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

Options

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`

Answer

A

Explanation

`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))`