Question

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

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

Answer 1

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



Resolving forces in vertical direction

$\begin{array}{}N=Mg-F\sin \theta & \text{(1)}\end{array}$
Resolving forces in horizontal direction:
$\begin{array}{}F\cos \theta =f_k=\mu N& \text{(2)}\end{array}$
From eq. (1) and (2)
$F\cos \theta =\mu [Mg-F\sin \theta ]$
$F\cos \theta +\mu F\sin \theta =\mu Mg$
$F=\frac{\mu Mg}{\cos \theta +\mu \sin \theta }$
As work done $W=Fd\cos \theta$
substituting the value of $F$ in above equation we get,
$W=\frac{\mu Mgd\cos \theta }{\cos \theta +\mu \sin \theta }$