Question

A large slab of mass $5kg$ lies on a smooth horizontal surface, with a block of mass $4kg$ lying on the top of it, the coefficient of friction between the block and the slab is $0.25$. If the block is pulled horizontally by a force of $F=6N$, the work done by the force of friction on the slab between the instants $t=2s$ and $t=3s$ is

• A. 2.4 J
• B. 5.55 J
• C. 4,44 J
• D. -10 J

Answer 1

Answer: (B)

Maximum frictional force between the slab and the block
$f_{\max }=\mu N=\mu mg=\frac{1}{4}\times 4\times 10=10N$

Evidently, $f\propto f_{\max }$
So, the two bodies will move together as a single unit.
If a be their combined acceleration, then
$a=\frac{F}{m+M}=\frac{6}{4+5}=\frac{2}{3}m{s}^{-2}$
Therefore, frictional force acting can be obtained as
$f=Ma=\frac{2}{3}\times 5=\frac{10}{3}N$
Using $S=\frac{1}{2}a{t}^2$
$S(2)=\frac{1}{2}\times \frac{2}{3}(2{)}^2=\frac{4}{3}$
and $S(3)=\frac{1}{2}\times \frac{2}{3}\times (3{)}^2=3$
Therefore, work done by friction $=F[S(3)-S(2)]$
$\because =\frac{10}{3}\left[3-\frac{4}{3}\right]=\frac{60}{9}=5.55J$