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A small mass slides down an inclined pla...

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

A

`(1)/(mu_(0)tan theta`

B

`(4)/(mu_(0))tan theta`

C

`(1)/(2mu_(0))tan theta`

D

`(1)/(mu_(0))tan theta`

Text Solution

Verified by Experts

The correct Answer is:
A
`F_("net")mg sin theta -mu mg cos theta`
`=mg sin theta -mu_(0)xg cos theta`
`a=(F_("net"))/m =g sin theta-mu_(0)xg cos theta`
`:. V.(dv)/(dx) = g sin theta-mu_(0)xg cos theta`
or `int_(0)^(0)vdv=int_(0)^(xm) (g sin theta-mu_(0)xg cos theta) dx`
Solving this equation we get,
`x_(m) =(2)/(mu_(0))`
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