A particle of mass `m` is given a velocity `u` on a rough horizontal surface of friction coefficient `mu`. The average power imparted by friction until it stops is

Options

A. zero
B. `(1)/(2)mumgu`
C. `mumgu`
D. `2mumgu`

Answer

B

Explanation

`(1)/(2)mu^2=W_f`
Retardation `a=(mumg)/(m)=mug`
The time after which the block stops
`0=u-atimpliest=(u)/(a)=(u)/(mug)`
`overlineP=(W_f)/(t)=((1)/(2)m u^2)/((u)/(mug))=(1)/(2)m umug`