.59 N=-2.79 N = fnet -2.79 N/3 kg= -.93 m/s^2
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01:02:57 In this case the frictional force would still have magnitude .59 Newtons, but would be directed opposite to the motion, or down the incline. If the direction down the incline is still taken as positive, the net force must be net force = 2.1 Newtons + .59 Newtons = 2.7 Newtons (approx). The cart would then have acceleration a = Fnet / m = 2.7 Newtons / 3 kg = .9 m/s^2.
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RESPONSE --> okay, I designated down as negative simply to show that the cart would decelerate.....
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08:30:36 `q012. Assuming a very long incline, describe the motion of the cart which is given an initial velocity up the incline from a point a few meters up from the lower end of the incline. Be sure to include any acceleration experienced by the cart.
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RESPONSE --> the cart will move up the incline accelerating at -.9 m/s^2 until it slows to a stop. then, since 2.1 N downward force is greater than the frictional force of .59 N against it, the cart will begin accelerating down the incline at .5 m/s^2.
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08:32:16 The cart begins with a velocity up the incline, which we still taken to be the negative direction, and an acceleration of +.9 m/s^2. This positive acceleration tends to slow the cart while it is moving in the negative direction, and the cart slows by .9 m/s every second it spends moving up the incline. Eventually its velocity will be 0 for an instant, immediately after which it begins moving down the incline as result of the acceleration provided by the weight component parallel to the incline. {}As soon as it starts moving down the incline its acceleration decreases to +.5 m/s^2, but since the acceleration and velocity are now parallel the cart speeds up, increasing its velocity by .5 m/s every second, until it reaches the lower end of the incline.
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RESPONSE --> okay, the negative and positive here confuse the living daylights out of me..... I think I am describing the motion of the cart, but to me, a slowing cart should have negative acceleration.
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I talked to you about this one......"
course phy 201 yPy饨̲Student Name:
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19:53:46 `q001. The accepted value of the acceleration of gravity is approximately 980 cm/s^2 or 9.8 m/s^2. This will be the acceleration, accurate at most places within 1 cm/s^2, of any object which falls freely, that is without the interference of any other force, near the surface of the Earth. If you were to step off of a table and were to fall 1 meter without hitting anything, you would very nearly approximate a freely falling object. How fast would you be traveling when you reached the ground?
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RESPONSE --> 2ad= vf^2 where v0 = 0 2* 9.8m^2/s^2= vf^2 19.6 m/s^2= vf^2 vf= 4.43 m/s 4.43m/s/2= vave= 2.22 m/s vave * t = d 1 m/2.22m/s= t= .45 seconds So, I would land .45 seconds after I jumped, traveling 4.43 m/s ( or 443 cm/s)
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19:55:11 You would have an initial vertical velocity of 0, and would accelerate at 9.8 m/s^2 in the same direction as your 1 meter vertical displacement. You would also have a slight horizontal velocity (you don't step off of a table without moving a bit in the horizontal direction, and you would very likely maintain a small horizontal velocity as you fell), but this would have no effect on your vertical motion. So your vertical velocity is a uniform acceleration with v0 = 0, `ds = 1 meter and a = 9.8 m/s^2. The equation vf^2 = v0^2 + 2 a `ds contains the three known variables and can therefore be used to find the desired final velocity. We obtain vf = +- `sqrt( v0^2 + 2 a `ds) = +- `sqrt ( 0^2 + 2 * 9.8 m/s^2 * 1 m)= +- `sqrt ( 19.6 m^2 / s^2) = +- 4.4 m/s, approx. Since the acceleration and displacement were in the direction chosen as positive, we conclude that the final velocity will be in the same direction and we choose the solution vf = +4.4 m/s.
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RESPONSE --> ok
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20:06:48 `q002. If you jump vertically upward, leaving the ground with a vertical velocity of 3 m/s, how high will you be at the highest point of your jump? Note that as soon as you leave the ground, you are under the influence of only the gravitational force. All the forces that you exerted with your legs and other parts of your body to attain the 3 m/s velocity have done their work and are no longer acting on you. All you have to show for it is that 3 m/s velocity. So as soon as you leave the ground, you begin experiencing an acceleration of 9.8 m/s^2 in the downward direction. Now again, how high will you be at the highest point of your jump?
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RESPONSE --> You are asking at what point my initial velocity of 3m/s is cancelled out by my acceleration of -9.8 m/s^2. at the highest point in my jump, my velocity is 0. so my change in velocity is vf- v0= - 3m/s acceleration (-9.8 m/s^2)= dv/t -9.8m/s^2 = -3m/s/t t= -3m/s/-9.8m/m/s^2= .306 seconds. so I would be jumping up for .306 seconds at an average velocity of (3m/s + 0)/2 or 1.5 m/s. I would jump up 1.5 m/s * .306 seconds = .459 m or 45.9 cm above the ground.
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20:08:52 From the instant the leave the ground until the instant you reach your highest point, you have an acceleration of 9.8 m/s^2 in the downward direction. Since you are jumping upward, and since we can take our choice of whether upward or downward is the positive direction, we choose the upward direction as positive. You might have chosen the downward direction, and we will see in a moment how you should have proceeded after doing so. For now, using the upward direction as positive, we see that you have an initial velocity of v0 = + 3 m/s and an acceleration of a = -9.8 m/s^2. In order to use any of the equations of motion, each of which involves four variables, you should have the values of three variables. So far you only have two, v0 and a. {}What other variable might you know? If you think about it, you will notice that when objects tossed in the air reach their highest point they stop for an instant before falling back down. That is precisely what will happen to you. At the highest point your velocity will be 0. Since the highest point is the last point we are considering, we see that for your motion from the ground to the highest point, vf = 0. Therefore we are modeling a uniform acceleration situation with v0 = +3 m/s, a = -9.8 m/s^2 and vf = 0. We wish to find the displacement `ds. Unfortunately none of the equations of uniformly accelerated motion contain the four variables v0, a, vf and `ds. This situation can be easily reasoned out from an understanding of the basic quantities. We can find the change in velocity to be -3 meters/second; since the acceleration is equal to the change in velocity divided by the time interval we quickly determine that the time interval is equal to the change in velocity divided by the acceleration, which is `dt = -3 m/s / (-9.8 m/s^2) = .3 sec, approx.; then we multiply the .3 second time interval by the 1.5 m/s average velocity to obtain `ds = .45 meters. However if we wish to use the equations, we can begin with the equation vf = v0 + a `dt and solve to find `dt = (vf - v0) / a = (0 - 3 m/s) / (-9.8 m/s^2) = .3 sec. We can then use the equation `ds = (vf + v0) / 2 * `dt = (3 m/s + 0 m/s) / 2 * .3 sec = .45 m. This solution closely parallels and is completely equivalent to the direct reasoning process, and shows that and initial velocity of 3 meters/second should carry a jumper to a vertical height of .45 meters, approximately 18 inches. This is a fairly average vertical jump. If the negative direction had been chosen as positive then we would have a = +9.8 m/s^2, v0 = -3 m/s^2 (v0 is be in the direction opposite the acceleration so if acceleration is positive then initial velocity is negative) and again vf = 0 m/s (0 m/s is the same whether going up or down). The steps of the solution will be the same and the same result will be obtained, except that `ds will be -.45 m--a negative displacement, but where the positive direction is down. That is we move .45 m in the direction opposite to positive, meaning we move .45 meters upward.
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RESPONSE --> ok
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21:09:26 `q003. If you roll a ball along a horizontal table so that it rolls off the edge of the table at a velocity of 3 m/s, the ball will continue traveling in the horizontal direction without changing its velocity appreciably, and at the same time will fall to the floor in the same time as it would had it been simply dropped from the edge of the table. If the vertical distance from the edge of the table to the floor is .9 meters, then how far will the ball travel in the horizontal direction as it falls?
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RESPONSE --> so we need to know how long the ball is in the air. vertical distance is .9 m initial vertical velocity is 0 so, .9m = 1/2 (9.8m/s^2)t^2 1.8 m= 9.8m/s^2(t^2) .18m^2/s^2= t^2 .43 seconds in the air. now horizontal velocity * time= horizontal distance. 3 m/s * .43 seconds = 1.29 meters
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21:11:29 A ball dropped from rest at a height of .9 meters will fall to the ground with a uniform vertical acceleration of 9.8 m/s^2 downward. Selecting the downward direction as positive we have `ds = .9 meters, a = 9.8 m/s^2 and v0 = 0. Using the equation `ds = v0 `dt + .5 a `dt^2 we see that v0 = 0 simplifies the equation to `ds = .5 a `dt^2, so `dt = `sqrt( 2 `ds / a) = `sqrt(2 * .9 m / (9.8 m/s^2) ) = .42 sec, approx.. Since the ball rolls off the edge of the table with only a horizontal velocity, its initial vertical velocity is still zero and it still falls to the floor in .42 seconds. Since its horizontal velocity remains at 3 m/s, it travels through a displacement of 3 m/s * .42 sec = 1.26 meters in this time.
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RESPONSE --> we had a difference in significant figures.....
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}xƝYΐk` Student Name: ambler assignment #008
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z{jצҵEɜԪwӘ Student Name: assignment #009
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22:48:41 `q001. Note that there are 10 questions in this set. .You have done the Introductory Force Experiment in which you used rubber bands and bags of water, and you understand that, at least in the vicinity of the Earth's surface, gravity exerts downward forces. You have also seen that forces can be measured in units called Newtons. However you were not given the meaning and definition of the Newton as a unit of force. You are also probably aware that mass is often measured in kilograms. Here we are going to develop in terms of an experiment the meaning of the Newton as a force unit. Suppose that a cart contains 25 equal masses. The cart is equal in mass to the combined total of the 25 masses, as indicated by balancing them at equal distances from a fulcrum. The cart is placed on a slight downward incline and a weight hanger is attached to the cart by a light string and suspended over a low-friction pulley at the end of the ramp. The incline is adjusted until the cart, when given a slight push in the direction of the hanging weight, is observed to move with unchanging, or constant, velocity (and therefore zero acceleration). The masses are then moved one at a time from the cart to the hanger, so that the system can be accelerated first by the action of gravity on one of the masses, then by the action of gravity onto of the masses, etc.. The time required for the system to accelerate from rest through a chosen displacement is observed with one, two, three, four, five, ... ten of the masses. The acceleration of the system is then determined from these data, and the acceleration is graphed vs. the proportion of the total mass of the system which is suspended over the pulley. It is noted that if the entire mass of the system, including the cart, is placed on the weight hanger, there will be no mass left on the incline and the entire weight will fall freely under the acceleration of gravity. Suppose the data points obtained for the 5 of the first 10 trials were (.04, 48 cm/s^2), (.08, 85 cm/s^2), (.12, 125 cm/s^2), (.16, 171 cm/s^2), (.20, 190 cm/s^2). Sketch these points on an accurate graph of acceleration vs. proportion of weight suspended and determine the slope and y-intercept of the line. What is your slope and what is the y intercept? What is the equation of the line?
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RESPONSE --> I did not get perfectly linear data (although it came very close). My best fit line had a y intercept of 15 cm/s^2. the slope was (190-48)/(2.0-.04)= (142/.16) 887.5 therefore the line is y= 887.5 mp + 15 cm/s^2 to check, using Mass Proportion .12 y = 887.5 * (.12) + 15 cm/s^2 This gives me a y value of 121.5 cm/s^2 this is in fact a point on my best fit line.
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23:02:52 Since there are 25 equal masses and the mass of the cart is equivalent to another 25 of these masses, each mass is 1/50 = .02 of the total mass of the system. Thus the first 10 data points should might have been something like (.02, 21 cm/s^2), (.04, 48 cm/s^2), (.06, 55 cm/s^2), (.08, 85 cm/s^2), (.10, 101 cm/s^2), (.12, 125 cm/s^2), (.14, 141 cm/s^2), (.16, 171 cm/s^2), (.18, 183 cm/s^2), (.20, 190 cm/s^2). The data given in the problem would correspond to alternate data points. The slope of the best-fit line is 925 x + 12.8, indicating a slope of 925 and a y intercept of 12.8. The 967.5 is in units of rise / run, or for this graph cm/s^2. If you calculated the slope based on the points (.04, 48 cm/s^2) and (.20, 190 cm/s^2) you would have obtained 151 cm/s^2 / (.16) = 950 cm/s^2. Whether this is close to the best-fit value or not, this is not an appropriate calculation because it uses only the first and last data points, ignoring all data points between. The idea here is that you should sketch a line that fits the data as well as possible, then use the slope of this line, not the slope between data points.
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RESPONSE --> 190 - 142= 48 ..... my best fit line included my first data point, last data point and my (.08, 85 cm^2) data point... using two middle points, I get (155-85)/(.16- .08)= 70/.08= 875 for my slope...... I still do not get a slope > 900......
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23:07:49 `q002. Do the points seem to be randomly scattered around the straight line or does there seem to be some nonlinearity in your results?
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RESPONSE --> it actually seems to increase in a decreasing manner at the end.
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23:16:16 The slope of your line should probably be somewhere between 900 cm/s^2 and 950 cm/s^2. The points should be pretty much randomly scattered about the best possible straight line. Careful experiments of this nature have shown that the acceleration of a system of this nature is to a very high degree of precision directly proportional to the proportion of the weight which is suspended.
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RESPONSE --> ok, I can see the randomness, and by redrawing a best fit line I get a slope of 916.7 cm/s^2
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23:20:40 `q003. If the acceleration of the system is indeed proportional to the net force on the system, then your straight line should come close to the origin of your coordinate system. Is this the case?
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RESPONSE --> My y intercept is still at about 11 cm/s^2..... what are you considering close? compared to 190 cm/s^2 it is :>.
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23:21:25 If the acceleration of the system is proportional to the net force, then the y coordinate of the straight line representing the system will be a constant multiple of the x coordinate--that is, you can always find the y coordinate by multiplying the x coordinate by a certain number, and this 'certain number' is the same for all x coordinates. The since the x coordinate is zero, the y coordinate will be 0 times this number, or 0. Your graph might not actually pass through the origin, because data inevitably contains experimental errors. However, if experimental errors are not too great the line should pass very close to the origin. In the case of this experiment the y-intercept was 12.8. On the scale of the data used here this is reasonably small, and given the random fluctuations of the data points above and below the straight-line fit the amount of deviation is consistent with a situation in which precise measurements would reveal a straight line through the origin.
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RESPONSE --> like I said, how do you define close!
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15:55:52 `q003. What is it that causes the system to accelerate more when a greater proportion of the mass is suspended?
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RESPONSE --> the suspended mass is accelerating due to gravity, while the mass that is not suspended is acting as an opposing force.
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15:56:57 The gravitational forces exerted on the system are exerted on the suspended masses and on the cart and the masses remaining in it. The supporting force exerted by the ramp counters the force of gravity on the cart and the masses remaining in it, and this part of the gravitational force therefore does not affect the acceleration of the system. However there is no force to counter the pull of gravity on the suspended masses, and this part of the gravitational force is therefore the net force acting on the mass of the entire cart-and-mass system. The force exerted by gravity on the suspended masses is proportional to the number of suspended masses--e.g, if there are twice as many masses there is twice the force. Thus it is the greater gravitational force on the suspended masses that causes the greater acceleration.
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RESPONSE --> i think i understand....
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23:07:15 `q004. This results of this sort of experiment, done with good precision, support the contention that for a given mass the acceleration of a system is indeed proportional to the net force exerted on the system. Other experiments can be done using rubber bands, springs, fans and other nongravitational sources of force, to further confirm this result. In another sort of experiment, we can see how much force is required on different masses to obtain a certain fixed acceleration. In such experiments we find for example that if the mass is doubled, it requires twice the force to achieve the same acceleration, and that in general the force required to achieve a given acceleration is proportional to the amount of mass being accelerated. In a certain experiment using the same cart and masses as before, plus several additional identical carts, a single cart is accelerated by a single suspended mass and found to accelerate at 18 cm/s^2. Then a second cart is placed on top of the first and the two carts are accelerated by two suspended masses, achieving an acceleration of 20 cm / s^2. Then a third cart is placed on top of the first to and the three carts are accelerated by three suspended masses, achieving and acceleration of 19 cm/s^2. A fourth cart and a fourth suspended mass are added and an acceleration of 18 cm/s^2 is obtained. Adding a fifth cart in the fifth suspended mass an acceleration of 19 cm/s^2 is obtained. All these accelerations are rounded to the nearest cm/s^2, and all measurements are subject to small but significant errors in measurement. How well do these results indicate that to achieve a given acceleration the amount of force necessary is in fact proportional to the amount of mass being accelerated?
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RESPONSE --> assuming identical carts and masses m1= 18 cm/s^2 2m1 20 cm/s^2 3m1 = 19 cm/s^2 4m1 = 18 cm/s^2 5m1= 19 cm/s^2 the average acceleration is 18.8 cm/s^2 or 19 cm/s/s. for each unit x being accelerated a mass of unit y must be added to keep acceleration consistant. this would indicate that force must be proportional to mass for consistant acceleration values.
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23:07:41 The accelerations obtained are all about the same, with only about a 10% variation between the lowest and the highest. Given some errors in the observation process, it is certainly plausible that these variations are the result of such observation errors; however we would have to have more information about the nature of the observation process and the degree of error to be expected before drawing firm conclusions. If we do accept the conclusion that, within experimental error, these accelerations are the same then the fact that the second through the fifth systems had 2, 3, 4, and 5 times the mass of the first with 2, 3, 4, and 5 times the suspended mass and therefore with 2, 3, 4, and 5 times the net force does indeed indicate that the force needed to achieve this given acceleration is proportional to the mass of the system.
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RESPONSE --> ok
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23:18:05 `q005. Now we note again that the force of gravity acts on the entire mass of the system when an entire system is simply released into free fall, and that this force results in an acceleration of 9.8 m/s^2. If we want our force unit to have the property that 1 force unit acting on 1 mass unit results in an acceleration of 1 m/s^2, then how many force units does gravity exert on one mass unit?
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RESPONSE --> If 1 force unit accelerates 1 mass unit by 1 m/s^2, and gravity accelerates 1 mass unit by 9.8 m/s^2, then f*m=a m= 1 a= 9.8m/s^2 so gravity is exerting 9.8 force units on the mass.
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23:18:17 Since gravity gives 1 mass unit an acceleration of 9.8 m/s^2, which is 9.8 times the 1 m/s^2 acceleration that would be experienced from 1 force unit, gravity must exerted force equal to 9.8 force units on one mass unit.
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RESPONSE --> yes!!
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23:18:49 `q006. If we call the force unit that accelerates 1 mass unit at 1 m/s^2 the Newton, then how many Newtons of force does gravity exert on one mass unit?
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RESPONSE --> 9.8 newtons of force is exerted by gravity on 1 mass unit
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23:18:57 Since gravity accelerates 1 mass unit at 9.8 m/s^2, which is 9.8 times the acceleration produced by a 1 Newton force, gravity must exert a force of 9.8 Newtons on a mass unit.
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RESPONSE --> yes
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23:20:04 `q007. The mass unit used here is the kilogram. How many Newtons of force does gravity exert on a 1 kg mass?
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RESPONSE --> again, mass * acceleration= force. acceleration is 9.8 m/s^2 * 1 kg mass= 9.8 newtons force
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23:20:12 Gravity exerts a force of 9.8 Newtons on a mass unit and the kg is the mass unit, so gravity must exert a force of 9.8 Newtons on a mass of 1 kg.
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RESPONSE --> yes
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23:23:19 `q008. How much force would gravity exert on a mass of 8 kg?
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RESPONSE --> 8 kg * 9.8 m/s^2= 78.4 kg*m/s^2 (newtons)
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23:23:38 Gravity exerts 8 times the force on 8 kg as on 1 kg. The force exerted by gravity on a 1 kg mass is 9.8 Newtons. So gravity exerts a force of 8 * 9.8 Newtons on a mass of 8 kg.
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RESPONSE --> ok
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23:25:34 `q009. How much force would be required to accelerate a mass of 5 kg at 4 m/s^2?
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RESPONSE --> 5kg * 4 m/s^2= force 20 newtons
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23:26:16 Compared to the 1 Newton force which accelerates 1 kg at 1 m/s^2, 2e have here 5 times the mass and 4 times the acceleration so we have 5 * 4 = 20 times the force, or 20 Newtons. We can formalize this by saying that in order to give a mass m an acceleration a we must exert a force F = m * a, with the understanding that when m is in kg and a in m/s^2, F must be in Newtons. In this case the calculation would be F = m * a = 5 kg * 4 m/s^2 = 20 kg m/s^2 = 20 Newtons. The unit calculation shows us that the unit kg * m/s^2 is identified with the force unit Newtons.
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RESPONSE --> I noted earlier that a newton is kg*m/s^2 ......
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23:27:27 `q010. How much force would be required to accelerate the 1200 kg automobile at a rate of 2 m/s^2?
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RESPONSE --> 1200 kg * 2m/s^2 (mass * acceleration...) = x newtons 1400 kg * m/s^2 = force 1400 newtons
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23:28:03 This force would be F = m * a = 1200 kg * 2 m/s^2 = 2400 kg * m/s^2 = 2400 Newtons.
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RESPONSE --> I wrote down the right equation, I just didn't do it!!
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bUIĎqX Student Name: assignment #011
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13:37:06 `q001. A cart on a level, frictionless surface contains ten masses, each of mass 2 kg. The cart itself has a mass of 30 kg. A light weight hanger is attached to the cart by a light but strong rope and suspended over the light frictionless pulley at the end of the level surface. Ignoring the weights of the rope, hanger and pulley, what will be the acceleration of the cart if one of the 2 kg masses is transferred from the cart to the hanger?
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RESPONSE --> okay, the force going forward is 2kg*9.8m/s^2, or 19.6 N (that is on the mass suspended). the original mass of the cart is 30 kg + 10*2 kg or 50 kg, therefore, when we move one 2 kg mass it becomes 48 kg. since force = m* a, then f/m = acceleration in this case 19.6 n/50 kg (altogether) = .392 m/s^2
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13:38:09 At the surface of the Earth gravity, which is observed here near the surface to accelerate freely falling objects at 9.8 m/s^2, must exert a force of 2 kg * 9.8 m/s^2 = 19.6 Newtons on the hanging 2 kg mass. This force will tend to accelerate the system consisting of the cart, rope, hanger and suspended mass, in the 'forward' direction--the direction in which the various components of the system must accelerate if the hanging mass is to accelerate in the downward direction. This is the only force accelerating the system in this direction. All other forces, including the force of gravity pulling the cart and the remaining masses downward and the normal force exerted by the level surface to prevent gravity from accelerating the cart downward, are in a direction perpendicular to the motion of these components of the system; furthermore these forces are balanced so that they add up to 0. The mass of the system is that of the 30 kg cart plus that of the ten 2 kg masses, a total of 50 kg. The net force of 19.6 Newtons exerted on a 50 kg mass therefore results in acceleration a = Fnet / m = 19.6 Newtons / (50 kg) = 19.6 kg m/s^2 / (50 kg) = .392 m/s^2.
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RESPONSE --> yes!!! that took a few minutes to figure out even using your stuff AND the book.
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13:51:05 `q002. Two 1-kg masses are suspended over a pulley, one on either side. A 100-gram mass is added to the 1-kg mass on one side of the pulley. How much force does gravity exert on each side of the pulley, and what is the net force acting on the entire 2.1 kg system?
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RESPONSE --> think of 1 kg mass without addition as side a the 1 kg mass with additional .1 kg is side b gravity is 9.8 m/s^2 and Force is measured in Newtons which is kg*m/s^2 side a has a net force due to gravity of 9.8m/s^2 * 1 kg, or 9.8 N side b has a net force due to gravity of 9.8 m/s^2 * 1.1 kg or 10.78 N If we think of this as an atwood machine, then the net force is simply fnet = fnetb-fneta or 1.02 newtons (- newtons if we think of down as negative.......)
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13:52:18 The 1-kg mass experiences a force of F = 1 kg * 9.8 m/s^2 = 9.8 Newtons. The other side has a total mass of 1 kg + 100 grams = 1 kg + .1 kg = 1.1 kg, so it experiences a force of F = 1.1 kg * 9.8 Newtons = 10.78 Newtons. {}{}Both of these forces are downward, so it might seem that the net force on the system would be 9.8 Newtons + 10.78 Newtons = 20.58 Newtons. However this doesn't seem quite right, because when one mass is pulled down the other is pulled up so in some sense the forces are opposing. It also doesn't make sense because if we had a 2.1 kg system with a net force of 20.58 Newtons its acceleration would be 9.8 m/s^2, and we know very well that two nearly equal masses suspended over a pulley won't both accelerate downward at the acceleration of gravity. So in this case we take note of the fact that the to forces are indeed opposing, with one tending to pull the system in one direction and the other in the opposite direction. {}{}We also see that we have to abandon the notion that the appropriate directions for motion of the system are 'up' and 'down'. We instead take the positive direction to be the direction in which the system moves when the 1.1 kg mass descends. {}{}We now see that the net force in the positive direction is 10.78 Newtons and that a force of 9.8 Newtons acts in the negative direction, so that the net force on the system is 10.78 Newtons - 9.8 Newtons =.98 Newtons. {}{}The net force of .98 Newtons on a system whose total mass is 2.1 kg results in an acceleration of .98 Newtons / (2.1 kg) = .45 m/s^2, approx.. Thus the system accelerates in the direction of the 1.1 kg mass at .45 m/s^2.
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RESPONSE --> again, that adding and subtracting thing......
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14:02:43 `q003. If in the previous question there is friction in the pulley, as there must be in any real-world pulley, the system in the previous problem will not accelerate at the rate calculated there. Suppose that the pulley exerts a retarding frictional force on the system which is equal in magnitude to 1% of the weight of the system. In this case what will be the acceleration of the system, assuming that it is moving in the positive direction (as defined in the previous exercise)?
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RESPONSE --> so the retarding force is 1% of the entire weight of the system, which is .01 * 2.1kg or .021N Now the force on the system is 10.78 N - .021N - 9.8N= .959 N .959N/2.1kg = .46 m/s^2
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14:06:27 We first determine the force exerted a friction. The weight of the system is the force exerted by gravity on the mass of the system. The system has mass 2.1 kg, so the weight of the system must be 2.1 kg * 9.8 m/s^2 = 20.58 Newtons. 1% of this weight is .21 Newtons, rounded off to two significant figures. This force will be exerted in the direction opposite to that of the motion of the system; since the system is assume to be moving in the positive direction the force exerted by friction will be frictional force = -.21 Newtons. The net force exerted by the system will in this case be 10.78 Newtons - 9.8 Newtons - .21 Newtons = .77 Newtons, in contrast with the .98 Newton net force of the original exercise. The acceleration of the system will be .77 Newtons / (2.1 kg) = .35 m/s^2, approx..
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RESPONSE --> I see, I did not read the question correctly-- when you said force was equal in magnitude to 1% of weight, I assumed you meant the percentage WAS the force.
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14:08:22 `q004. Why was it necessary in the previous version of the exercise to specify that the system was moving in the direction of the 1.1 kg mass. Doesn't the system have to move in that direction?
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RESPONSE --> The system does not have to move in the direction of the heavier mass if other forces are being applied to the system........
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14:08:45 If the system is released from rest, since acceleration is in the direction of the 1.1 kg mass its velocity will certainly always be positive. However, the system doesn't have to be released from rest. We could give a push in the negative direction before releasing it, in which case it would continue moving in the negative direction until the positive acceleration brought it to rest for an instant, after which it would begin moving faster and faster in the positive direction.
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RESPONSE --> ok
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22:27:59 `q005. If the system of the preceding series of exercises is initially moving in the negative direction, then including friction in the calculation what is its acceleration?
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RESPONSE --> The mass and acceleration due to gravity is still the same, as is the force applied by friction. It will still be .35 m/s^2 it's just that at first it will be negative acceleration....
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22:36:57 Its acceleration will be due to the net force. This net force will include the 10.78 Newton force in the positive direction and the 9.8 Newton force in the negative direction. It will also include a frictional force of .21 Newtons in the direction opposed to motion. Since motion is in the negative direction, the frictional force will therefore be in the positive direction. The net force will thus be Fnet = 10.78 Newtons - 9.8 Newtons + .21 Newtons = +1.19 Newtons, in contrast to the +.98 Newtons obtained when friction was neglected and the +.77 Newtons obtained when the system was moving in the positive direction. , The acceleration of the system is therefore , 1.19 N / (2.1 kg) = .57 m/s^2.
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RESPONSE --> oh, ok..... I didn't take the directions of force into account-----
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23:27:56 `q006. If friction is neglected, what will be the result of adding 100 grams to a similar system which originally consists of two 10-kg masses, rather than the two 1-kg masses in the previous examples?
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RESPONSE --> Again, this is an atwood machine where net force is determined by subtracting force on mass a from force on mass b. the force being applied by the acceleration due to gravity on the mass. f(a)= 9.8 m/s^2 * 10.1 kg= 98.98 n f(b) = 9.8m/s^2 * 10 = 98 n fnet = (9.8 m/s^2* 10.1 kg) - (9.8 m/s^2 * 10 kg) 98.98 N - 98 N = .98 n since the mass of the system is 20.1kg, then Acc = .98 N/20.1 kg= .049 m/s^2 Obviously this is a slower acceleration than in the first system. This makes since, as a percentage, the mass added to the second side is much smaller than that of the first system. 9.8 N will be the net force on the system. since the total mass is 20.1 kg
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23:28:21 In this case the masses will be 10.1 kg and 10 kg. The force on the 10.1 kg mass will be 10.1 kg * 9.8 m/s^2 = 98.98 Newtons and the force on the 10 kg mass will be 10 kg * 9.8 m/s^2 = 98 Newtons. The net force will therefore be .98 Newtons, as it was in the previous example where friction was neglected. We note that this.98 Newtons is the result of the additional 100 gram mass, which is the same in both examples. The mass in this case will be 20.1 kg, so that the acceleration of the system is a = .98 Newtons / 20.1 kg = .048 m/s^2, approx.. The same .98 Newton net force acting on the significantly greater mass results in a significantly smaller acceleration.
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RESPONSE --> yes.
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23:42:32 `q007. If friction is not neglected, what will be the result for the system with the two 10-kg masses with .1 kg added to one side? Note that by following what has gone before you could, with no error and through no fault of your own, possibly get an absurd result here, which will be repeated in the explanation then resolved at the end of the explanation.
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RESPONSE --> friction is an opposing force to the downward acceleration of the system (well, down on one side, up on the other......) If friction is once again 1 % of system weight, then 20.1 kg * 9.8 m/s^2= 196.98 N * 1% = 1.97 N force applied due to friction. So the final net force would be 98.98 N - 1.97 N - 98 N= - .99 N This would seemingly end in the lighter side accelerating downward...... however, if we remember that friction opposes ANY movement, then in this case we simply conclude that the force on the system is not enough to overcome the opposing force of friction and that the system does not accelerate due to gravity in this instance.
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00:00:16 If friction is still equal to 1% of the total weight of the system, which in this case is 20.1 kg * 9.8 m/s^2 = 197 Newtons, then the frictional force will be .01 * 197 Newtons = 1.97 Newtons. This frictional force will oppose the motion of the system. For the moment assume the motion of the system to be in the positive direction. This will result in a frictional force of -1.97 Newtons. The net force on the system is therefore 98.98 Newtons - 98 Newtons - 1.97 Newtons = -.99 Newtons. This net force is in the negative direction, opposite to the direction of the net gravitational force. If the system is moving this is perfectly all right--the frictional force being greater in magnitude than the net gravitational force, the system can slow down. Suppose the system is released from rest. Then we might expect that as a result of the greater weight on the positive side it will begin accelerating in the positive direction. However, if it moves at all the frictional force would result in a -.99 Newton net force, which would accelerate it in the negative direction and very quickly cause motion in that direction. Of course friction can't do this--its force is always exerted in a direction opposite to that of motion--so friction merely exerts just enough force to keep the object from moving at all. Friction acts as though it is quite willing to exert any force up to 1.97 Newtons to oppose motion, and up to this limit the frictional force can be used to keep motion from beginning. In fact, the force that friction can exert to keep motion from beginning is usually greater than the force it exerts to oppose motion once it is started.
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RESPONSE --> exactly
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00:14:31 `q008. A cart on an incline is subject to the force of gravity. Depending on the incline, some of the force of gravity is balanced by the incline. On a horizontal surface, the force of gravity is completely balanced by the upward force exerted by the incline. If the incline has a nonzero slope, the gravitational force (i.e., the weight of the object) can be thought of as having two components, one parallel to the incline and one perpendicular to the incline. The incline exerts a force perpendicular to itself, and thereby balances the weight component perpendicular to the incline. The weight component parallel to the incline is not balanced, and tends to accelerate the object down the incline. Frictional forces tend to resist this parallel component of the weight and reduce or eliminate the acceleration. A complete analysis of these forces is best done using the techniques of vectors, which will be encountered later in the course. For now you can safely assume that for small slopes (less than .1) the component of the gravitational force parallel to the incline is very close to the product of the slope and the weight of the object. [If you remember your trigonometry you might note that the exact value of the parallel weight component is the product of the weight and the sine of the angle of the incline, that for small angles the sine of the angle is equal to the tangent of the angle, and that the tangent of the angle of the incline is the slope. The product of slope and weight is therefore a good approximation for small angles or small slopes.] What therefore would be the component of the gravitational force acting parallel to an incline with slope .07 on a cart of mass 3 kg?
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RESPONSE --> It has been a while since I took trig, but this does make sense..... 3 kg * 9.8 m/s^2= 29.4 kg m/s^2 or 29.4 N, so the gravitational component of the force acting on the cart would be .07 (slope) * 29.4 N (weight) = 2.06
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00:16:30 The gravitational force on a 3 kg object is its weight and is equal to 3 kg * 9.8 m/s^2 = 29.4 Newtons. The weight component parallel to the incline is found approximately as (parallel weight component) = slope * weight = .07 * 29.4 Newtons = 2.1 Newtons (approx.).. STUDENT COMMENT: It's hard to think of using the acceleration of 9.8 m/s^2 in a situation where the object is not free falling. Is weight known as a force measured in Newtons? Once again I'm not used to using mass and weight differently. If I set a 3 kg object on a scale, it looks to me like the weight is 3 kg.INSTRUCTOR RESPONSE: kg is commonly used as if it is a unit of force, but it's not. Mass indicates resistance to acceleration, as in F = m a. {}{}eight is the force exerted by gravity. {}{}The weight of a given object changes as you move away from Earth and as you move into the proximity of other planets, stars, galaxies, etc.. As long as the object remains intact its mass remains the same, meaning it will require the same net force to give it a specified acceleration wherever it is.{}{}An object in free fall is subjected to the force of gravity and accelerates at 9.8 m/s^2. This tells us how much force gravity exerts on a given mass: F = mass * accel = mass * 9.8 m/s^2. This is the weight of the object. **
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RESPONSE --> ok. actually, my problem is units--- what units were we measuring slope in? cm/cm? cm/m? how would I note that as a product with newtons?
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00:20:26 `q009. What will be the acceleration of the cart in the previous example, assuming that it is free to accelerate down the incline and that frictional forces are negligible?
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RESPONSE --> Since the mass is 3 kg and the force acting on the cart (parrallel to the slope) is 2.1 kg/m/s^2 (I used your answer), then acceleration is 2.1 kg/m/s^2/ 3kg (f/m=a) the acceleration would be about .7 m/s^2
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00:20:39 The weight component perpendicular to the incline is balanced by the perpendicular force exerted by the incline. The only remaining force is the parallel component of the weight, which is therefore the net force. The acceleration will therefore be a = F / m = 2.1 Newtons / (3 kg) = .7 m/s^2.
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RESPONSE --> ok
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00:27:08 `q010. What would be the acceleration of the cart in the previous example if friction exerted a force equal to 2% of the weight of the cart, assuming that the cart is moving down the incline? [Note that friction is in fact a percent of the perpendicular force exerted by the incline; however for small slopes the perpendicular force is very close to the total weight of the object].
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RESPONSE --> 3 kg = mass weight = 9.8 m/s^2 * 3= 29.4 N friction force = 2% * 29.4= .59N 29.4 n * .07 = 2.06 Newtons parallel to cart , perpendicular force cancels out. 2.06 newons down slope - .59 n friction = 1.47 N down slope. acceleration = f/m 1.47 kg/m/s^2/ 3kg= .49 m/s^2
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00:29:06 The weight of the cart was found to be 29.4 Newtons. The frictional force will therefore be .02 * 29.4 Newtons = .59 Newtons approx.. This frictional force will oppose the motion of the cart, which is down the incline. If the downward direction along the incline is taken as positive, the frictional force will be negative and the 2.1 Newton parallel component of the weight will be positive. The net force on the object will therefore be net force = 2.1 Newtons - .59 Newtons = 1.5 Newtons (approx.). This will result in an acceleration of a = Fnet / m = 1.5 Newtons / 3 kg = .5 m/s^2.
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RESPONSE --> Ok, I did not round up to .5
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01:02:01 `q011. Given the conditions of the previous question, what would be the acceleration of the cart if it was moving up the incline?
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RESPONSE --> these are the questions that get to me.....! acceleration is still due to net force. in this case gravitational forces are negative (away from the current motion) since the gravitational forces are still exerted DOWN the incline at 2.1 newtons . this time, however, friction is ALSO a negative in opposition to upward movement. -2.1 + -.59 N=-2.79 N = fnet -2.79 N/3 kg= -.93 m/s^2
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01:02:57 In this case the frictional force would still have magnitude .59 Newtons, but would be directed opposite to the motion, or down the incline. If the direction down the incline is still taken as positive, the net force must be net force = 2.1 Newtons + .59 Newtons = 2.7 Newtons (approx). The cart would then have acceleration a = Fnet / m = 2.7 Newtons / 3 kg = .9 m/s^2.
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RESPONSE --> okay, I designated down as negative simply to show that the cart would decelerate.....
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08:30:36 `q012. Assuming a very long incline, describe the motion of the cart which is given an initial velocity up the incline from a point a few meters up from the lower end of the incline. Be sure to include any acceleration experienced by the cart.
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RESPONSE --> the cart will move up the incline accelerating at -.9 m/s^2 until it slows to a stop. then, since 2.1 N downward force is greater than the frictional force of .59 N against it, the cart will begin accelerating down the incline at .5 m/s^2.
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08:32:16 The cart begins with a velocity up the incline, which we still taken to be the negative direction, and an acceleration of +.9 m/s^2. This positive acceleration tends to slow the cart while it is moving in the negative direction, and the cart slows by .9 m/s every second it spends moving up the incline. Eventually its velocity will be 0 for an instant, immediately after which it begins moving down the incline as result of the acceleration provided by the weight component parallel to the incline. {}As soon as it starts moving down the incline its acceleration decreases to +.5 m/s^2, but since the acceleration and velocity are now parallel the cart speeds up, increasing its velocity by .5 m/s every second, until it reaches the lower end of the incline.
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RESPONSE --> okay, the negative and positive here confuse the living daylights out of me..... I think I am describing the motion of the cart, but to me, a slowing cart should have negative acceleration.
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I talked to you about this one......"