91Ó°ÊÓ

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, $\mathrm{μ}$

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

It is expressed as follows,

$$\begin{gathered} \mathbf{F}\mathbf{=}\mathbf{μ\pm ·} \\ \mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\left(}\mathbf{1}\mathbf{\right)} \end{gathered}$$

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

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

$$\begin{gathered} F=\mathrm{μ}N \\ =\mathrm{μ}Mg \end{gathered}$$

Here, N is the normal force on the box,$\mathrm{μ}$ 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,

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

…(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.

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

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

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

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, $\frac{F}{M}=\mathrm{μ}g$and u=0,

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

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

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$d$ .

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.