A block of mass m is moving with a constant acceleration a on a rough horizontal plane. If the coefficient of friction between the block and ground is `mu`, the power delivered by the external agent in a time interval t from the beginning is equal to:

Options

A. `ma^(2)y`
B. `mu` mgat
C. `mu m ( a+ mu g)` gt
D. `m(a + mu g) `at

Answer

D

Explanation

Instantaneous power delivered `=P=vec(F).vec(v)=Fv`
where, `F-f=ma rArr F = g + mg rArr P=(f+ma)v`
Put `f=mu mg`
`:.P=(mu mg+ma)v=m(a+mu g).at`
Hence (D) is correct .