MCQ
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 surface is $\mu$. If the block travels at a uniform velocity , then work-done during its displacement $d$ is
  • A
    $\frac{{\mu \,mg\,d}}{{\sin\, \theta }}$
  • B
    $\frac{{\mu \,mg\,d}}{{\cos \,\theta }}$
  • C
    $\frac{{\mu \,mg\,d}}{{\cos \,\theta  - \mu \sin \,\theta }}$
  • $\frac{{\mu \,mg\,d}}{{\cos\, \theta  + \mu \sin \,\theta }}$

Answer

Correct option: D.
$\frac{{\mu \,mg\,d}}{{\cos\, \theta  + \mu \sin \,\theta }}$
d
$\mathrm{N}=\mathrm{mg}-\mathrm{Fsin} \theta$

The motion will be uniform if

$\mathrm{F} \cos \theta=\mu(\mathrm{mg}-\mathrm{F} \sin \theta)$

$\therefore \mathrm{F}=\frac{\mu \mathrm{mg}}{\cos \theta+\mu \sin \theta}$

$w=F \times d=F=\frac{\mu \mathrm{mgd}}{\cos \theta+\mu \sin \theta}$

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