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 surface is `mu`. If the block travels at a uniform velocity, find the work done by this applied force during a displacement d of the block

Options

A. `(mumgd)/(cos theta+mu sin theta)`
B. `(mumgd cos theta)/(cos theta+mu sin theta)`
C. `(mumgd sin theta)/(cos theta+mu sin theta)`
D. `(mumgd cos theta)/(cos theta-mu sin theta)`

Answer

B

Explanation

(b) Normal force on the block `N=mg-F sin theta`
Block moves with uniform velocity. Hence net force = 0
or, `" " F cos theta=muN=mu(mg-F sin theta)`
`:. " " F=(mu mg)/(cos theta+mu sin theta)`
`W=Fs cos theta =(mu mgd cos theta)/(cos theta +mu sin theta)`