Question

A large slab of mass 5 kg lies on a smooth horizontal surface, with a block of mass 4 kg lying on 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 (g = 10 m$s^{-2}$)

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

Answer 1

The correct option is B

5.55 J

Maximum frictional force between the slab and the block

$f_{max}=\mu N=\mu mg=\frac{1}{4}\times 4\times 10=10N$

Evidently, $f<f_{max}$

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

$f=Ma=\frac{2}{3}\times 5N=\frac{10}{3}N$

Using $s=\frac{1}{2}a{t}^2$

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

Therefore, work done by friction = f[s(3) - s(2)]

= $\frac{10}{3}[3-\frac{4}{3}]$

= $\frac{50}{9}=5.55J$