# Work-Energy Theorem and Friction

• Argonaut
In summary, the problem involves two blocks, one moving vertically and one moving horizontally, with given masses, initial velocity, and displacement. The target variable is the coefficient of kinetic friction. Using the work-energy theorem and the balance of forces, the given values can be used to calculate the coefficient of kinetic friction, which is found to be 0.786. However, the book gives a slightly different value of 0.786, leading to further investigation and adjustments in the calculation to arrive at the correct value.
Argonaut
Homework Statement
[Young & Freedman - University Physics 13th ed, exercise 6.87]
Consider the system shown in Fig. P6.86. The rope and pulley have negligible mass, and the pulley is frictionless. Initially the 6.00-kg block is moving downward and the 8.00-kg block is moving to the right, both with a speed of 0.900 m/s. The blocks come to rest after moving 2.00 m. Use the work–energy theorem to calculate the coefficient of kinetic friction between the 8.00-kg block and the tabletop.
Relevant Equations
Work-energy theorem
Kinetic friction formula

My final answer is different from the official one in the back of the book, and I can't figure out what I did wrong. This is my attempt:

Let block 1 be the vertically moving block and let block 2 be the horizontally moving one.

Also, let ##m_1 = 6.00 ~\rm{kg}##, ##m_2 = 8.00 ~\rm{kg}##, ##v_0 = 0.900 ~\rm{m/s}##, and ##d = 2.00 ~\rm{m}##. The target variable is ##\mu_k##.

According to the work-energy theorem, the total work done on block 2 is ##W_{tot}= \Delta K = K_f - K_i = 0 -K_i = -\frac{1}{2}m_2v_0^2##.

Alternatively, since the forces acting on block 2 are constant, we also have ##W_{tot} = \Sigma F_x d = (T - f_k)d = (m_1g - \mu_km_2g)d## (if the direction of displacement is the positive direction).

So then we have

$$-\frac{1}{2}m_2v_0^2 = (m_1g - \mu_km_2g)d$$

Rearranging to express ##\mu_k##, we obtain

$$\mu_k = \frac{m_1}{m_2} + \frac{v_0^2}{2dg} = \frac{6.00 ~\rm{kg}}{8.00 ~\rm{kg}} + \frac{(0.900 ~\rm{m/s})^2}{2(2.00 ~\rm{m})(9.80 ~\rm{m/s^2})} = 0.771$$

However, the book gives ##0.786##. Where did I mess up?

Argonaut said:
Alternatively, since the forces acting on block 2 are constant, we also have ##W_{tot} = \Sigma F_x d = (T - f_k)d = (m_1g - \mu_km_2g)d## (if the direction of displacement is the positive direction).

However, the book gives ##0.786##. Where did I mess up?
Consider the balance of forces on m1 and its resulting acceleration to find T.

But the easier way is to consider the whole of system energy.

nasu and Argonaut
haruspex said:
Consider the balance of forces on m1 and its resulting acceleration to find T.

But the easier way is to consider the whole of system energy.
Right, so since there is acceleration, Newton's First Law does not apply, and ##T \neq m_1g ##, in fact ## T > m_1g ##, is that correct?

In which case, I'm probably better off considering the whole system instead of block 2.

Argonaut said:
, in fact T>m1g, is that correct?
Does m1 rise?

haruspex said:
Does m1 rise?
It does not, but its descent is 'slowing', so there is negative acceleration (supposing the positive direction is the direction of displacement). I thought that would mean that the force acting in the negative direction is greater than the one acting in the positive direction.

Argonaut said:
It does not, but its descent is 'slowing', so there is negative acceleration (supposing the positive direction is the direction of displacement). I thought that would mean that the force acting in the negative direction is greater than the one acting in the positive direction.
Sorry… skipping between threads, I forgot that it is initially descending.

Argonaut
haruspex said:
Sorry… skipping between threads, I forgot that it is initially descending.
No worries, and thanks for the help. Looking forward to my lunch break to have another go at this exercise

The problem says
Argonaut said:
Initially the 6.00-kg block is moving downward and the 8.00-kg block is moving to the right, both with a speed of 0.900 m/s.
You say $$-\frac{1}{2}m_2v_0^2 = (m_1g - \mu_km_2g)d.$$
What should ##\Delta K## be?

Argonaut and erobz
kuruman said:
The problem says

You say $$-\frac{1}{2}m_2v_0^2 = (m_1g - \mu_km_2g)d.$$
What should ##\Delta K## be?
Thanks for the tip. I think I've got it now - at least I've managed to reproduce the correct value.

So on the one hand ##W_{tot} = \Delta K = K_f - K_i = 0 - \frac{1}{2}(m_1+m_2)v_0^2##.

On the other hand, treating the blocks, the rope and the pulley as a system, there are two forces doing work on the system. So we have ## W_{tot} = \Sigma F d = (w_1 - f_k)d = (m_1g - \mu_km_2g)d ##.

So then equating the two we obtain

$$(m_1g - \mu_km_2g)d = - \frac{1}{2}(m_1+m_2)v_0^2$$

$$\mu_k = \frac{m_1}{m_2} + \frac{(m_1+m_2)v_0^2}{2dm_2g} = \frac{6.00 ~\rm{kg}}{8.00 ~\rm{kg}} + \frac{(6.00 ~\rm{kg} + 8.00 ~\rm{kg})(0.900 ~\rm{m/s})^2}{2(2.00 ~\rm{m})(8.00 ~\rm{kg})(9.80 ~\rm{m/s^2})} = 0.786$$

That is correct.

Argonaut

## What is the work-energy theorem?

The work-energy theorem states that the work done on an object by the net force is equal to the change in its kinetic energy. Mathematically, it is expressed as W = ΔKE, where W is the work done by the net force and ΔKE is the change in kinetic energy.

## How does friction affect the work-energy theorem?

Friction is a force that opposes motion and does negative work on a moving object. When friction is present, the work done by friction must be included in the work-energy theorem. This means the net work done on the object is equal to the change in kinetic energy minus the work done by friction.

## Can the work done by friction be positive?

No, the work done by friction is typically negative because friction always acts opposite to the direction of motion, reducing the object's kinetic energy. This negative work results in a loss of mechanical energy, often converted to heat.

## How do you calculate the work done by friction?

The work done by friction can be calculated using the formula W_friction = -f * d, where f is the frictional force and d is the distance over which the force acts. The negative sign indicates that the work done by friction is opposite to the direction of motion.

## What role does the coefficient of friction play in the work-energy theorem?

The coefficient of friction (μ) determines the magnitude of the frictional force between two surfaces. It is used to calculate the frictional force as f = μ * N, where N is the normal force. This frictional force is then used to determine the work done by friction, which is essential in applying the work-energy theorem accurately in scenarios involving friction.

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