course Phy 201 gÚͧúÇ—ï˜rõ}ꨥõìÙ¬ï“Student Name:
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12:30:31 `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 --> slope = 925cm/s^2 y-intercept = 12.8 equation = 925x+12.8
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12:31:58 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 --> ok
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12:33:16 `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 --> The points are all in a straight line, with the last two (.16, 171) and (.2, 190) being slightly off of the line, but still very linear.
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12:36:33 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, so the weight suspended from the object is what determine's it's acceleration. That makes sense, I just want to make sure I interpretted the statement correctly. I'm not sure if this question means much, but a weight suspended from the object means that no ramp incline is needed... so is it safe to say that the acceleration isn't dependent on the incline, even if there is one?
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12:37:17 `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 --> Yup! It passes directly through the origin.
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12:39:31 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 --> ok
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12:41:15 `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 --> My guess is that gravitational pull is greater on the heavier mass of the object, so it's acceleration would be greater. And I'm thinking also that the acceleration is proportional to the mass.
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12:42:42 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 --> ok
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12:46:22 `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 --> If the amount of force needed didn't depend on the mass of the carts being accelerated, then the acceleration data take would not be so similar. But since each additional cart came with the same amount of weight suspended, the acceleration remained constant, showing that the force is proportional to the mass.
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12:47:06 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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12:51:40 `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 --> I'm not sure. I think that gravity would also exert 1 force unit onto the one mass unit. I say this because 1 force unit acts on 1mass unit and results in an acceleration of 1m/s^2, and they all seem to be proportional to each other, then they would all be proportional to the gravity exerted also...
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12:52:36 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 --> Oops. Sadly, that thought crossed my mind but I was thinking of the proportionality stuff from one of the previous assignments.
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12:53:39 `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 --> 9.8newtons?
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12:53:48 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 --> ok
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12:54:10 `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 --> shouldn't it still be 9.8 newtons...
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12:54:19 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 --> ok
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12:54:59 `q008. How much force would gravity exert on a mass of 8 kg?
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RESPONSE --> 9.8n * 8kg = 78.4n/kg.
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12:55:17 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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12:56:46 `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 * 4m/s^2 = 20n
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12:57:19 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 --> ok
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12:58:02 `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 --> F = m*a F = 1200kg * 2m/s^2 = 2400kg*m/s^2 F = 2400Newtons
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12:58:06 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 --> ok ....
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ëź´¢„Æxã÷åҶÊÎǺØf¸…þŸí Student Name: assignment #010
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21:01:32 `q001. If a block of mass 10 kg accelerates at 2 m/s^2, then what net force is acting on the block?
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RESPONSE --> F = m*a F = 10kg * 2m/s^2 = 20Newtons
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21:01:39 The net force on the block is the product F = m * a of its 10 kg mass and its 2 m/s^2 acceleration. The net force is therefore F = 10 kg * 2 m/s^2 = 20 kg * m / s^2. The unit of force, which is the product of a quantity in kg and another quantity in m/s^2, is just the algebraic product kg * m/s^2 of these two units. This unit, the kg * m / s^2, is called a Newton. So the net force is 20 Newtons.
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RESPONSE --> ok
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21:03:08 `q002. How much force must be exerted by someone pulling on it to accelerate a 10 kg object at 2 m/s^2?
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RESPONSE --> Force exerted also equals 20newtons
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21:04:19 This depends on what forces might be resisting the acceleration of the object. If the object is accelerating on a surface of some type, then there is a good chance that a frictional force is opposing the motion. If the object is being pulled upward against the force of gravity, then more force is required then if it is sliding along a low-friction horizontal surface. If it is being pulled downhill, the force exerted by gravity has a component in the direction of motion and perhaps even less force is required. However, in every case the net force, which is the sum of all the forces acting on the object, must be 20 Newtons. The person pulling on the object must exert exactly enough force that the net force will be 20 Newtons.
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RESPONSE --> I knew that the force exerted had to equal something, but I couldn't figure out if it was equal to net forces or what.
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21:07:10 `q003. If friction exerts a force of 10 Newtons in the direction opposite the motion of a 10 kg object, then how much force must be exerted by someone pulling on it to accelerate the 10 kg object at 2 m/s^2, with the acceleration in the same direction as the motion?
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RESPONSE --> Ok I know that the force in the direction of the motion is 20Newtons, but with the friction exerting a force of 10Newtons in the opposite direction, I think that the overall force exerted should be 10 Newtons, but I'm not sure.
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21:08:16 Since the 10 Newton frictional force is in the direction opposite to motion, and since the acceleration is in the same direction as the motion, the frictional force is opposed to the accelerating force. If the direction of motion is taken as positive, then the frictional force will be in the negative direction and can be denoted fFrict = - 10 Newtons. To achieve the given acceleration the net force on the object must be net force = 10 kg * (+2 m/s^2) = +20 kg * m/s^2 = +20 Newtons. In order to achieve the +20 Newton net force when there is already a frictional force of -10 Newtons, it should be clear that a force of +30 Newtons is required. This can be thought of as 10 Newtons to overcome friction and another 20 Newtons to achieve the required net force.
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RESPONSE --> Ok I see how 30+Newtons is required.
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21:10:01 `q004. How can we write an equation to solve this problem? Hint: What equation would relate the net force Fnet, the force F exerted by the person and the force fFrict of friction?
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RESPONSE --> F = m*a Fnet = F + fFrict ?
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21:10:27 If Fnet is the net force and F the force actually exerted by the person, then Fnet = F + fFrict. That is, the net force is the sum of the force exerted by the person and the frictional force. We know that Fnet is +20 Newtons and fFrict is -10 Newtons, so we have the equation 20 Newtons = F + (-10 Newtons). Solving for F we see that F = 20 Newtons + 10 Newtons = 30 Newtons.
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RESPONSE --> Ok cool
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21:11:19 `q005. If a constant net force of 12 Newtons acts on a cart of mass 6 kg, then at what rate does the velocity of the cart change?
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RESPONSE --> F = m * a 12N = 6kg * a a = 12N/6kg = 2meters
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21:12:30 The velocity of the cart will change at a rate a which is related to the net force and the mass by Fnet = m * a. Thus a = Fnet / m = 12 Newtons / (6 kg) = 12 kg * m/s^2 / (6 kg) = 2 m/s^2. We note that the force unit Newtons is broken down to its fundamental units of kg * m / s^2 in order to perform the unit calculation. Dividing kg * m / s^2 by kg we have (kg / kg) * m/s^2 = m/s^2. It is important to always do the unit calculations. This habit avoids a large number of errors and also can be used to reinforce our understanding of the relationships in a problem or situation.
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RESPONSE --> I got my units wrong, i knew it had to be m/s^2 for acceleration, but I forgot that Newtons equals kg*m/s^2.
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21:19:00 `q006. If a force of 50 Newtons is exerted in the direction of the object's motion by a person, on a 20 kg object, and if friction exerts a force of 10 Newtons opposed to the direction of motion, then what will be the acceleration of the object?
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RESPONSE --> Fnet = 50N; m = 20kg; fFrict = -10 Fnet = F + fFrict 50N = F + -10 F = 60Newtons So a = F/m = 60N / 20kg = 3m/s^2
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21:23:00 The object will accelerate at a rate determined by Newton's Second Law, Fnet = m * a. The acceleration will therefore be a = Fnet / m. The net force on the object will be the sum of the 50 Newton force in the direction of motion and the 10 Newton force opposed to the direction of motion. If we take the direction of motion as positive, then the net force is Fnet = 50 N - 10 N = 40 N. It follows that the acceleration is a = Fnet / m = 40 N / (20 kg) = 40 kg m/s^2 / (20 kg) = 2 m/s^2.
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RESPONSE --> Ok. So F is when a force is exerted by someone. I did the same method as the previous problem, which is why I used the 50Newtons for Fnet instead of F.
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21:25:28 `q007. If a force of 50 Newtons is exerted opposite to the direction of the object's motion by a person, on a 20 kg object, and if friction exerts a force of 10 Newtons opposed to the direction of motion, then what will be the acceleration of the object?
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RESPONSE --> Fnet = 50 Newtons m = 20kg fFrict = -10Newtons Hopefully I got it right and we're looking for F... Fnet = F + fFrict F = Fnet - fFrict = 50 - -10 = 60Newtons a = F/m = 60Newtons / 20kg = 3m/s^2
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21:26:17 If we take the direction of motion to be positive, then since both the 50 Newton force and the 10 Newton force are opposed to the direction of motion the net force must be net force = -50 Newtons - 10 Newtons = -60 Newtons. The acceleration of the object will therefore be a = Fnet / m = -60 Newtons / (10 kg) = -60 kg * m/s^2 / (20 kg) = -3 m/s^2. The fact that the acceleration is opposed to the direction of motion indicates that the object will be slowing down. The force exerted by the person, being in the direction opposite to that of the motion, is seen to be a retarding force, as is friction. So in this case the person is aided by friction in her apparent goal of stopping or at least slowing the object.
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RESPONSE --> Well I had everything right with the exception of the Fnet = -50 since it is against the motion of the object.
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21:48:44 `q008. If a 40 kg object is moving at 20 m/s, then how long will a take a net force of 20 Newtons directed opposite to the motion of the object to bring the object to rest?
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RESPONSE --> m = 40kg Fnet = -20newtons v = 20m/s 'dt = ? a = F/m = -20N/40kg = -0.5m/s^2 I'm not sure if the 20m/s is v0, and the vf = 0. All the answers I'm getting have a negative time limit and something tells me that's not possible. I'm not sure what I need to do to get the 'dt. Maybe this: vf = v0+dv dv = vf - v0 = 0 - 20 = -20 'dt = 'dv / a = -20 / -.5 = 40seconds?
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21:51:09 The force on the object is in the direction opposite its motion, so if the direction of motion is taken to be positive the force is in the negative direction. We therefore write the net force as Fnet = -20 Newtons. The acceleration of the object is therefore a = Fnet / m = -20 Newtons / 40 kg = -20 kg * m/s^2 / (40 kg) = -.5 m/s^2. We can therefore describe uniformly accelerated motion of the object as v0 = 20 m/s, vf = 0 (the object comes to rest, which means its velocity ends up at 0), a = -.5 m/s^2. We can then reason out the time required from the -20 m/s change in velocity and the -.5 m/s^2 acceleration, obtaining `dt = 40 seconds. We can confirm this using the equation vf = v0 + a `dt: Solving for `dt we obtain `dt = (vf - v0) / a = (0 m/s - 20 m/s) / (-.5 m/s^2) = -20 m/s / (-.5 m/s^2) = 40 m/s * s^2 / m = 40 s.
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RESPONSE --> Awesome. I tried using the vf = v0+a'dt equation to solve for 'dt, but I kept getting negative answers b/c I wasn't doing the math right.
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21:54:54 `q009. If we wish to bring an object with mass 50 kg from velocity 10 m/s to velocity 40 m/s in 5 seconds, what constant net force would be required?
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RESPONSE --> m = 50kg v0 = 10m/s vf = 40m/s 'dt = 5s a = (vf - v0)/'dt = (40m/s - 10m/s) / 5s = 6m/s^2. Fnet = m*a = 50kg * 6m/s^2 = 300Newtons
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21:55:05 The net force would be Fnet = m * a. The acceleration of the object would be the rate which its velocity changes. From 10 m/s to 40 m/s the change in velocity is +30 m/s; to accomplish this in 5 seconds requires average acceleration 30 m/s / (5 s) = 6 m/s^2. Thus the net force required is Fnet = 50 kg * 6 m/s^2 = 300 kg m/s^2 = 300 Newtons.
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RESPONSE --> ok
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21:58:06 `q010. If a constant net force of 50 Newtons brings an object to rest in four seconds from an initial velocity of 8 meters/second, then what must be the mass of the object?
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RESPONSE --> Fnet = -50Newtons 'dt = 4s vf = 0 v0 = 8m/s kg = ? a = (vf - v0) / 'dt = (0 - 8m/s) / 4s = -2m/s^2 Fnet = m*a m = Fnet / a = -50Newtons / -2m/s^2 = 25kg
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21:58:13 We know the net force and we have the information required to calculate the acceleration. We will therefore be able to find the mass using Newton's Second Law Fnet = m * a. We first find the acceleration. The change in velocity from 8 m/s to rest is -8 m/s, and this occurs in 4 seconds. The acceleration is therefore -8 m/s / (4 s) = -2 m/s^2. The 50 Newton net force must be in the same direction as the acceleration, so we have Fnet = -50 Newtons. We obtain the mass by solving Newton's Second Law for m: m = Fnet / a = -50 N / (-2 m/s^2) = -50 kg m/s^2 / (-2 m/s^2) = 25 kg.
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RESPONSE --> ok
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è~߬w÷ÀÚÊ„±}†åôÓÏ„ assignment #009 °~—¼”ȪÁs²ý¦J»©®Ž‘£Êíäì½¥– Physics I 10-11-2005
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13:03:25 Introductory prob set 3 #'s 1-6 If we know the distance an object is pushed and the work done by the pushing force how do we find the force exerted by the object?
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RESPONSE --> dW = Fn * 'ds Since we know 'ds and dW, then in order to find the force, solve the equation for Fn. Fn = dW / 'ds
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13:03:32 ** Knowing the distance `ds and the work `dW we note that `dW = F * `ds; we solve this equation and find that force is F=`dw/`ds **
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RESPONSE --> ok
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13:04:51 If we know the net force exerted on an object and the distance through which the force acts how do we find the KE change of the object?
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RESPONSE --> We know Fn and 'ds, then KE change of the object is KE = Fn * 'ds
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13:10:32 **`dW + `dKE = 0 applies to the work `dW done BY the system and the change `dKE in the KE OF the system. The given force acts ON the system so F `ds is work done ON the system. The work done BY the system against that force is `dW = -F * `ds. When you use the energy equation, this is the work you need--the work done BY the system. **
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RESPONSE --> In the notes, it says that the work done on the object will be the product of force and distance... which is how I decided that KE = Fn * 'ds. So is dW + dKE = 0 the equation we need or is it the other one? I get that the dW + dKE = 0 is used when the work is done by the system, not on the system...
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13:15:33 Why is KE change equal to the product of net force and distance?
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RESPONSE --> Isn't net force and distance proportional to each other? That would make them proportional to the KE change also wouldn't it?
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13:17:19 ** It comes from the equation vf^2 = v0^2 + 2 a `ds. Newton's 2d Law says that a = Fnet / m. So vf^2 = v0^2 + 2 Fnet / m `ds. Rearranging we get F `ds = 1/2 m vf^2 - 1/2 m v0^2. Defining KE as 1/2 m v^2 this is F `ds = KEf - KE0, which is change in KE. **
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RESPONSE --> Ok none of this is in the problem set that I have. I couldn't get the link to problem number 4 to work so I'm assuming that this method is shown in that problem. It makes sense but I need to study it more to really make the concept concrete.
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13:18:42 When I push an object with a constant force, why is KE change not equal to the product of the force I exert and the distance?
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RESPONSE --> The object must do work equal and opposite to the work done on it in order for the change in KE to be equal... (I think)
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13:20:16 ** Change in KE is equal to the work done by the net force, not by the force I exert. When I push an object in the real world, with no other force 'helping' me, there is always at least a little force resisting my push. So the net force in this case is less than the force I exert, in which case the change in KE would be less than the product of the force I exert and the distance. If another force is 'helping' me then it's possible that the net force could be greater than the force I exert, in which case the change in KE would be greater than the product of the force I exert and the distance. It is actually possible for the 'helping' force to exactly balance the resisting force, but an exact balance would be nearly impossible to achieve. ANOTHER WAY OF LOOKING AT IT: If I push in the direction of motion then I do positive work on the system and the system does negative work on me. That should increase the KE of the system. However if I'm pushing an object in the real world and there is friction and perhaps other dissipative forces which tend to resist the motion. So not all the work I do ends up going into the KE of the object. **
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RESPONSE --> ok
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13:20:44 Query pendulum force proportionalities (video exp #9) Describe how we conclude that for small displacements the force tending to restore a pendulum to equilibrium is proportional to the displacement
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RESPONSE --> Video exp #9 is not currently assigned.
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13:21:01 ** Since displacement over length equals the ratio of the effective suspended mass over mass of pend., and this ratio is equal to Force over weight, then displacement is also proportionate to the force. **
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RESPONSE --> ok
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13:21:08 Explain how your common sense leads you to believe in the plausibility of the hypothesis that the force tending to restore a pendulum to its equilibrium position is in fact in the same proportion to the weight of the pendulum as the displacement to the length of the pendulum.
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RESPONSE --> ok
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13:21:20 ** Since weight is the measurement of gravity acting on mass, then the weight implies some force, and the length of pendulum implies some distance over which gravity works, then there should be a set proportion. **
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RESPONSE --> ok
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13:21:33 Describe how we conclude that for small displacements the force tending to restore a pendulum to equilibrium is in the same proportion to the weight of the pendulum as the displacement to the length of the pendulum.
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RESPONSE --> ok
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13:22:07 ** Intuitively, the further you pull back the greater the restoring force. As you pull back the ratio x / L increases. As you pull back the ratio F / (M g) = restoring force / pendulum weight also increases. The proportionality between the two says, basically, that if you pull the pendulum back a distance equal to 5% of its length then the restoring force is 5% of the pendulum's weight. This proportionality applies very well to small displacements. As displacement gets larger the linearity of the relationship deteriorates. More rigorously: The force m * g exerted on the suspended mass m is the force exerted by gravity, and with a small correction for friction is equal to the displacing force. This force is equal and opposite to the restoring force. The weight of the pendulum is the force M * g exerted by gravity on its mass M. If x / m = L / M then x / L = m / M = m * g / M * g = displacing force / weight. **
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RESPONSE --> ok
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Week 4 Quiz #2 Version #16 An object is given an unknown initial velocity up a long ramp on which its acceleration is known to have magnitude of 20cm/s^2. 0.142 seconds later it passes, for the first time, a point 6.5cm up the ramp from its initial position. · What is its initial velocity, and how much longer will it be before it passes this point again? a = 20cm/s^2 ‘dt = .142s ‘ds = 6.5cm v0 = ? ‘ds = v0*’dt + 0.5a’dt^2 6.5cm = v0*.142s + .5(20cm/s^2) (.142^2) 6.5cm = v0*.142s + .20164cm/s^2 6.5cm - .20164cm/s^2 = v0*.142s 6.29836cm/s^2 / 0.142s = v0 v0 = 44.355cm/s a = 20cm/s^2 v0 = 44.355cm/s