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

Answer 1

Answer: (B)

Because the block moves with a uniform velocity, the resultant force is zero. Resolving $F$ into horizontal component $F\cos \theta$ and vertical component $F\sin \theta$, we get
$R+Fsin\theta =mg$ or $R=mg-F\sin \theta$
Also $f=\mu R=\mu (mg-Fsin\theta )$
But $Fcos\theta =f$
or $F\cos \theta =\mu (mg-F\sin \theta )$
or $F(\cos \theta +\mu sin\theta )=\mu mg$
$\therefore F=\frac{\mu mg}{\cos \theta +\mu sin\theta }$
Work $W=Fscos\theta$
$\therefore W=\frac{\mu mgd\cos \theta }{\cos \theta +\mu sin\theta }$
$\left(\because s=d\right)$