Question

A block of mass m is kept over another block of mass 2m and the system rests on a smooth horizontal surface. The coefficient of friction between the blocks is 0.50. Find the work done by the force of friction on the smaller block by the bigger block during a displacement d the system, when a force mg is applied to the lower block.

• A. $\frac{mgd}{3}$
• B. mgd
• C. $\frac{mFd}{2(M+m)}$
• D. Zero

Answer 1

The correct option is C $\frac{mFd}{2(M+m)}$

Given that acceleration$,a=F/2(M+m)$

Now for $\left(a\right):-$ Find the coefficient of Kinetic Friction $.$

Weight of the system $=(M+m)g=N$ $\$Where\$N\$isthenormalforce\$$

The force of Friction $=\mu N$

Also given$,a=F/2(m+M)$

Now for equation of force$,$ along its surface

$F-\mu N=(m+M)a$

$F-\mu (m+M)g=(m+M)F/2(M+m)$

$F-\mu (m+M)g=F/2$

$2F-2\mu g(m+M)=F$

$\mu ==\frac{F}{2g\left(m+M\right)}$

Hence the coefficient of Kinetic friction is $\frac{F}{2g\left(m+M\right)}$

Now for $\left(b\right):-$ Frictional force $:-$

Let$,$ the frictional force be f acting on the small block

so the acceleration must be $,a=F/2(m+M)$

So the force $=m\times a$

$f=m\left[F/2\right(m+M)$

$f=mF/2(m+M)$

Hence the frictional force is $mF/2(m+M)$

Now for $\left(c\right):-$ Work done $:-$

The velocity of the block during displacement $d,$

$v^2=u^2+2ad$

$v^2=2ad${$since$u =0}$

$v^2=2Fd/2(m+M)$

$v^2=Fd(m+M)$

Now the final Kinetic energy $K.E=1/2(m{v}^2$)

$K.E=mFd/2(m+M)$

So$,$ the work done on smaller block is larger than bigger block by the frictional force $,$

so change in $K.E$ is $mFd/2(m+M)$

Hence,

option $\left(C\right)$ is correct answer.