course Phy 121 Physics I02-10-2008
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17:18:48 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 --> vf^2 = v0^2 + 2 a `ds vf = +- `sqrt( v0^2 + 2 a `ds) which equals vf = +- `sqrt ( 0^2 + 2 * 9.8 m/s^2 * 1 m) which equals +- `sqrt ( 19.6 m^2 / s^2) = +- 4.4 m/s self critique assessment: 3
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17:23:59 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 --> `dt = (vf - v0) / a = (0 - 3 m/s) / (-9.8 m/s^2) = .3 sec. `ds = (vf + v0) / 2 * `dt = (3 m/s + 0 m/s) / 2 * .3 sec = .45 m. Which means we have jumped .45 meters up. self critique assessment: 2
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17:27:20 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 --> `ds = v0 `dt + .5 a `dt^2 `sqrt(2 * .9 m / (9.8 m/s^2) ) = .42 sec So it travels through a displacement of 3 m/s * .42 sec = 1.26 meters in this time. self critique assessment: 2
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÷UXѮⲌ| assignment #008 008. Using the Acceleration of Gravity Physics I 02-10-2008 _ӢU~zٲ assignment #009 009. Forces exerted by gravity Physics I 02-10-2008
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20:14:12 `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 --> First you would calculate slope by dividing rise by run. I used the points (.04, 48) and (.20, 190). The equation would be (190-48)/(.20 -.04) = 887.5 Now we find the y-int by using the equation y = mx + b and substituting one point and the slope. It would be calculated as follows using the point (.02, 48): 48 = 887.5(.04) + b which equals 48 = 35.5 + b. Now divide 48 by 35.5 which equals 1.35 which equals y-int. The equation for a line would then be: y = 887.5x + 1.35 confidence assessment: 3
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20:16:37 `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 --> They are randomly scattered around the line. confidence assessment: 3
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20:17:18 `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 --> Yes confidence assessment: 3
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20:18:37 `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 --> Gravity confidence assessment: 3
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20:34:43 `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 --> I don't feel like this would be that precise. confidence assessment: 1
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20:38:09 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 --> I didn't understand the quesiton at first, but it is now obvious. self critique assessment: 3
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20:39:13 `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 confidence assessment: 2
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20:40:29 `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 --> 9.8 kg of force confidence assessment: 1
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20:42:08 `q008. How much force would gravity exert on a mass of 8 kg?
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RESPONSE --> 9.8 * 8 = 78.4 Newtons on a mass of 8kg. confidence assessment: 3
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20:44:12 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 --> F = m * a = 5 kg * 4 m/s^2 = 20 kg m/s^2 = 20 Newtons self critique assessment: 3
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20:45:16 `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 --> You would multiply the mass by the rate which is equal to 1200kg * 2m/s^2 = 2400 Newtons of Force. confidence assessment: 3