Question

A block of mass $m$ is pulled slowly by a minimum constant force on a horizontal surface through a distance $x$. The coefficient of kinetic friction is $\mu$. Find the work done by force $F$.

Diagram

• A. $\frac{2l}{\sqrt{V^2-v^2}}$
• B. $\frac{2vl}{V^2-v^2}$
• C. $\frac{\mu \text{mgx}}{1+{\mu }^2}$
• D. $\frac{2Vl}{V^2-v^2}$

Answer 1

The correct option is C $\frac{\mu \text{mgx}}{1+{\mu }^2}$

Draw FBD for the given system.

Diagram

Find minimum value of the force.
From FBD,
$F\cos \theta =f\dots \left(i\right)$

$N+F\sin \theta =mg\dots \left(ii\right)$

And force of friction,
$f=\mu N\dots \left(iii\right)$

Using equation $\left(ii\right)$ and $\left(iii\right)$ in equation $\left(i\right)$,

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

$F\cos \theta +\mu F\sin \theta =\mu mg$

$F=\frac{\mu mg}{\cos \theta +\mu \sin \theta }$

for $F$ to be minimum $cos\theta +\mu sin\theta$ must be maximum

let
$\frac{dF\left(\theta \right)}{d\left(\theta \right)}=0=-\sin \theta +\mu \cos \theta$

$\tan \theta =\mu$

For $\theta ={\tan }^{-1}\mu$, $F$ is minimum.

Find work done by force F.
Formula used: $W=Fs\cos \theta$

Given, distance travelled by block, $=x$
So, work done by force $F$,

$W=Fx\cos \theta$

$W=\frac{\mu mg}{\cos \theta +\mu \sin \theta }\left(x\cos \theta \right)$

$W=\frac{\mu mgx}{1+\mu \tan \theta }$

$W=\frac{\mu mgx}{1+{\mu }^2}$