A block of mass M is pulled along a horizontal surface by applying a force at angle `theta` with the horizontal. The friction coefficient between the block and the surfasce is `mu`. If the block travels at a uniform velocity, find the work donen by this applied force during a displacement d of the blcok.

Options

A. `(mgdsintheta)/(costheta+musintheta)`
B. `(mgdsintheta)/(sintheta+mucostheta)`
C. `(mumgdcostheta)/(costheta+musintheta)`
D. `(mugdcostheta)/(costheta+mucostheta)`

Answer

C

Explanation

`uarr:N+Fsintheta=mg`
`N=mg-Fsintheta`
`f=muN=mu(mg-Fsintheta)`
Since the block is moving with a uniform velocity
`Fcostheta=f=mu(mg-Fsintheta)`
`F=(mumg)/(costheta+musintheta)`
Work done by the applied force is
`Fcosthetad=(mumgcosthetad)/(costheta+musintheta)`