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A block of mass m is pulled along a hori...

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

`(mu mgd)/(cos theta + mu sin theta)`

B

`(mu m g d cos theta)/(cos theta + mu cos theta)`

C

`(mu m g d sin theta)/(cos theta + mu sin theta)`

D

`(mu m g d cos theta)/(cos theta - mu sin theta)`

Text Solution

Verified by Experts

The correct Answer is:
B

Because the block moves with a uniform velocity, the resultant force is zero. Resolving F into horizontal components `F cos theta` and vertical component `F sin theta,` we get
`R + F sin theta = mg or R = mg - F sin theta`
Also , `f = mu R = mu(mg - F sin theta)`
But `f = F cos theta`
`:. F cos theta = mu (mg - F sin theta)` or `F (cos theta + mu sin theta) = mu mg`
or `F = (mu mg)/(cos theta + mu sin theta theta)`
work done, `W = Fs cos theta, W = (mu m g d cos theta)/(cos theta + mu sin theta) ( :. s = d)`
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Knowledge Check

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