The coefficient of friction between a particle moving with some velocity `V_(0)` and the rough horizontal surface is `((V_(0))/(2 g t_(0)))`. Find how much kinetic energy is lost in time `t_(0)` due to friction :

Options

A. `1//4`
B. `1//2`
C. `3//4`
D. `2//3`

Answer

C

Explanation

`v^(1)=v_(0)-(mu g) t_(0)`
=`v_(0)-((v_(0))/(2g t_(0)))g t_(0) = v_(0)//2` ltMbrgt velocity left `= v_(0)//2`
`K.E` left =` (1)/(2) m ((v_(0))/(2))^(2) = (1)/(4)((1)/(2) mv_(0)^(2))`
`=(1)/(4)` of initial `K.E`.