Question

A aprtical moves on a rough horizontal ground with same initial velocity say `v_(0)`. If `(3//4)th` of its kinetic energy is lost in friction in time `t_(0)` , then coefficient of friction between the partical and the ground is:
A. `(v_(0))/(2 gl_(0))`
B. `(v_(0))/(4 gl_(0))`
C. `(3 v_(0))/(4 gl_(0))`
D. `(v_(0))/(gl_(0))`

Answer 1

`3//4^(th)` energy of lost, i.e., `1//4^(th)` kinetic energy is left . Hence, its velocity becomes `v_(0)//2` under a retardation of mmg in time `t_(0)`
`(v_(0))/(2) = v_(0) - mu g t_(0)` or `mu = (v_(0))/(2g t_(0))`