Question

A large slab of mass $5Kg$ lies on a smooth horizontal surface, with a block of mass $4Kg$ 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 $F = 6N$ , 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

We know when a body moves, it experiences an opposing force, and this force is calculated by multiplying frictional coefficient with the mass of the object. In this question we can obtain the frictional force, then using the equations of motion and work done we can obtain work done by the force.

Complete Step-By-Step Solution:
We know frictional force is defined as the force that opposes the movement of the body. Frictional force exists at every contact point.
Now, the frictional force between the slab and the block can be calculated as:
$f = \mu mg$
Where,
$f$is the frictional force
$\mu$ is the frictional coefficient
$m$is the mass of the BLOCK
Thus, we obtain:
$f = \dfrac{1}{4} \times 4 \times 10 = 10N$
Now, considering the block and the slab as an single unit, the total mass becomes$= (4 + 5)Kg = 9Kg$
Now, we can apply the equations of motion and obtain the acceleration of unit as:
$a = \dfrac{F}{{mass}}$
In the question, it is given that the force applied is $6N$
Thus, we get:
$a = \dfrac{6}{9} = \dfrac{2}{3}m{s^{ - 2}}$
Thus, the frictional force acting on the slab is obtained as:
$f = ma = 5 \times \dfrac{2}{3} = \dfrac{{10}}{3}N$
We will now find the distance covered using:
$S = ut + \dfrac{1}{2}a{t^2}$
Distance travelled at $t = 2s$:
$S = 0 + \dfrac{1}{2} \times \dfrac{2}{3} \times {(2)^2} = \dfrac{4}{3}m$
Distance travelled at $t = 3s$:
$S = 0 + \dfrac{1}{2} \times \dfrac{2}{3} \times {(3)^2} = 3m$
We know work done is calculated as the product of force and displacement.
Now, we can find work done as:
$W = f \times [S(3) - S(2)]$
We obtain:
$W = \dfrac{{10}}{3} \times (3 - \dfrac{4}{3}) = 5.5J$
This is the required amount of work done.

Hence, option (B) is correct.

Note:
The amount of work done is stored as energy. We can say that when there is a kinetic energy of a body, there is a positive amount of work done by it. The amount of work done depends on the amount of force applied to bring about the work done.