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 (as shown in above 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.

Find the total work performed by the friction forces in this process. Given reduce mass of system $\mu =\frac{mM}{m+M}$.

• A. $\mu gh$
• B. $-\mu gh$
• C. $2\mu gh$
• D. $-2\mu gh$

Answer 1

The correct option is B $-\mu gh$

As per the problem at the top of hill the disc is at rest hence,
$K_1=V=mgh$
where, $K_1$ is kinetic energy and $V$ is potential energy of disc.

When the disc is sliding down a smooth hill it acquires kinetic energy (equals to the potential energy)

let, v be the velocity of the disc while sliding down the hill, and
$K_2=\frac{1}{2}m(v{)}^2$

$\Rightarrow \frac{1}{2}m(v{)}^2=mgh$

$\therefore (v{)}^2=2gh$

$\therefore v=\sqrt{2gh}$

The disc is now gets onto the plank and then beginning with a certain moment, moves in one piece with the plank. Let, $v^{\prime }$ be the velocity of plank.

Hence,
$mv=(m+M)v^{\prime }$

$v^{\prime }=\frac{mv}{(m+M)}$

$v^{\prime }=\frac{m\left(\sqrt{2gh}\right)}{m+M}$

The kinetic energy of the plank is

$K_2=\frac{1}{2}(m+M)v^{\prime 2}$

$K_2=\frac{1}{2}(m+M)(\frac{m\left(\sqrt{2gh}\right)}{m+M}{)}^2$

$K_2=\frac{m^{2}gh}{m+M}$

Now, the total work performed by the friction forces in this process is given by change in kinetic energy.

$W_F=K_2-K_1$
$W_F=\frac{m^{2}gh}{m+M}-mgh$

$W_F=\frac{gh{m}^2-gh{m}^2-ghmM}{m+M}$

$W_F=-\frac{ghmM}{m+M}$

$W_F=-\mu gh$
where, $\mu =\frac{mM}{m+M}$