Question

A body of mass $m$ was slowly hauled up the hill by a force which at each point was directed along the trajectory. Find the work performed by this force, if the height of this hill is $h$, the length of its base is $l$ and the coefficient of friction is $\mu$.

• A. $mg(\mu l+h)$
• B. $mg(l+h)$
• C. $mg(\mu h+l)$
• D. $mg({\mu }^{2}l+h)$

Answer 1

The correct option is A $mg(\mu l+h)$

The free body diagram of the body showing forces acting on the body is as shown below.
Resolving the weight of the body $mg$ along the surface and perpendicular to the surface as shown below.
For the body to be in equilibrium, force can be equated as,
$\begin{array}{}F=mgsin\theta +\mu N& \text{(1)}\end{array}$
$\begin{array}{}N=mgcos\theta & \text{(2)}\end{array}$
Substituting the value of $N$ in equation (1), we get
$F=mgsin\theta +\mu mgcos\theta$
$\begin{array}{}F=mg(sin\theta +\mu cos\theta )& \text{(3)}\end{array}$
Taking an element of length $ds$ on the trajectory and let $dW$ be the work done by force $F$ to move the body by displacement $ds$. Then
$dW=F.ds$
Substituting $F$ from (3),
$dW=\left(mg\right(sin\theta +\mu cos\theta \left)\right).ds$
Since force is directed along the trajectory, therefore angle between $F$ and $ds$ will be $0^{\circ }$. Therefore,
$dW=\left(mg\right(sin\theta +\mu cos\theta \left)\right)\times ds$
$\begin{array}{}dW=mg(dssin\theta +\mu dscos\theta )& \text{(4)}\end{array}$
The elemental length $ds$ can be resolved into two components i.e horizontal $\left(dl\right)$ and vertical $\left(dh\right)$ as shown below.
Therefore, $dssin\theta =dh$
$dscos\theta =dl$
Putting these values in (4),
$dW=mg(dh+\mu dl)$
Integrating the above equation,
$\int dW=\int \left(mg\right(dh+\mu dl\left)\right)$
$\Rightarrow W=mg\int dh+mg\mu \int dl$
Now $\int dh=h$ and $\int dl=l$
Hence $W=mgh+mg\mu l$
$\Rightarrow W=mg(h+\mu l)$

For detailed solution watch next video.