Question

A body of mass $m$ was slowly hauled up the hill by a force $F$ as shown in the figure, which at each point was directed along a tangent to the trajectory. Find the work performed by this force, if the height of the hill is $h$, the length of its base is $l$ and the coefficient of friction is $\mu .$ (Given acceleration due to gravity $=g$)

• A. $W_F=mgh+\mu mgl$
• B. $W_F=mgh-\mu mgl$
• C. $W_F=\mu mgl-mgh$
• D. $0$

Answer 1

The correct option is A $W_F=mgh+\mu mgl$

Four forces are acting on the body:
1. weight ($mg$)
2. normal reaction ($N$)
3. friction ($f$) and
4. the applied force ($F$)

Using work-energy theorem

$W_{net}=\Delta KE$
or $W_{mg}+W_N+W_f+W_F=0$

Here, $\Delta KE=0,because{K}_i=0=K_f$

1. $W_{mg}=-mgh$

2. $W_N=0$
(as normal reaction is perpendicular to displacement at all points)

3. $W_f$ can be calculated as under:

$f=\mu mg\cos \theta$

$\therefore {\left(d{W}_{AB}\right)}_f=-fds$

$=-\left(\mu mg\cos \theta \right)ds$

$=-\mu mg\left(dl\right)(asds\cos \theta =dl)$

$\therefore f=-\mu mg\Sigma dl$

$=-\mu mgl$

Substituting these values in Eq. (i), we get

$W_F=mgh+\mu mgl$

Note: if we want to solve this problem without using work-energy theorem we will first find magnitude of applied force $\overrightarrow{F}$ at different locations and then integrate $dW(=\overrightarrow{F}\cdot \overrightarrow{dr})$ with proper limits.