Question

A block of mass M is pulled along a horizontal surface by applying a force at an angle 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. $\frac{\mu Mgdsin\theta }{cos\theta +\mu sin\theta }$
• B. $\frac{2\mu Mgdcos\theta }{cos\theta +\mu sin\theta }$
• C. $\frac{\mu Mgdcos\theta }{cos\theta +\mu sin\theta }$
• D. $\frac{\mu Mgdcos\theta }{sin\theta +\mu cos\theta }$

Answer 1

The correct option is C $\frac{\mu Mgdcos\theta }{cos\theta +\mu sin\theta }$


Forces on the block are
(i) Its weight Mg,
(ii) The normal force N
(iii) The applied force F and
(iv) The kinetic friction $\mu N$.
The forces are shown in figure. As the block moves with a uniform velocity, the forces add up to zero. Taking horizontal and vertical components,
$Fcos\theta =\mu NandFsin\theta +N=Mg$
Eliminating N from these equations,
$Fcos\theta =\mu (Mg-Fsin\theta )$
or, $F=\frac{\mu Mg}{cos\theta +\mu sin\theta }$.
The work done by this force during a displacement d is
$W=Fdcos\theta =\frac{\mu Mgdcos\theta }{cos\theta +\mu sin\theta }$