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A small mass slides down an inclined pla...

A small mass slides down an inclined plane of inclination `theta` with the horizontal the coefficient of friction is `mu = mu_(0)x`, where x is the distance through which the mass slides down. Then the distance covered by the mass before it stops is

A

`(2)/(mu_(o)) tan theta`

B

`(4)/(mu_(o)) tan theta`

C

`(1)/(2 mu_(o)) tan theta`

D

`(1)/(mu_(o)) tan theta`

Text Solution

Verified by Experts

The correct Answer is:
A
`a = g sin theta -(mu_(o)g cos theta)x :. int_(o)^(v) vdv = int_(o)^(x_(max)) [g sin theta - (mu_(o)gcos theta)x]dx`
After solving we get `x_(max) = (2 tan theta)/(mu_(o))`
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