A particle moves move on the rough horizontal ground with some initial velocity `V_(0)`. If `(3)/(4)` of its kinetic enegry lost due to friction in time `t_(0)`. The coefficient of friction between the particle and the ground is.

Options

A. `(v_0)/(2 g t_0)`
B. `(v_0)/(4 g t_0)`
C. `(3v_0)/(4 g t_0)`
D. `(v_0)/(g t_0)`

Answer

A

Explanation

`a = - g mu " Final velocity " = sqrt(((1 - 3/4)1/2 mv_0^2)/(1//2 m))= (v_0)/(2) " " implies " "(v_0)/2 = v_0 - g mu implies t = (v_0)/(2 g mu)`.