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

Answer 1

Let the applied force be $F.$

As the force $F$ makes an angle $\theta$ with the horizontal, so the component of the force in the horizontal direction is $Fcos\theta$

This force tends to move the block inn the forward direction.

The other force that is acting on the block is the frictional force which acts opposite to the applied force.

The frictional force is given as:

$f=\mu RThereactionforce,R=mg-Fsin\theta Therefore,f=\mu (mg-Fsin\theta )Forthehorizontalmotion,Fcos\theta \ge f\Rightarrow Fcos\theta \ge \mu (mg-Fsin\theta )\Rightarrow F\ge \frac{\mu mg}{cos\theta +\mu sin\theta }Therefore,theworkdoneinmovingtheblockbydistanced,W=\overrightarrow{F}\cdot \overrightarrow{d}=Fdcos\theta =\frac{\mu mgdcos\theta }{cos\theta +\mu sin\theta }$