Boxes are transported from one location to another in a warehouse by means of a conveyor belt that moves with a constant speed of 0.50 m/s. At a certain location the conveyor belt moves for 2.0 m up an incline that makes an angle of $\mathbf{10}\mathbf{°}$with the horizontal, then for 2.0 m horizontally, and finally for 2.0 m down an incline that makes an angle of $\mathbf{10}\mathbf{°}$ith the horizontal. Assume that a 2.0 kg box rides on the belt without slipping. At what rate is the force of the conveyor belt doing work on the box as the box moves (a) up the $\mathbf{10}\mathbf{°}$incline, (b) horizontally, and (c) down the $\mathbf{10}\mathbf{°}$incline?

It is given that,

Mass of the crate is m= 2 kg

Velocity of the box is v =0.25 m/s

The angle between incline and horizontal is$\theta =10°$

The problem deals with the instantaneous velocity. It is the limit of an averaged velocity as the elapsed time approaches to zero. To solve this problem,the dot product of force and instantaneous velocity of the object can be used.

Formula:

P =F.v

=F v cos$\varphi$

where,F is force, vis velocity and Pis the power.

Now,

$P_v=F_{g}v\cos \varphi$

The force on the incline is equal to a combination of normal and friction force. This will prevent the box from sliding down. As the angle between the horizontal and belt is $\theta =10°$, the angle between the belt and weight of the box is $\varphi =80°$.

Therefore,

$P=F.v\cos \varphi$

$$\begin{gathered} F=mg \\ =19.6\mathrm{N} \end{gathered}$$

Hence, the power when the box moves up is$P=1.7\mathrm{W}$

Now, the angle between the gravitational force and displacement x is $\varphi$$=90°$,

$$\begin{gathered} P=F_{g}v\cos 90 \\ P=0 \end{gathered}$$

Hence, the power when the box moves horizontals islocalid="1657178556093" $P=0\mathrm{W}$.

For this case, the angle between the gravitational force and displacement x is

$$\begin{gathered} P=F.v\cos 100 \\ P=-1.7\mathrm{W} \end{gathered}$$

Hence, the power when the box moves down is$P=-1.7\mathrm{W}$