Question

A large slab of mass $5\, kg$ lies on a smooth horizontal surface, with a block of mass $4\, kg$ 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 $F=6\, N$, the work done by the force of friction on the slab, between the instants $t=2\, s$ and $t=3\, s$, is $\left(g=10\, m s ^{-2}\right)$

• 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
$f_{\max } =\mu N=\mu m g$
$=\frac{1}{4} \times 4 \times 10=10\, N$
Evidently, $f < f_{\max }$
So, the two bodies will move together as a single unit. If $a$ be their combined acceleration, then
$a=\frac{F}{m+M}=\frac{6}{4+5}=\frac{2}{3} m s ^{-1}$
Therefore, frictional force acting can be obtained as
$f=M a=\frac{2}{3} \times 5 N =\frac{10}{3} N$
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}(3)^{2}=3$
Therefore, work done by friction $=f[s(3)-s(2)]$
$=\frac{10}{3}\left[3-\frac{4}{3}\right]=\frac{50}{9}=5.55\, J$