At `t = 0`, a block (of mass m) is projected with speed v on a rough long horizontal plank (of mass 3m) placed on smooth ground, as shown. Work done by friction on the block between `t = 0` and `t = oo` equals :

Options

A. `(15 mv^(2))/(32)`
B. `(-15 mv^(2))/(32)`
C. `- (3mv^(2))/(8)`
D. `(- mv^(2))/(16)`

Answer

B

Explanation

`vec(P)_(i) = vec(P)_(f)` ( `F_("ext") = 0` for the system )
`mv = 4 mv'`
`v' = (v)/(4)`
for `m, W_(f) = Delta KE = (1)/(2) m ((v)/(4))^(2) - (1)/(2) mv^(2)`
`= - (15)/(32) mv^(2)`