Question

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`

Answer 1

Correct Answer - 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))`