course PHY121
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18:05:07 `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 --> a = F / m 30kg + 10(2) = 50kg cart. 50kg - 2kg = 48kg cart. a = 9.8kg/s^2 / 48kg = 0.204 kg/s^2 I'm not really sure, this is just a guess.
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18:07:55 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 --> I wasn't completely wrong in my process, I just didn't know in the beginning to multiply 2 kg * 9.8 m/s^2 = 19.6 Newtons. And even though the washer was relocated, the weight of the entire system remained the same so my process of subtracting the 2kg weight was off completely.
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18:21:30 `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 --> F = m * a F = 1.1kg * 9.8m/s^2 = 10.78Newtons for the side of the pulley with the 100g mass. F = 1kg * 9.8m/s^2 = 9.8Newtons for the side without the extra 100g mass. F = 2.1kg * 9.8m/s^2 = 20.58Newtons for the entire system
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18:25:43 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 --> I couldn't figure out the Fnet. I know the equation for Fnet = F - Ffrict, but I couldn't figure out how the Ffrict applied to this problem.
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18:36:29 `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 --> the system weighs 2.1kg * .01 = .021.
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18:42:17 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 --> Ok so the frictional force is taken from the overall weight of the system, not just from the mass like I did. Then I use the Fnet = F - Ffrict equation to find the net force and then divide it by the mass of the system to get the acceleration. Seeing the answer written out, it makes sense but for some reason, I still doubt that I'd remember all this if I needed to do it again.
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18:43:40 `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 --> If the system is heavier on that one side, then there's more work being done on it to push it downward. I'd think the system would have to move in that direction if it's up to just the gravitational forces to move it.
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18:45:08 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 --> So my idea was right. If something (in this case, us pushing it in the opposite direction) other than the gravity affects the system, then it won't always move downward. But when it is just gravitational force being considered, then it will move in the direction with the heaviest weight.
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18:46:57 `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 --> I have no idea. I know the system will have to pause and have an acceleration of 0 for a second before it starts to fall back toward the heavier side of the pulley, but as for figuring out the acceleration of that entire process, I'm lost.
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18:50:54 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 --> Easy enough. My question is that if the system was being pulled in the opposite direction (towards the side with the least amount of weight), then wouldn't the math be -10.78N + 9.8N -.21N = -1.19Newtons?
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18:55:01 `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 --> I'm assuming the 100g will be added to one side of the system... So F = 10.1kg * 9.8m/s^2 = 98.98Newtons F = 10kg * 9.8m/s^2 = 98Newtons 10+10.1kg = 20.1kg 98.98 - 98 = 0.98 Newtons / 20.1kg = .049 m/s^2
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18:55:29 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 --> ok cool.
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18:59:21 `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 --> So the friction is 1% of the total weight, which in this case is 20.1 * .01 = 0.201, 20.1kg - 0.201 = 19.899kg for the entire system. So the Fnet = 98.98N - 98.0N - 0.201N = 0.779Newtons. 0.779N / 20.1kg = .039 m/s^2 for the acceleration
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19:04:01 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 --> So again, I calculated the 1% wrong. I think I made the exact same mistake that I did previously, only while I was calculating, I thought I'd done the right thing. I just keep thinking that the 20.1kg is the weight, when it's only the mass. That's my problem, I keep assuming they're the same or something. Other than that I did everything else right, just with the wrong numbers.
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19:08:17 `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 --> "" 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."" So by this, I'd say that the weight of the object is 3kg * 9.8m/s^2 = 29.4Newtons. Then the slope of 0.7 * 29.4N = 20.58newtons as the gravitational force.
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19:15:07 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 --> So by saying that I weigh 150lbs, we're only looking at the mass (that is, if converted to kg.) Technically my weight is 150 lb / 2.20lb = 68kg 68kg * 9.8m/s^2 = 668.2Newtons. Saying I weight 668Newtons sounds much worse than saying 150 pounds...
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19:20:15 `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 --> The weight is 29.4N... So, Fnet = 29.4N - 9.8N = 19.6N 19.6N / 3kg = 6.5m/s^2 I'm not sure if this is right or not.
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19:21:46 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 so if there's no friction and we only have the 29.4N to work with, then we simply do the a = F / m equation to figure out acceleration.
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19:30:34 `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 --> 29.4N * .02 = 0.588N is the friction. 29.4N - 0.588N = 28.812N 28.81N / 3kg = 9.6m/s^2 Something tells me that this isn't right either.
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19:32:01 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 couldn't figure out where the 2.1Newtons fit into this equation. I'm doing everything right I just can't ever figure out which weights to use in the equations.
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19:33:48 `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 --> -2.1N - .59N = -2.69N a = -2.69N / 3kg = -0.89 m/s^2
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19:34:19 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 --> ok
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19:39:24 `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 --> I have no idea what this is even asking. So the cart is moving up the ramp, but it's starting a few meters from the bottom of the ramp? Does this mean that if the ramp is 20m long, then the initial velocity is taken at, say, 3m? I think the way it's worded has just confused me... In order to move the cart up the incline, it has to have some form of weight pulling or pushing it, so I'm just guessing and saying that the acceleration is constant, but I don't have any clue.
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19:41:16 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 --> Where's the .9m/s^2 and the .5m/s come from?
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zcaJhʢjװdϏ assignment #010 ~ȪsJ콥 Physics I Class Notes 10-18-2005
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16:03:39 Why does it make sense that the work done by gravity on a set of identical hanging washers should be proportional to the product of the number of washers and the distance through which they fall?
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RESPONSE --> It is proportional to the number of washers and the distance they fall because the more washers used and the longer the distance becomes, the more work that gravity must do to push the heavier weight downward.
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16:03:55 ** When force is parallel to displacement, work is force * distance, so the proportionality with distance follows immediately. The force comes from the gravitational interaction of the masses of the washers with the Earth's gravitational field. More washers imply more force, and the force is proportional to the number of washers. **
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RESPONSE --> ok
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16:05:05 Why is this consistent with the idea that the work done on a given cart on an incline is proportional to the vertical distance through which the cart is raised?
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RESPONSE --> It's the same concept. The higher the vertical distance, the more work that must be done in order for it to climb the incline.
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16:06:00 ** The steeper the incline the more washers are required to pull the cart up the incline. In fact for small inclines and correcting for friction the number of washer is proportional to the slope of the incline. The vertical distance, being the rise through which the cart is raised, is also proportional to the slope of the incline. So for a given cart the number of washers is proportional to the incline. Since work is proportional to the number of washers we conclude the work is proportional to the vertical distance through which the cart is raised. **
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RESPONSE --> ok so the relationship between the work done, the vertical distance and the number of washers are all proportional to each other.
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16:10:34 How does the work done against friction of the cart-incline-pulley-washer system compare with the work done by gravity on the washers and the work done to raise the cart?
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RESPONSE --> A certain number of washers are needed to overcome the friction of the cart and the pulley. In order to pull the car up the incline, the force of gravity needed is equal to the leftover washers that are not used to overcome friction.
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16:10:42 ** The work against friction plus the work done to raise the cart are all provided by the gravitational force acting on the descending weight. **
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RESPONSE --> ok
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16:15:39 How do we establish, using a cart-pulley-washer system on a constant-velocity incline, that the acceleration of cart is proportional to the net force on the cart?
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RESPONSE --> The function a('dt) = 98/dt^2 showed a graph that seemed to be linear and also passing through the origin, which shows that acceleration and net force are proportional. Also, since the force depends on the weight of the washers and the incline, that means that it also is proportional to the acceleration of the object.
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16:16:18 ** STUDENT ANSWER: In order to keep the velocity constant the amount of force should be constant but it must overcome the effects of gravity and therefore it must increase at a constant rate . The increase in the ramp slope will correspond to the increase in the force. INSTRUCTOR RESPONSE: Good start. This is half of it. This indicates how the force on the cart is proportional to acceleration. The other half is that the acceleration of the cart is proportional to the effective slope. Since the force and the acceleration of the cart are both proportional to the effective slope, they are proportional to each other. **
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RESPONSE --> ok
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E،ےwc assignment #011 ~ȪsJ콥 Physics I Class Notes 10-18-2005
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16:22:35 How do the experiment with acceleration vs. number of washers and experiment with acceleration vs. ramp slope convince us that for a given cart there is a net force down the ramp which is proportional to the slope in excess of the constant-velocity slope (the slope excess)?
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RESPONSE --> I'm not sure, but I think that since the ramp slope is proportional to the acceleration of the object, then the net force on the object will be greater for a higher slope of the ramp.
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16:23:08 ** A graph of acceleration vs. the number of washers in excess of the number required to maintain a constant velocity is consistent with a straight-line graph through the origin. **
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RESPONSE --> oh ok. I didn't think to mention the graph characteristics.
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16:24:09 If for small slopes we find that the net force exerted on a cart is proportional to the slope excess, then why does Newton's Second Law ensure us that for small slopes the acceleration of the cart should in fact be proportional to the slope excess?
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RESPONSE --> Since the acceleration is proportional to the slope excess, then the net force will all be proportional to the slope excess, which menas the acceleration is proportional to the net force.
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16:24:18 ** Newton's Second Law tells us that for a fixed mass the acceleration is proportional to the force, so if for small slopes force is proportional to slope excess, then acceleration must also be proportional to slope excess. **
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RESPONSE --> ok
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16:25:52 How is the slope of the net force vs. acceleration graph for an object reveal the mass of the object? Why should a greater mass result in a greater slope?
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RESPONSE --> The greater mass results in a greater slope because more force is needed to accelerate it. mass is the force / acceleration, so on the graph, it is the rise/run of the line, in which case is the slope.
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16:25:56 ** Since F = m a, when we graph F vs. a the slope is m. **
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RESPONSE --> ok
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16:30:17 How do we obtain our definitions of work and kinetic energy in terms of the equation vf^2 = v0^2 - 2 a `ds and Newton's Second Law?
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RESPONSE --> the vf^2 = v0^2 - 2 a `ds equation becomes vf^2 = v0^2 + 2(F/m) 'ds since a = F/m. Then F'ds = 1/2 m vf^2 - 1/2 mv0^2 and 1/2mv^2 is KE. Therefore, the work done by the net force on the object is equal to the KE change by the object.
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16:30:35 ** Substituting a = Fnet / m into vf^2 = v0^2 + 2 a `ds we get vf^2 = v0^2 + 2 Fnet / m * `ds, which we easily rearrange to get Fnet `ds = 1/2 m vf^2 - 1/2 m v0^2. Fnet `ds is defined as work, 1/2 m v^2 is defined as KE, and our equation tells us that the work done by the net force is equal to the change in KE. **
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RESPONSE --> ok
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Week 5 Quiz #2 Version #17 If the slope of a graph of number of paper clips needed to maintain equilibrium v/s ramp slope is 50clips/unit of ramp slope, and if the slope of a graph of the acceleration of a cart v/s the number of paper clips attached by a string and suspended over a pulley is 20cm/s^2 / clip, then how many cm/s^2 of acceleration should correspond to 1 unit of ramp slope? If we require 45 clips to match the mass of the cart, then if we could apply this force to the cart without the extra mass of all those clips, what would be the acceleration of the cart? 1 unit of ramp slope = 50paper clips, and each paper clip = 20cm/s^2, then wouldnt the units of acceleration be 50 * 20cm/s^2 = 1000cm/s^2 units of acceleration needed for 1 unit of ramp slope?