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 instant $t=2s$ and $t=3s$ is :

• A. $2.4J$
• B. $5.55J$
• C. $4.44J$
• D. $10J$

Answer 1

The correct option is B $5.55J$

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 $S=\frac{1}{2}a{t}^2$

$S\left(2\right)=\frac{1}{2}\times \frac{2}{3}{\left(2\right)}^2=\frac{4}{3}$

and $S\left(3\right)=\frac{1}{2}\times \frac{2}{3}\times {\left(3\right)}^2=3$

Therefore, work done by friction $=F[S\left(3\right)-S\left(2\right)]$

$=\frac{10}{3}[3-\frac{4}{3}]=\frac{50}{9}=5.55J$