Question

A block of mass $M$ is pulled along a horizontal surface by applying a force at an angle $\theta$ with horizontal. Coefficient of friction between block and surface is $\mu$. If the block travels with uniform velocity, find the work done by this applied force during a displacement $d$ of the block.

• A. $\frac{\mu Mgd\cos \theta }{\cos \theta +\mu \sin \theta }$
• B. $\frac{\mu Mgd\sin \theta }{\cos \theta +\mu \sin \theta }$
• C. $\frac{Mgd\cos \theta }{\cos \theta +\mu \sin \theta }$
• D. $\frac{Mgd\sin \theta }{\cos \theta +\mu \sin \theta }$

Answer 1

The correct option is A $\frac{\mu Mgd\cos \theta }{\cos \theta +\mu \sin \theta }$

The forces acting on the block are shown in the figure,
As the block moves with uniform velocity the resultant force on it is zero.

$\therefore F\cos \theta =\mu N.........\left(1\right)$

Applying equation of equilibrium in vertical direction

$F\sin \theta +N=Mg........\left(2\right)$

Eliminating $N$ from equations $\left(1\right)$ and $\left(2\right)$,

$F\cos \theta =\mu (Mg-F\sin \theta )$

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

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

Work done by this force during a horizontal displacement $d$ will be

$W=\left(F\cos \theta \right)d=\frac{\mu Mgd\cos \theta }{\cos \theta +\mu \sin \theta }$

Therefore, option $\left(a\right)$ is correct.