Question

It is sometimes claimed that friction forces always slow an object down, but this is not true. If you place a box of mass Mon a moving horizontal conveyor belt, the friction force of the belt acting on the bottom of the box speeds up the box. At first there is some slipping, until the speed of the box catches up to the speed vof the belt. The coefficient of friction between box and belt is$\mu$. (a) What is the distance d(relative to the floor) that the box moves before reaching the final speed v? Use energy arguments, and explain your reasoning carefully. (b) How much time does it take for the box to reach its final speed? (c) The belt and box of course get hot. Is the effective distance through which the friction force acts on the box greater than or less than d? Give as quantitative an argument as possible. You can assume that the process is quick enough that you can neglect transfer of energyQ due to a temperature difference between the belt and the box. Do not attempt to use the results of the friction analysis in this chapter; rather, apply the methods of that analysis to this different situation. (d) Explain the result of part (c) qualitatively from a microscopic point of view, including physics diagrams.E

Short answer

a) $\frac{v^2}{2\mu g}$

b) $\frac{v}{\mu g}$

c) The effective distance is bigger than the distance d.

d) Due to the frictional force, the effective distance is greater than the distance d.

Step-by-step solution

Identification of the given data

The given data is listed below as-

• The mass of the box is, M
• The speed of the belt is, V
• The coefficient of friction between the belt and the box is, $\mu$

Significance of the frictional force and kinetic energy

The frictional force is described as the product of the coefficient of friction and normal force.

It is expressed as follows,

$$\begin{gathered} \mathbf{F}\mathbf{=}\mathbf{\mu N} \\ \mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \end{gathered}$$

The Kinetic energy is described as half of the product of the mass and the square of the velocity.

Determination of the distance moved by the box before it reaches to its maximum speed (a)

The expression for the frictional force exerted on the box is expressed as,

$$\begin{gathered} F=\mu N \\ =\mu Mg \end{gathered}$$

Here, N is the normal force on the box,$\mu$ is the coefficient of the frictional force, is the mass of the box and g is the acceleration due to gravity.

From the energy principle, the equation of the translational kinetic energy is expressed as,

$$\begin{gathered} ∆K_{trans}=Fd \\ ∆K_{trans}=\mu Mgd \\ d=\frac{∆K_{trans}}{\mu Mg} \end{gathered}$$

…(1)

Here, is the distance moved by the box.

Another expression for the translational kinetic energy is expressed as,

$$\begin{gathered} ∆K_{trans}=\frac{1}{2}M\left(v^{2-u^2}\right) \\ =\frac{1}{2}M{v}^2 \end{gathered}$$

Here, is the initial speed of the box which is zero as the box is not moving initially, and is the final speed of the box.

$\mathrm{For}∆K_{trans}=\frac{1}{2}M{v}^2\mathrm{in}\mathrm{equation}\left(1\right)$

$$\begin{gathered} d=\frac{{\displaystyle \frac{1}{2}}M{v}^2}{\mu Mg} \\ =\frac{v^2}{2\mu g} \end{gathered}$$

Thus, the distance that the box moves before reaching the final speed is $\frac{v^2}{2\mu g}$.

Determination of the time taken by the box to reach the final speed (b)

The expression for the force is as follows,

$$\begin{gathered} F=Ma \\ a=\frac{F}{M} \end{gathered}$$

Here, is the acceleration of the box.

From the first equation of motion, final velocity of the box is written as,

$$\begin{gathered} v=u+at \\ =u+\frac{F}{M}t \end{gathered}$$

For, $\frac{F}{M}=\mu g$and u=0,

$$\begin{gathered} v=\mu gt \\ t=\frac{v}{\mu g} \end{gathered}$$

Thus, the time the box takes to reach its final speed is $\frac{v}{\mu g}$.

Determination of the effective distance of the box (c)

The friction is in the opposite direction of the direction of the actual force.

Hence, as the friction force acts on the object, then the equation for the change in the energy can be written as,

$$\begin{gathered} ∆K_{trans}+∆E_{internal}=W_{fric} \\ ∆K_{trans}+∆E_{internal}=Fd \\ ∆K_{trans}+W_{fric}-∆K_{trans} \\ =Fd-=∆K_{trans} \end{gathered}$$

Here,$∆E_{internal}$ is the internal energy of the system and$W_{fric}$ is the work done by the frictional force that is the product of the force and the distance

Here, there is no other force acting on the box, but as the temperature increases, the object is accelerating at a constant speed.

Thus, the effective distance is bigger than the distance$d$ .

Explanation of the results of part c (d)

The free body diagram of the box has been drawn below,

Here, in the diagram, the box is producing a force on the conveyor and the conveyor is also exerting an equal and opposite force on the box. The translational and the internal energy are combined to produce the work done due to friction, which is one of the main reasons that the effective distance is bigger than the distance d.

Thus, due to the frictional force, the effective distance is greater than the distance d.