OpenQuery26

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course Phy 241

415pm 11/29/2011

If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

026. `query 26

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Question: `qgen phy and principles of phy 4.36: If the coefficient of kinetic friction is .30, how much force is required to push a 35 kg crate across the floor at

constant speed?

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Your solution:

confidence rating #$&*:

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Given Solution:

`aIf the crate is moving then the force exerted by friction is .30 times the normal force between it and the floor. If the push is horizontal, then the only

horizontal forces acting on the crate are the downward force of gravity and the upward force exerted by the floor. Since the crate is not accelerating in the

vertical direction, these forces are equal and opposite so the normal force is equal to the weight of the crate.

The weight of the crate is 35 kg * 9.8 m/s^2 = 340 N, approx.

The frictional force is therefore

f = .30 * 340 N = 100 N, approx..

If the crate moves at constant speed, then its acceleration is zero, so the net force acting on it is zero.

The floor exerts its normal force upward, which counters the gravitational force (i.e., the weight).

The frictional force acts in the direction opposite motion; if net force is zero an equal and opposite force is required, so you must push the box with a force of 100 N in the direction of motion.

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qgen phy 4.55 18 kg box down 37 deg incline from rest, accel .27 m/s^2. what is the friction force and the coefficient of friction?

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Your solution:

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Given Solution:

The coordinate system is rotated from its traditional vertical-horizontal orientation in such a way that the positive x axis points down the incline, below the horizontal. As the system is rotated, the negative y axis 'swings away' from a vector which remains in the vertical downward direction, and that vector ends up in the fourth quadrant. The result will be something like the figure below, though the rotation in this figure is only 20 degrees and won't completely agree with the rotation in this problem:

The weight vector being vertical in the downward direction is therefore in the fourth quadrant, at an angle of 37 degrees with respect to the negative y axis.

Thus the weight vector makes angle 270 deg + 37 deg = 307 deg with the positive x axis and its x and y components are

wtx = 18 kg * 9.8 m/s^2 * cos(307 deg) = 106 N and

wty = 18 kg * 9.8 m/s^2 * sin(307 deg) = -141 N.

The figure below depicts a fourth-quadrant vector and its components. As before the angle of rotation in this figure is only 20 degrees, as opposed to the 37 degrees of this problem, so the x in the problem component is relatively smaller and the y component relatively larger than depicted below:

The indented discussion of the right-angle trigonometry of the situation is in a font different from the rest of the solution, and may be skipped without loss of continuity.

You get the same results using the sin and cos of the 37 deg angle. The following assumes some familiarity with right-triangle trigonometry, which is assumed for General College Physics students and should be very familiar to University Physics students.

In the figure below we have constructed a right triangle whose hypotenuse coincides with the weight vector, with one of the legs along the negative y axis.

If theta is the angle at the vertex located at the origin, then theta is the angle through which the axes are rotated. Theta therefore coincides with the angle of the original incline.

The x and y components of the vector are congruent with the legs of the triangle, the x component with the leg opposite the angle theta and the y component with the leg adjacent to theta.

Thus the 'opposite' leg is hypotenuse * sin(theta), and the 'adjacent' leg is hypotenuse * cos(theta).

The vector in this example represents the weight of the object, so the hypotenuse represents the weight.

The x component is positive and the y component is negative. We conclude that

x component = + weight * sin(theta) and

y component = - weight * cos(theta),

where again theta is the angle of the incline.

For the mass of the object in this problem, the weight is 18 kg * 9.8 m/s^2, and we have

x component = 18 kg * 9.8 m/s^2 * sin(theta) and

y component = 18 kg * 9.8 m/s^2 * cos(theta)

Common point of confusion:

The x component in one model is given by the cosine, and in the other by the sine. The y components are also given by two different functions.

The key idea is that the circular-model definitions lead us to an x-y coordinate system in which the weight vector is at angle 307 degrees relative to the positive x axis, while the right-triangle analysis can be done in terms of a triangle containing the 37-degree incline. The sine of one angle is the cosine of the other, and vice versa, except for + and - signs.

The + and - signs arise naturally from the circular model, which is the advantage of this model.

The triangular model also has its advantages.

Detailed resolution of point of confusion:

The angles used in the two models are different.

The angle used in the 'circular' model is 307 deg, the angle of the weight vector with the positive x axis of our coordinate system.

The angle used in the 'triangle' model is 37 deg, the angle of the incline with horizontal.

The sines and cosines of these angles are related:

cos(307 deg) = sin(37 deg)

sin(307 deg) = - cos(37 deg)

The only difference is that the circular model, if set up correctly, automatically gives you the correct signs of the components (i.e., x component is positive, y component negative). If you use the triangle you have to think a little bit more, and put the correct signs on the angles. In this respect the circular model is simpler.

However it's very important in advanced applications to 'see' the triangles involved, so the triangle model is also essential for General College Physics and especially for University Physics students.

Principals of Physics students who are not familiar with trigonometry generally do better with the circular model.

The given solution continues below

Having found the components of the weight we find the normal force:

The only other force in the y direction, other than the y component of the weight, is the normal force (which is the elastic or compressive force exerted by the incline in response to the y component of the weight).

Since the mass does not accelerate in the y direction, we know that the net force in the y direction is zero. It follows that

normal force + y component of weight = 0, so that

normal force = - y component of weight = - ( -141 N ) = 141 N.

The frictional force is

f_Frict = coefficient of friction * normal force, or in symbols

f_Frict = mu * F_normal.

The acceleration of the system is .27 m/s^2 down the incline.

The component of the gravitational force down the incline is 106 N.

The net force along the incline is therefore 106 N - mu * F_normal = 106 N - mu * 141 N.

The net force is also F_net = m a = 18 kg * .27 m/s^2 = 5 N, approx..

Thus we have the equation

F_net = wt_x - mu * F_normal. Solving for mu:

mu = -(F_net - wt_x) / (F_normal) = - (5 N - 106 N) / (141 N) = 101 N / (141 N) = .7, approx.

It's best to write and solve the equation for mu before substituting, but it doesn't hurt to represent the relationship using the actual quantities

F_net = wt_x - mu * F_normal. Substituting the quantities we have already calculated this becomes

5 N = 106 N - mu * 141 N so that

mu = (5 N - 106 N) / (-141 N) = .7.

STUDENT QUESTION

What does mu stand for?

INSTRUCTOR RESPONSE

mu is the coefficient of friction. Generally the lowercase Greek letter mu is the symbol used to represent the coefficient of friction.

The symbol looks like this: m

The meaning is implicit in the lines below, quoted from the given solution:

The frictional force is

f_Frict = coefficient of friction * normal force, or in symbols

f_Frict = mu * F_normal.

`aGOOD STUDENT SOLUTION USING RIGHT-ANGLE TRIGONOMETRY: (I don't know why, but I was hoping you would pick an odd numbered problem here)Here goes.....For an 18kg box on an incline of 37 degrees with an acceleration of .270 m/s/s, I first drew out a diagram showing the forces involved. Next the forces had to be derived.

First, to find the force associated with the weight component parrallel to the inline moving the box downward....Fp=sin 37 deg(18kg)(9.8m/s/s)=106N.

Next, the Normal force that is counter acting the mg of the box is found by..

Fn=cos 37 deg. (18kg)(9.8 m/s/s) = 141N.

The frictional force can be found by using F=(mass)(acceleration) where (Net Force)-(frictional coeffecient*Normal Force)=(m)(a) so that...

106N - (141N * Friction Coeff.) = (18kg)(.270 m/s/s) where by rearranging, the frictional coeffecient is seen to be .717.

INSTRUCTOR COMMENT:

Note that you should specify the direction of your positive x axis. The simplest choice will be an x axis oriented down the incline, so that the acceleration will be positive.

You either used triangle trigonometry or quoted formulas from the text (hopefully the former, so you will understand the picture that goes with this problem). Either way you got the correct result and did a good job.

My solutions generally formulate vectors in the context of the circular definition of trigonometric functions. The triangle-based solution is probably more standard, but requires more understanding of right-angle trigonometry, and is more prone to confusion regarding signs.

STUDENT COMMENTS

I don’t know if I know and was trying the circular method or the triangular method. I read through the Given Solution and I am still confused. I don’t understand in the student’s solution why the net force was 106.2 N. That was determined by the sin product, and isn’t that the y-component??

And I thought I remember reading that friction works on the y-component. And I have no idea why your diagram puts this incline in the 4th quadrant.

By picturing a 37 degree ramp with it ascending up towards the right, it’s clear that the incline should be in the 1st quadrant. Please don’t tell me that there are times when we switch the sin and cos when solving a problem....

INSTRUCTOR RESPONSE

The student's solution used the angle of the incline rather than the angle with the positive x axis.

The given solution then put the problem in the context of the circular definition.

An incline at 37 deg from horizontal could be depicted rising from left to right, or as descending from left to right. It has to be one or the other, and either choice is equally valid. The choice illustrated in my solution is the latter.

When using the circular definition, you set your coordinate system, and the sines and cosines follow the definitions. They don't switch.

The frictional force is proportional to the normal force, which is often (as it is in the given solution) in the y direction.

The normal force is in the direction perpendicular to the surface, whatever that direction is (again, in this problem the y axis is perpendicular to the incline, so the normal force is in the y direction).

When using the trigonometric definitions, the choice of sine or cosine for a given quantity is also clear in terms of those definitions.

STUDENT SOLUTION (COMMON ERROR: CONFUSING ANGLE OF INCLINE WITH ANGLE OF WEIGHT VECTOR)

Weight = 18kg * 9.8m/s^2 = 176.4N

Parallel = 176.4 cos 37 = 141N

Perpendicular = 176.4 sin 37 = 106N = normal force

INSTRUCTOR RESPONSE

The magnitude of the parallel component is of the weight clearly less than that of the perpendicular component for an angle less than 45 degrees.

The weight vector does not make a 37 degree angle with the positive x axis. 37 degrees is the angle of elevation of the incline.

If you use cos and sin, the angle has to be between the positive x axis and the weight vector. In this case that angle would be either 233 deg or 307 deg, depending on whether the incline is drawn sloping upward or downward as you move from left to right.

In the 233 deg case you would get parallel component = 176.4 cos(233 deg) = -106 N, perpendicular component 176.4 sin(233 deg) = -141 N.

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Self-critique (if necessary):

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Self-critique Rating:

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Question: `qUniv. 5.90 (5.86 10th edition).

4 kg and 8 kg blocks, 30 deg plane, coeff .25 and .35 resp. Connected by string. Accel of each, tension in string.

What if reversed?

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Your solution:

The question is number 5.92 in the 12th edition

m1 = 4 kg

m2 = 8kg

u_k1 = 0.25

u_k2 = 0.35

g = 9.81 m/s^2

theta = 30°

N_1 = 4kg*9.81 m/s^2 * cos(30) = 33.98N

N_2 = 8kg*9.81m/s^2 * cos(30) = 67.97N

F_frict1 = N_1 * 0.25 = 8.495N

F_frict2 = N_2 * 0.35 = 23.79N

F_1x = m1*g*sin(30) = 19.62N

F_2x = m1*g*sin(30) = 39.24N

Draw two seperate free-body diagrams

Set masses at origin of x-y axis

FBD m1

Sum of all the masses in the x-direction

0 = -19.62N + 8.495 + T1

T1 = 11.25N

FBD m2

Sum of all the forces in x-direction

0 = -39.24N - T2 + 23.79N

T2 = 15.45N

The tensions are not equal, therefore they are moving at different speeds. Mass2 has a force in the negative x direction equal to the T2 and Mass1 has a force in the negative x-direction equal to T1

So the forces of the two blocks down the incline are the negative tension in the string. There is no tension though, because the 2nd block is moving down the incline at a faster speed.

a1 = T1/m

= 11.25N/4kg

= 2.8125 m/s^2

a2 = T2/m

= 15.45N/8kg

= 1.93 m/s^2

There would be tension in the string if the blocks were switched and m2 was closer to the bottom of the incline. If that were the case, we would need to draw new FBD find the sum of all the forces

in the x-direction for each block. The tension in the string for the mass1 replaces T1, becuase the larger mass accelerates the smaller mass.

m1

0 = -T_string - 19.62N + 8.495N

T_string = -22.375

The negative is simply direction. The mass1 is being pulled in the negative x direction by the bigger mass2

m2

0 = -F - 39.24N + 22.75 + 23.79N

F = 7.3N

This force is the new force of m2, which is pulling m1.

The acceleration of m2 is a = 7.3N/12kg = 0.61 m/s^2

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Given Solution:

`a** We will use the direction down the incline as the positive direction in all the following:

The normal forces on the two blocks are 4 kg * 9.8 m/s^2 * cos(30 deg) = 34 N, approx., and 8 kg * 9.8 m/s^2 * cos(30 deg) = 68 N, approx. If sliding the 4 kg block

will therefore experience frictional resistance .25 * 34 N = 8.5 N, approx. and the 8 kg block a frictional resistance .35 * 68 N = 24 N, approx.

The gravitational components down the incline are 4 kg * 9.8 m/s^2 * sin(30 deg) = 19.6 N and 8 kg * 9.8 m/s^2 * sin(30 deg) = 39.2 N.

If the blocks were separate the 4 kg block would experience net force 19.6 N - 8.5 N = 11.1 N down the incline, and the 8 kg block a

net force of 39.2 N - 24 N = 15.2 N down the incline. The accelerations would be 11.1 N / (4 kg) = 2.8 m/s^2, approx., and 15.2 N / (8 kg) = 1.9 m/s^2, approx.

If the 4 kg block is higher on the incline than the 8 kg block then the 4 kg block will tend to accelerate faster than the 8 kg block and the string will be unable

to resist this tendency, so the blocks will have the indicated accelerations (at least until they collide).

If the 4 kg block is lower on the incline than the 8 kg block it will tend to accelerate away from the block but the string will restrain it, and the two blocks

will move as a system with total mass 12 kg and net force 15.2 N + 11.1 N = 26.3 N down the incline. The acceleration of the system will therefore

be 26.3 N / (12 kg) = 2.2 m/s^2, approx..

In this case the net force on the 8 kg block will be 8 kg * 2.2 m/s^2 = 17.6 N, approx.. This net force is the sum of the tension T, the gravitational

component m g sin(theta) down the incline and the frictional resistance mu * N:

Fnet = T + m g sin(theta) - mu * N

so that

T = Fnet - m g sin(theta) + mu * N = 17.6 N - 39.2 N + 24 N = 2.4 N approx.,

or about 2.4 N directed down the incline.

The relationship for the 4 kg mass, noting that for this mass T 'pulls' back up the incline, is

Fnet = m g sin(theta) - T - mu * N so that

T = -Fnet + m g sin(theta) - mu * N = -8.8 N + 19.6 N - 8.5 N = -2.3 N. equal within the accuracy of the mental approximations used here to the result obtained by considering the 8 kg block and confirming that calculation. **

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Self-critique (if necessary):

I don't understnad how the string will restrain the 4 kg block when it is lower on the incline. The 8kg block will quickly catch up to the 4 kg block and the two blocks will meet and create their own force, down the incline.

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Self-critique Rating: 2

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The 8 kg block cannot catch up with the 4-kg block. If the string wasn't there the 4 kg block would accelerate at a greater rate than the 8 kg block; if the string was slack it would not exert any forces and the blocks would accelerate as if it wasn't there. However that would mean that the separation between the blocks would increase, taking the slack out of the string. So the string would never be able to go slack.

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Self-critique (if necessary):

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Self-critique rating:

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Self-critique (if necessary):

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Self-critique rating:

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&#Good work. See my notes and let me know if you have questions. &#

course Phy 121

November 29, 2011 at 7:21 p.m.

This short version is currently a working draft. Please ask for clarification when that is necessary. Modifications will be made to this document in response to questions, which will also be answered in the usual manner.Note that this is an alternative short version of the experiment intended for

• Principles of Physics students

• General College Physics students whose lab goal in only a passing lab average

This is not intended for University Physics students.

See also the short video at

http://www.vhcc.edu/dsmith/genInfo/qa_query_etc/EnergyConIncline_DialUp200.wmv

which demonstrates a finger delivering a quick impulse to the ball, which coasts to a stop as it travels up the ramp, then coasts with increasing speed down the ramp. The finger 'pokes' the ball at the very beginning of the video.

Note: You may assume for the purposes of this writeup that the ball has a mass of 40 grams. This probably isn't accurate, but it will serve the purpose of the experiment.

Report a preliminary run of the experiment

I suggest that before actually running the experiment you read through the instructions, set up the system, get a few preliminary timings and submit them with a brief description of what you did. I'll be able to tell you if your results make sense, and might make a suggestion or two. No need to do any calculations, and no need for a detailed description. 15 minutes should do it for the preliminary observations. I'll also be glad to clarify anything you think requires clarification.

Goal of the experiment

The experiment concerns a ball which coasts up a ramp, stops, and coasts back down. We are trying to detect the difference between the magnitude of the acceleration going up and the acceleration going down. It is suggested that to get an initial 'feel' for the system you take the

ball, the ramp and a domino, set the ramp up with a fairly small slope, and use your finger to 'bump' the ball in this manner. Again, the ball

just has to go up and come back. This will give you a point of reference for further instructions.

Basic instruction

The basic instruction is this:

There is a difference between the time required for a ball, given an initial velocity at the bottom of the ramp, to roll up a ramp and the

time required for it to roll back down to its initial position.

• This is easiest to detect on a ramp whose slope is just enough that the ball will 'turn around' and roll back down.

• It is necessary to start the ball with an initial impulse in the form of a 'bump' from your finger or from an object, as opposed to a prolonged push. This is because the presumably uniform acceleration up the ramp does not begin until the ball loses contact with the source of the impulse.

• You can set the system up with too much slope, which will make the difference in time up and time down undetectable, or with too little slope, which will not result in uniform acceleration. It is up to you to determine the optimal slope, but if your time intervals are less than a couple of seconds you won't be able to time them with sufficient accuracy.

• You also need to 'bump' the ball hard enough, but not too hard. You won't always get it right; simply disregard the trials that don't result in sufficient time intervals.

• If the 'bump' isn't strong enough the ball won't go very far and you'll be timing a short interval.

• If the 'bump' is too strong the ball will keep going right off the high end of the ramp.

More specific instructions

More specifically:

Too little slope causes problems:

• If the ramp has too little slope it will be difficult to give the ball an initial impulse that causes it to travel most of the length of the

ramp without rolling off the 'high' end.

• If you don't have quite enough ramp slope, once the ball comes to rest it might simply stay at rest.

Also, if there is too little slope, the small irregularities in the track interfere with the uniformity of the acceleration and throw off the results.

Too much slope causes problems:

• There are unavoidable uncertainties in timing. If the slope is too great the time intervals will be short, and the resulting percent uncertainty will be too high to make an accurate distinction between the 'up' and 'down' intervals.

• A ramp with a rise of a single domino is probably steeper than necessary. It is suggested that you use coins or shims. You won't need an accurate measurement of the slope.

Suggestions for experimental technique:

• You want to start the ball rolling up the ramp, using a sudden impulse rather than a sustained push, giving it enough velocity that it travels

20 cm or more before coming to rest for an instant and then traveling back down.

• At the instant of the 'bump' you need to start the TIMER,

and you need to operate the TIMER in such a way as to determine as accurately as possible the time up the ramp and the time back down.

• You also need to observe, with reasonable accuracy, the point at which the ball comes to rest before rolling back down.

• In the event of a mis-strike (e.g., too hard or too easy) it's simple enough to try again. After a couple of minutes' practice it's not difficult to do this at least well enough to get a good trial every minute or so.

• To get good results it's important to avoid backspin and overspin; if the ball is struck just a little higher than the middle, the ball will start out with about the right amount of spin.

You have a good trial when you have data that allows you to determine the acceleration of the ball up the ramp, and back down.

• You need at least half a dozen good trials.

You therefore need to get a reasonable number of trials, timing the ball from 'strike' to 'turnaround' then back to the original position. It's

important to try to detect and eliminate or correct for systematic errors in timing.

The goal is to try to detect the difference in acceleration between the ball as it travels up the incline and as it travels down. It is assumed

that this difference is independent of how far the ball travels, and also independent of the slope, as long as the slope is small.

It's up to you to find a slope that yields good results. As outlined above, too much slope is counterproductive, as is too little.

Analysis of data and interpretation

First give a synopsis of your setup and all relevant data.

I am using the 1 ft. piece of ramp on a flat corner. The left end is elevated by the threaded rod. I am using the marble to travel up and down the ramp. I have marked the ramp off in Cm. I have a domino at the high end to keep it from rolling off and on at the bottom held flat to strike the ball.

Then, for each trial, determine the acceleration of the ball as it travels up the ramp, and as it travels down the ramp. Show, using a couple of representative sample calculations, how your results were obtained from your data.

T1 .32 m/s^2, .14 m/s^2

T2 .20 m/s^2, .23 m/s^2

T3 .13 m/s^2, .16 m/s^2

T4 .18 m/s^2, .13 m/s^2

T5 .28 m/s^2, .14 m/s^2

Obtained by getting vAve ~ds / `dt, multiplying by 2 to get vF since it started from rest, divided by `dt for acceleration.

The acceleration of the ball results mainly from two forces, one being the component of the gravitational force parallel to the incline, the other the force of rolling friction between the ball and the ramp. The difference in the accelerations is due to two facts:

• The gravitational component is in the direction opposite the ball's velocity during one phase of the motion, and in the same direction during the other phase.

• The frictional force is in the direction opposite the ball's velocity.

In the absence of friction the only force would be that of the gravitational component parallel to the incline, which is the same for motion up the incline as for motion down the incline.

• How then does rolling friction result in a difference in the two accelerations?

The friction has a force countering the force resulting in motion against gravity. This frictional force decreases the force against gravity until it decelerates the ball to a stop. At which point gravity’s force kicks in again causing the ball to travel back down the ramp. Friction does not stop the ball in this case because the ball is traveling with the grictional force.

According to your data, what is the magnitude of the acceleration due to the frictional force on the rolling ball, and what is this acceleration as a percent of the acceleration of gravity?

• More appropriately, you might choose to give upper and lower bounds for the magnitude of the acceleration due to friction (e.g., the magnitude of the acceleration is at least ___, and at most ___).

Energy conservation states that `dW_NC_ON = `dKE + `dPE. How do your results illustrate this law?

More detailed questions about interpretation

If you have answered the above questions with some degree of confidence you don't need to answer the following at this point. Depending on the answers you submit I might ask you to look at these questions.

You don't need to do any more calculations, except perhaps a couple of additions or subtractions, but answer the following. Don't spend hours thinking through your answers--just think about what is going on with this system and give the best answers you can in, say, 30 minutes. Many of the answers are pretty obvious. When I get your answers I'll be able, if necessary, to help clarify some of the more difficult points.

What forces act on the ball as it rolls freely up or down the ramp? List the forces.

Which of these listed forces are identical both for motion up the ramp and for motion down?

Which of the listed forces change as the ball reverses direction?

How does your answer to these questions help explain why the magnitudes of the two accelerations should be different?

What do you think the acceleration of the system would be in the absence of frictional forces?

Don't actually calculate any of the quantities in the subsequent questions related to work and energy:

What happens to the gravitational PE of the ball as it goes up the incline, and what happens as it goes down the incline?

What happens to the KE of the ball as it goes up the incline, and as it goes down?

How does the PE change up the ramp compare to the PE change down the ramp?

As it rolls up the ramp, how does the PE change of the ball compare with the KE change?

As it rolls down the ramp, how does the PE change of the ball compare with the KE change?

In which case is the magnitude of the ratio of PE change to KE change greater, and why?

Do any nonconservative forces act along the line of the ball's motion as it rolls up the ramp, and as it rolls down the ramp?

How does the action of the nonconservative forces explain the answers to some of these questions?

Between the end of the 'bump' and the ball's return to the same position

• Does its KE increase, decrease or remain the same?

• Does it PE increase, decrease or remain the same?

• Do nonconservative forces do positive work on the ball, negative work or no work?

In what ways do the results of this experiment and support the conservation-of-energy equation `dW_NC_ON = `dKE + `dPE?

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&#Your work on this lab exercise is good. Let me know if you have questions. &#