course Phy 121 ?m????{?D?????assignment #011
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11:36:51 set 3 problems 15-19. Explain the difference between a conservative and a nonconservative force, and give an example of each.
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RESPONSE --> P. 148 in textbook Forces such as gravity, for which the work done doesn not depend on the path taken but only on the initial and final positions, are called conservative forces. Friction, on the other hand, is a nonconservative force since the work it does depends on the path. confidence assessment: 2
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11:37:13 ** A conservative force conserves energy--you can get your energy back. For example: Push something up a hill, climb back down the hill and turn your back and it will probably return your energy to you--all at once--after regaining it as it rolls back down. So you do work against gravity, and gravity can return the energy as it pulls the thing back to you. However, there is some friction involved--you do extra work against friction, which doesn't come back to you. And some of the energy returned by gravity also gets lost to friction as the object rolls back down the hill. This energy isn't conserved--it's nonconservative. **
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RESPONSE --> I think my answer was correct. self critique assessment: 3
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11:40:38 If a system does work W1 against a nonconservative force while conservative forces do work W2 on the system, what are the change in the KE and PE of the system? Explain your reasoning from a commonsense point of view, and include a simple example involving a rubber band, a weight, an incline and friction.
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RESPONSE --> Wnc = 'dKE - Wc I am not quite sure how to explain an example, but I know all of this has to do (at least in some way) with the work-energy principle. confidence assessment: 1
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11:42:24 ** `dKE is equal to the NET work done ON the system. The KE of a system changes by an amount equal to the net work done on a system. If work W1 is done BY the system against a nonconservative force then work -W1 is done ON the system by that force. `dPE is the work done BY the system AGAINST conservative forces, and so is the negative of the work done ON the system BY nonconservative forces. In this case then `dPE = - W2. PE decreases, thereby tending to increase KE. If work -W1 is done ON the system by a nonconservative force and W2 is done ON the system by a conservative force, the NET work done ON the system is -W1 + W2. The KE of the system therefore changes by `dKE = -W1 + W2. If the nonconservative force is friction and the conservative force is gravity, then since the system must do positive work against friction, W1 must be positive and hence the -W1 contribution to `dKE tends to decrease the KE. e.g., if the system does 50 J of work against friction, then there is 50 J less KE increase than if there was no friction. If the work done by the nonconservative force on the system is positive, e.g., gravity acting on an object which is falling downward (force and displacement in the same direction implies positive work), the tendency will be to increase the KE of the system and W2 would be positive. If W2 is 150 J and W1 is 50 J, this means that gravity tends to increase the KE by 150 J but friction dissipates 50 J of that energy, so the change in KE will be only 100 J. If the object was rising, displacement and gravitational force would be in opposite directions, and the work done by gravity would be negative. In this case W2 might be, say, -150 J. Then `dKE would be -150 J - 50 J = -200 J. The object would lose 200 J of KE (which would only be possible if it had at least 200 J of KE to lose--think of an object with considerable velocity sliding up a hill). **
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RESPONSE --> It makes sense to read this solution, but I'm not sure that I would have been able to come up with all of this using just my thought processes. How did you know where to start and what aspects to explain? self critique assessment: 2
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11:44:14 If the KE of an object changes by `dKE while the total nonconservative force does work W on the object, by how much does the PE of the object change?
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RESPONSE --> 'dKE = KE2 - KE1 Wnc = 'dKE + 'dPE Wnc - 'dKE = 'dPE P. 149 confidence assessment: 0
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11:45:11 ** We have `dKE + `dPE + `dWbyNoncons = 0: The total of KE change of the system, PE change of the system and work done by the system against nonconservative forces is zero. Regarding the object at the system, if W is the work done ON the object by nonconservative forces then work -W is done BY the object against nonconservative forces, and therefore `dWnoncons = -W. We therefore have `dKE + `dPE - W = 0 so that `dPE = -`dKE + W. **
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RESPONSE --> I think my answer may have been close to what it should have been. self critique assessment: 2
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11:45:54 Give a specific example of such a process.
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RESPONSE --> Push or pull by a person. confidence assessment: 0
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11:46:25 ** For example if I lift an object weighing 50 N and in the process the total nonconservative force (my force and friction) does +300 J of work on the object while its KE changes by +200 J then the 300 J of work done by my force and friction is used to increase the KE by 200 J, leaving 100 J to be accounted for. This 100 J goes into the PE of the object. **
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RESPONSE --> I see where I made my mistake. self critique assessment: 2
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18:20:42 Class notes #10. 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? 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 --> According to the class notes, the net force acting on the system is the force of gravity on the washers, and is hence proportional to the number of washers suspended. When data are corrected for frictional resistance, adn when the mass of the system is kept constant by keeping any unused washers in the cart, the graph of the ideal situation will be perfectly linear and will pass through the origin. This graph indicates that the acceleration of the cart is directly proportional to the number of washers and hence to the net force acting on the system. Also according to class notes: For any number of suspended washers, there exists a ramp slope such that the washers tend to pull the cart up the ramp without accelerating it. The slope of a particular graph tells us how many units of ramp slope correspond to one washer. From the first two graphs, where we get the acceleration change per washer, we should be able to determine how much change in ramp slope corresponds to addition of a washer. This last quantity is the slope of the third graph. If the change in ramp slope per additional washer is close to the slope of the third graph, then we can say that the results of the three experimetns are consistent. confidence assessment: 1
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18:20:50 ** The force exerted by gravity is the same on each clip, so the total gravitational force on the hanging clips is proportional to the number of clips. The work done is the product of the force and the displacement in the direction of the force, so the work done is proportional to product of the number of washers and the vertical displacement. To pull the cart up a slope at constant velocity the number of washers required is proportional to the slope (for small slopes), and the vertical distance through which the cart is raised by a given distance of descent is proportional to the slope, to the work done is proportional to the vertical distance thru which the cart is raised. **
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RESPONSE --> self critique assessment: 3
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18:51:45 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? Which is greatest? What is the relationship among the three?
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RESPONSE --> According to class notes: When a force is exerted in or opposite to the direction of an object's motion, we say that the force does work on the object. Work can raise an object against the force of gravity, or work can speed up or slow down an object. When the washers descend, we think of gravity as doing work on them. When the cart ascends, we think of work being done on it to raise it. Whe the cart moves up an incline at constant velocity, the work done on it serves only to raise it. The cart doesn't speed up, so none of the work done contributes to its speed. Since more washers and more distance imply more work, we regard the work done on the washers as the force exerted on the washers by gravity multiplied by the distance through which they descend. Since a heavier cart moved through a greater vertical distance would be associated with more work, we think of the work done on the cart as the product of its weight and distance. The work done by gravity is greatest because of conservative forces. Friction is a nonconservative force, and nothing can be gained back from this particular force. confidence assessment: 1
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18:53:00 ** The force exerted by gravity on the hanging weights tends to move the system up the incline. The force exerted by gravity on the cart has a component perpendicular to the incline and a component down the incline, and the force exerted by friction is opposed to the motion of the system. In order for the cart to move with constant velocity up the incline the net force must be zero (constant velocity implies zero accel implies zero net force) so the force exerted by gravity in the positive direction must be equal and opposite to the sum of the other two forces. So the force exerted by gravity on the hanging weights is greater than either of the opposing forces. So the force exerted by friction is less than that exerted by gravity on the washers, and since these forces act through the same distance the work done against friction is less than the work done by gravity on the washers. The work done against gravity to raise the cart is also less than the work done by gravity on the washers. The work friction + work against gravity to raise cart = work by gravity on the hanging weights. **
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RESPONSE --> I may have touched upon some of the points mentioned in the solution, but I think I may be missing the 'big picture'. What am I missing? self critique assessment: 2
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19:01:46 What is our evidence that the acceleration of the cart is proportional to the net force on the cart?
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RESPONSE --> To answer this question we must consider the definition of net force. Net force is the sum total of all the forces acting on a particular object in motion. Therefore, the addition or subtraction of any force on the cart (ex. acceleration) will automatically affect the net force. confidence assessment: 1
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19:02:13 ** the graph of acceleration vs. number of washers should be linear **
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RESPONSE --> self critique assessment: 3
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19:03:17 prin phy and gen phy prob 34: Car rolls off edge of cliff; how long to reach 85 km/hr?
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RESPONSE --> Shouldn't I have more information in order to answer this question properly? confidence assessment: 0
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19:03:57 We know that the acceleration of gravity is 9.8 m/s^2, and this is the rate at which the velocity of the car changes. The units of 85 km/hr are not compatible with the units m/s^2, so we convert this velocity to m/s, obtaining 85 km/hr ( 1000 m/km) ( 1 hr / 3600 sec) = 23.6 m/s. Common sense tells us that with velocity changing at 9.8 m/s every second, it will take between 2 and 3 seconds to reach 23.6 m/s. More precisely, the car's initial vertical velocity is zero, so using the downward direction as positive, its change in velocity is `dv = 23.6 m/s. Its acceleration is a = `dv / `dt, so `dt = `dv / a = 23.6 m/s / (9.8 m/s^2) = 2.4 sec, approx..
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RESPONSE --> Okay, I see where I made my mistake. self critique assessment: 3
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19:04:40 **** prin phy and gen phy problem 2.52 car 0-50 m/s in 50 s by graph How far did the car travel while in 4 th gear and how did you get the result?
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RESPONSE --> Wouldn't I need to know what gear the car was in to start with? confidence assessment: 0
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19:05:21 ** In 4th gear the car's velocity goes from about 36.5 m/s to 45 m/s, between clock times 16 s and 27.5 s. Its average velocity on that interval will therefore be vAve = (36.5 m/s + 45 m/s) / 2 = 40.75 m/s and the time interval is 'dt = (27.5s - 16s) = 11.5 s. We therefore have 'ds = vAve * `dt = 40.75 m/s * 11.5 s = 468.63 m. The area under the curve is the distance traveled, since vAve is represented by the average height of the graph and `dt by its width. It follows that the area is vAve*'dt, which is the displacement `ds. The slope of the graph is the acceleration of the car. This is because slope is rise/run, in this case that is 'dv/'dt, which is the ave rate of change of velocity or acceleration. We already know `dt, and we have `dv = 45 m/s - 36.5 m/s = 8.5 m/s. The acceleration is therefore a = `dv / `dt = (8.5 m/s) / (11.5 s) = .77 m/s^2, approx. **
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RESPONSE --> How do you know this initial infomation mentioned in the first sentence of the solution? self critique assessment: 2
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19:05:31 **** Gen phy what is the meaning of the slope of the graph and why should it have this meaning?
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RESPONSE --> Not applicable. confidence assessment: 0
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19:05:39 ** The graph is of velocity vs. clock time, so the rise will be change in velocity and the run will be change in clock time. So the slope = rise/run represents change in vel / change in clock time, which is acceleration. **
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RESPONSE --> self critique assessment: 3
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19:05:48 Gen phy what is the meaning of the area under the curve, and why should it have this meaning?
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RESPONSE --> Not applicable. confidence assessment: 0
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19:05:52 ** The area under the curve is the distance traveled. This is so because 'ds = vAve*'dt. 'dt is equal to the width of the section under the curve and vAve is equal to the average height of the curve. The area of a trapezoid is width times average height. Although this is not a trapezoid we can annalyze it as one for estimation puposes. **
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RESPONSE --> self critique assessment: 3
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19:06:06 Gen phy what is the area of a rectangle on the graph and what does it represent?
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RESPONSE --> Not applicable. confidence assessment: 0
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19:06:11 ** The area of a rectange on the graph represents a distance. **
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RESPONSE --> self critique assessment: 3
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19:06:19 univ phy problem 2.90 from 10th edition (University Physics students should solve this problem now). Superman stands on the top of a skyscraper 180 m high. A student with a stopwatch, determined to test the acceleration of gravity for himself, steps off the top of the building but Superman can't start after him for 5 seconds. If Superman then propels himself downward with some init vel v0 and after that falls freely, what is the minimum value of v0 so that he catches the student before that person strikes the ground?
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RESPONSE --> Not applicable. confidence assessment: 0
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19:06:29 univ phy what is Superman's initial velocity, and what does the graph look like (be specific)?
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RESPONSE --> Not applicable. confidence assessment: 0
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19:06:34 ``a** In time interval `dt after leaving the building the falling student has fallen through displacement `ds = v0 `dt + .5 a `dt^2, where v0 = 0 and, choosing the downward direction to be positive, we have a = -9.8 m/s^2. If `ds = -180 m then we have `ds = .5 a `dt^2 and `dt = sqrt(2 * `ds / a) = sqrt(2 * -180 m / (-9.8 m/s^2)) = 6 sec, approx.. Superman starts 5 seconds later, and has 1 second to reach the person. Superman must therefore accelerate at -9.8 m/s^2 thru `ds = -180 m in 1 second, starting at velocity v0. Given `ds, `dt and a we find v0 by solving `ds = v0 `dt + .5 a `dt^2 for v0, obtaining v0 = (`ds - .5 a `dt^2) / `dt = (-180 m - .5 * -9.8 m/s^2 * (1 sec)^2 ) / (1 sec) = -175 m/s, approx. Note that Superman's velocity has only about 1 second to change, so changes by only about -9.8 m/s^2, or about -10 m/s^2. **
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RESPONSE --> confidence assessment: 3
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19:06:45 ``qsketch a graph of Superman's position vs. clock time, and on the same graph show the student's position vs. clock time, with clock time starting when the person begins falling
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RESPONSE --> Not applicable. confidence assessment: 0
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19:06:50 ** If we start our clock at t = 0 at the instant the student leaves the top of the building then at clock time t the student's `dt will be just equal to t and his position will be x = x0 + v0 t + .5 a t^2 = .5 a t^2, with x0 = 180 m and a = -9.8 m/s^2. A graph of x vs. t will be a parabola with vertex at (0,180), intercepting the t axis at about t = 6 sec. For Superman the time of fall will be `dt = t - 5 sec and his position will be x = x0 + v0 (t-5sec) + .5 a (t-5sec)^2, another parabola with an unspecified vertex. A graph of altitude vs. t shows the student's position as a parabola with vertex (0, 180), concave downward to intercept the t axis at (6,0). Superman's graph starts at (5,180) and forms a nearly straight line, intercepting the t axis also at (6,0). Superman's graph is in fact slightly concave downward, starting with slope -175 and ending with slope -185, approx. **
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RESPONSE --> self critique assessment: 3
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14:56:25 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 --> If you consider the idea of net force, this at least partially explains why the variables in this situation would be related. Net force is the sum total of all the forces acting on an object in motion. The addition or subtraction of any force acting on an object will have an effect on the net force. Therefore in this situation, proportionally should be conclused because if the washers have the same mass and gravity is acting in the same way on all of these equal masses then the distance they fall should be equal as well.
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14:57:30 ** 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 --> Okay, I didn't explain my answer in terms of the formula for work (W = F * d), but I was at least on the right track.
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15:05:32 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 --> Consider the definition of work; work equals force times distance. In this situation we have work being done on a cart over a certain distance. Therefore force would have to be proportional to the amount of distance covered by the cart.
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15:06:34 ** 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 --> I think my answer was okay aside from neglecting to mention that the number of washers is proportional to the incline.
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15:12:42 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 --> Both systems still allow work to be done to an object, however, these systems do have important differences. For example, friction is a nonconservative force; that is to say that you can get work back out of a frictional force. Gravity is a conservative force because it is possible to get work back out of it.
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15:13:32 ** 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 --> I see where I made my mistake. I think all of the things I said were true, but not what the question was asking in this particular case.
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15:16:54 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 --> Consider the definition of net force: Net force is the sum total of all the forces acting on an object in motion. The definition of acceleration is the change in speed divided by time. Velocity (change in speed) can be quantified as a vector, therefore acceleration can be considered as a force acting upon this object that if added or subtracted from the mentioned situation would affect the overall net force.
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15:18:10 ** 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 --> I may have gotten the part about how the forces are related and proportional to one another, however I didn't mention how the acceleration of the cart is related to the slope.
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??????q????assignment #011 011. Note that there are 12 questions in this set. .Situations involving forces and accelerations. Physics I 07-24-2007
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15:23:20 `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 --> I am not entirely sure about how to properly answer this question. F = ma, but we don't know the force or acceleration. We don't know frictional force. We don't know the distance covered, however we do know the constant for gravity is 9.8m/s^2. I am sure you use gravity and the mass somehow, but I don't know the proper formula for the presented situation. confidence assessment: 0
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15:25:25 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 --> This solution makes sense. It's nice to know I was on the right track anyway. How did you know to first multiply 2kg times 9.8? self critique assessment: 2
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15:31:20 `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 = ma 1 kg * 9.8 = 9.8N 100g * (1000kg / 1 g) = 100000 kg 100001 kg * 9.8 = 980009 N This second set of calculations may not be entirely correct. The net force would be the sum total of the two calculated forces listed above, but since the second calculation is incorrect this answer is likely to be incorrect as well. 9.8N + 980009N = 980019N = net force confidence assessment: 0
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15:37:03 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 --> Okay my conversion on the second part was completely wrong. However the rest would have been correct once I had taken into account that the forces were opposing one another. self critique assessment: 2
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15:39:12 `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 --> Because I missed the problem, I haven't written down all the correct values; but, friction must be subtracted from the total net force because friction is a nonconservative force. confidence assessment: 0
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15:41:03 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 --> This solution makes sense. Had I written down all of the correct values mentioned in the solution to the previous problem I could have made a similar conclusion as mentioned here. self critique assessment: 2
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15:43:37 `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 --> We have to know what direction the system is moving so we know and/or can calculate the forces contributing to the overall net force. The system doesn't have to move in that direction, it depends on what forces are acting on the system and the masses involved. F = ma confidence assessment: 1
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15:45:00 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 --> I think I may have misunderstood what the question was asking here. I didn't consider whether the system was at rest, or if it was already in motion. self critique assessment: 2
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15:46:07 `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 --> .77 Newtons / (2.1 kg) = .35 m/s^2 confidence assessment: 0
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15:47:14 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 --> I made a miscalculation in summing (and subtracting) all of the forces here. Otherwise, I would have gotten the correct answer. self critique assessment: 2
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15:49:18 `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 am still working out how to do the correct conversion in these types of problems, however the approach would be the same as in the previous situation (neglecting friction) except you would use the addition of 100 grams. confidence assessment: 0
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15:50:08 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 --> If I would have done the conversion correctly I would have gotten the correct answer. Could you show me step by step how to properly do this conversion? self critique assessment: 2
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15:51:37 `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 --> I would have needed to have gotten the correct answer to the previous question to answer this one correctly, but I am sure the approach is the same as it was in a previously presented situation. confidence assessment: 0
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15:51:51 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 --> This solution makes sense. self critique assessment: 2
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15:53:12 `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. The 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 --> I have no idea how to do this, or even where to start. Out of that entire long question, how do I begin to sort out which bits are important to get a good start? confidence assessment: 0
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15:53:20 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 --> self critique assessment: 0
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15:54:05 `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 --> I would have needed to get the correct answer to the previous question to answer this one. confidence assessment: 0
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15:54:45 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 --> This solution makes sense. Had I known how to do the previous question I would have had no trouble with this one. self critique assessment: 2
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15:55:37 `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 --> I would have needed to answer the first question correctly in this particular series of questions to get the correct answer here. confidence assessment: 0
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15:56:21 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 --> This solution makes sense. I am just not sure how to get started when attempting to answer this sort of question. self critique assessment: 2
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15:56:37 `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 --> a = F / m confidence assessment: 0
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15:57:13 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 --> I had the correct formula, and I could have answered the question correctly if I had had the correct values. self critique assessment: 2
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16:00:18 `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 --> F = ma I don't know the mass of the cart or the mass of the object pulling the cart up the long incline. But, the acceleration of the cart is related to gravity and the initial velocity mentioned in the problem. Also, frictional force will contribute here and should be subtracted from the total net force. I would say acceleration of the cart should increase over time if the frictional force isn't too great. confidence assessment: 0
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16:01:55 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 what I thought was the positive direction is actually the negative direction. How do you know which is which? Also, where did the +.9m/s^2 come from? Any clarification would be most appreciated. self critique assessment: 2
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