The following Activities are for the Static and Kinetic Friction applet. Make sure you know how the applet functions by consulting Help, Assumptions, and ShowMe under Applet Help on the applet's Help menu.

Activity 1. The purpose of Activity 1 is to use the CM-work theorem to predict the final speed of a block subject to an applied force and to friction.

Exercise 1. RESET the applet. Set the static coefficient of friction to 0.30 and the kinetic coefficient of friction to 0.20. Set the applied force to a magnitude of 20.0 N and keep its direction horizontal. Keep the initial velocity equal to 0 and the block's mass equal to 5.0 kg.

PLAY the block's motion until the block has undergone a displacement of 2.00 m. You may want to use the STEP button to complete the motion when the block's displacement gets near the 2.00-m value.

Use the CM-work theorem to predict the speed of the block for the displacement Δx = 2.00 m. Remember that there are two external forces acting on the block, the applied force and the friction force. Both are doing CM-work on the block, and the amounts of CM-work done by these two forces must be added to give the net CM-work done on the block. The net CM-work is required in the CM-work theorem,

W_CM,net = Δ [M/2 V²], (1)

where M is the mass of the block and V the speed of the block's center of mass (CM). Let's call V the speed of the block.

F_fric = μ_k N (2)

where N is the magnitude of the normal force. In the present case, N = Mg where g the magnitude of the acceleration due to gravity, which the applet takes as 9.8 m/s². Display the data box to compare your value for the friction.

In the present case, in which the initial kinetic energy is 0, the CM-work theorem directly gives you the block's (translational) kinetic energy M/2 V² corresponding to the 2.00-m displacement. Compare the value you obtain for the kinetic energy with that in the data box. The value in the data box may differ slightly from yours because of round-off error.

Exercise 2. REWIND the applet, and change the block's initial velocity to v_x = 2.0 m/s. Otherwise, use the same settings for the applet as in Exercise 1. Again, use the CM-work theorem to predict the block's kinetic energy when the block has undergone a displacement of 2.00 m and from that the block's speed at this point.

Equ.(1) gives the change in the block's kinetic energy. Since the block's initial kinetic energy is not 0 in this case, one more step than in Exercise 1 will be required in the calculation of the block's kinetic energy at Δx = 2.00 m. Again, PLAY the motion and compare your answers with those in the applet's data box.

Exercise 3. REWIND the applet, and repeat Exercise 2, but with the initial velocity set to v_x = -2.0 m/s. Note that the block first moves left, comes to a stop momentarily, and then moves right. When the block's displacement is Δx = 2.00 m, the block is at the same final point as in the previous Exercise. Is the block's speed at this point the same as in Exercise 2? If not, in a few sentences explain why not. (Hint: Compare the directions of the friction force in Exercises 2 and 3. Watch the free-body diagram as the motion progresses.)

Exercise 4. Repeat Exercises 1 to 3 with different settings for the kinetic coefficient of friction, initial velocity, applied force, and mass of the block. In particular, include a run without friction.

If you choose an applied force that forms a non-zero angle with the horizontal, pay attention to the fact that the magnitude N of the normal force needed in Equ.(2) is not equal to Mg. In general, N must be calculated as follows.

Apply Newton's second law in the y-direction (vertical direction) to the block. The gravitational force, normal force, and applied force have non-zero y-components in general. Since a_y = 0, Newton's seond law applied to the block implies that

N + F_appl,y - Mg = 0. (3)

N = Mg - F_appl sin θ(4)

where θ is the angle between the applied force and the horizontal.

Activity 2. The purpose of Activity 2 is to investigate the difference between CM-work and work in relation to energy conservation.

Exercise 1. RESET the applet. Set the static coefficient of friction to 0.30 and the kinetic coefficient of friction to 0.20. Set the initial velocity equal to 2.0 m/s. Keep the block's mass equal to 5.0 kg.

Adjust the magnitude of the applied force so that the block's acceleration is zero. This should be at F_appl = 9.8 N if the applied force is horizontal. Make sure the force is horizontal.

For these settings, calculate both the work W_appl and the CM-work W_CM,appl done by the applied force on the block if the block's displacement is Δx = 2.00 m. (W_CM,appl = W_appl because the CM has the same displacement as the point at which the applied force is acting.)

PLAY/STEP the motion until the block's displacement is Δx = 2.00 m, and compare your value of W_appl with that displayed in the data box. The (equal) value of the CM-work done by the applied force is not listed.

Exercise 2. What happens to the energy transferred when the work W_appl is done on the block under the conditions considered in Example 1? The block is not gaining any kinetic energy because the block's velocity is constant.

To answer this question, let us first consider the system consisting of both the block and the table. This system has two kinds of energy: translational kinetic energy of the block, KE, and internal energy of both the block and the table, U. Another word for the latter is thermal energy. The table remains at rest and has no translational kinetic energy. The work done by the applied force on this system is work done by an external force and therefore amounts to a transfer of energy to the system. Under the given conditions, the kinetic energy does not increase. How about the internal energy? Let's use the applet to find out.

REWIND the applet, and make no changes in the applet's settings. Display the box with the energy bars. Observe that initially the applied work W_appl and the change ΔU in the internal energy U of the block-table system are zero.

PLAY/STEP the motion until the block has the displacement Δx = 2.00 m, and observe how the bars representing W_appl and ΔU change during the motion. They should be rising steadily so that W_appl = ΔU at all times. Also observe that the block's kinetic energy KE remains equal to the initial kinetic energy KE(0).

REWIND the applet, and PLAY the motion again. This time display the data box and observe that W_appl = ΔU at all times while KE = KE(0) = 10.0 J. Record the values of W_appl, KE(0), KE, and ΔU for Δx = 2.00 m.

In a few sentences, summarize your observations and how they support the law of conservation of energy.

Example 3. The observations in Example 2 show that, as long as the velocity of the block remains constant, all of the energy transferred to the block-table system via W_appl is converted to internal energy of the system. Both the block and the table get warmer. None of this work is converted to translational kinetic energy. What if the block is accelerating and its velocity not constant?

REWIND the applet, and change the magnitude of the applied force to 20.0 N. Keep the direction of the force horizontal, and do not change any of the other settings. Again, PLAY/STEP the motion until the block's displacement is Δx = 2.00 m. Display the energy box, and observe the bars representing W_appl, KE, and ΔU.

REWIND the applet, and change none of its settings. PLAY/STEP the motion again, but this time with the data box displayed. Observe the values of W_appl, KE, and ΔU during the motion. Record the values of W_appl, KE(0), KE, and ΔU for Δx = 2.00 m.

Describe your observations in a few sentences, and explain how they support the law of energy conservation.

Exercise 4. Exercises 2 and 3 deal with the change ΔU in internal energy of the block-table system. The law of energy conservation applied to this system demands that

W_appl = ΔKE + ΔU.(5)

One can use this equation to calculate ΔU since W_appl and ΔKE can be calculated. Calculate W_appl and ΔKE for the settings in Examples 2 and 3. Then calculate the corresponding ΔU. Compare your values for ΔU with those you recorded when doing Examples 2 and 3.

Hint: ΔKE can be calculated using the CM-work theorem, i.e., Equ.(1). This is done in Activity 1, Exercise 2 for the settings in Example 3. W_appl should be calculated from the definition of work.

Exercise 5. What about the change in the internal energy of just the block or just the table? Can that be calculated similarly to the way the internal energy of the block-table system is calculated in Exercise 4?

To answer this question one needs to treat the block (or the table) as the system and needs to know how much work is done on this system by the external forces acting on it. Let us consider the block first. There are two external forces acting on it: the applied force and the friction force. (For the block-table system, friction is an internal force, but for the system consisting of just the table it is an external force.)

To take the simplest possible case, let F_appl = F_fric = 9.8 N so that the block's velocity is constant. (All other settings are assumed to be as in Examples 1-4.)

First calculate the CM-work W_CM,fric done on the block by friction in this process. Its value should be equal to -W_appl because the friction force is directed opposite to the applied force while its magnitude is equal to that of the applied force and because the CM's displacement is equal to that of the point at which the applied force is acting.

REWIND the applet, set F_appl = 9.8 N, but don't change any of the other settings. PLAY/STEP the motion until Δx = 2.00 m. Check your value for W_CM,fric against that for W_appl displayed in the data box.

Next calculate the work W_fric done on the block by friction. Do you expect it to be equal to W_CM,fric, just as W_appl is equal to W_CM,appl?

It would be tempting to answer YES to this question, but that is not the right answer. The truth is that W_fric can not be equal to F_fric,xΔx, which is equal to W_CM,fric. Δx, which is the displacement of the CM of the block is not the right displacement to take when calculating W_fric.

To see why not, suppose that W_fric = F_fric,xΔx = W_CM,fric. Then, since W_CM,fric = -W_appl, it would follow that W_fric = -W_appl. Therefore, the net work done by the external forces acting on the block, W_fric + W_appl, would be zero. This means no energy would be transferred to the block. Since the block's kinetic energy is constant in the given process, the block's internal energy would have to be constant as well, by the law of energy conservation. The block would not get warmer. This is contradicted by the observations and is also inconsistent with the the result obtained in Example 4. (Note that a similar argument would show that the internal energy of the table is zero as well. Carry out this argument, calculating the work done by friction on the table using the displacement of the table's CM which is zero.

The fact of life is that it is impossible to calculate the work done by friction on either the block or the table. The reason is that we cannot know the displacements of the points at which the friction force is acting. The friction force is acting at the many tiny surface irregularities at the contacting surfaces, and these irregularities bend elastically to a certain degree before snapping back into position while the two surfaces are moving relative to each other. The displacements of these irregularities is what one would need to know to be able to calculate W_fric. Note that the applet does not list values for W_fric in the data box.