Question

A crate of mass m is being pulled (by an engine), up on an inclined plane making an angle $\alpha$ with the horizontal. The coefficient of friction between crate and incline is $\mu$. The crate starts from rest and being, pulled by engine at constant acceleration so that it travels a distance (displacement) $s$ in time $t$. Determine the power delivered by engine to crate as a function of time? Take $t=0$ as the instant when crate is at rest.

• A. $mg(sin\alpha +\mu cos\alpha )\times \frac{2s}{t}$
• B. $\mu mgcos\alpha \times \frac{2s}{t}$
• C. $\frac{4m{s}^2}{t^3}$
• D. $\frac{2ms}{t}[\frac{2s}{t^2}+g(sin\alpha +\mu cos\alpha \left)\right]$

Answer 1

The correct option is D $\frac{2ms}{t}[\frac{2s}{t^2}+g(sin\alpha +\mu cos\alpha \left)\right]$

Free body diagram for the given scenario is shown in the attached figure.

Normal to the inclined plane,

$N=mg\cos \alpha$

Hence, frictional force is,

$F_r=\mu N=\mu mg\cos \alpha .........\left(i\right)$

Using the second equation of motion,

$s=ut+a{t}^2/2$

$⟹a=\frac{2s}{t^2}...........\left(ii\right)$

Parallel to the inclined plane,

$F-F_r-mg\sin \alpha =ma$

$F=\mu mg\cos \alpha +mg\sin \alpha +ma$ using $\left(i\right)$

Work done by engine as a function of distance,

$W={\int }_0^{s}F.ds$

$W={\int }_0^s(\mu mgcos\alpha +mg\sin \alpha +ma)ds$

$W=\mu mgs\cos \alpha +mgs\sin \alpha +mas$

$W=(\mu mg\cos \alpha +mg\sin \alpha +ma)\frac{a{t}^2}{2}$ using $\left(ii\right)$

$P=\frac{dW}{dt}$

$P=(\mu mg\cos \alpha +mg\sin \alpha +ma)at$

Substituting value of $a$ from $\left(ii\right)$, we get

$P=(\mu gcos\alpha +g\sin \alpha +\frac{2s}{t^2})\frac{2ms}{t}$