Question

A small disc of mass m slides down a smooth hill of height h without initial velocity and gets onto a plank of mass M lying on the horizontal plane at the base of the hill. (figure). Due to friction between the disc and the plank the disc slows down and, beginning with a certain moment, moves in one piece with the plank.
(1) Find the total work performed by the friction forces in this process.
(2) Can it be stated that the result of obtained does not depend on the choice of the reference frame?

Answer 1

(a) whan the disc slides down and comes onto the plank, then
`mgh=1/2mv^(2)`
`:. v=sqrt((2gh))`
Let `v_(1)` be the common velocity of both, the disc and plank when they move together. From law of conservation of linear momentum,
`mv=(M + m)v_(1)`
`:. v_(1)=(mv)/((M + m))`
Now, change in `KE=(K)_f-(K)_i=("work done")_("friction")`
`:. 1/2(M + m)v_(1)^(2)-(1)/(2)mv^(2)= ("work done" )_("friction")`
or `W_("fr")=1/2(M = m)[(mv)/(M + m]]^(2)-1/2m v^(2)`
`=1/2mv^(2)[m/(M + m)-1]`
as`(1)/(2)mv^(2)=mgh`
`W_(f r)=-mgh[M/(M+m)]`
(b) In part (a), we have calculated work done from the ground frame of reference. Now let us take plank as the reference frame.
.
Accleration of plank `a_(0)=f/m =(mu mg)/M`
Free body diagram of disc with respect to plank is shown in figure.
Here, `ma_(0)="pseudo force"`.
Retardation of disc `w.r.t.` plank.
`a_r=(f+ma_(0))/(m)=(mumg+(mum^(2)g)/(M))/(m)=mumg+(mumg)/(M)`
`=((M+m)/(M))mug`
The disc will stop after travalling a distance `S_(r)` relative to plank, where
`S_(r)=(v_(r)^(2))/(2a_(r))(Mgh)/((M + m)mug), (0=v_(r)^(2)-2a_(r)S_(r))`
`:.` Work done by friction in this frame of reference
`W_(f r)=-fS_(r)-(mumg)[(Mgh)/((M+m)mug)]`
`=-(Mmgh)/((M + m))`
which is same as part (a).