Question

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

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

Answer 1

The correct option is B $\frac{\mu \text{mgd cos}\theta }{\text{cos}\theta +\mu \text{sin}\theta }$

$N=mg-F\text{sin}\theta$
Block moves with uniform velocity. Hence net force = 0
or, $\text{F cos}\theta =\mu N=\mu (mg-\text{F sin}\theta )$
$\therefore F=\frac{\mu mg}{\text{cos}\theta +\mu \text{sin}\theta }$
$W=\text{Fs cos}\theta =\frac{\mu \text{mgd cos}\theta }{\text{cos}\theta +\mu \text{sin}\theta }$