Question

A body mass of 6kg is under a force which causes displacement in it given by = $\dfrac{{{t^2}}}{4}$ metres where t is the time. The work done by the force in 2 seconds is
A. 12 J
B. 9 J
C. 6J
D. 3J

Answer 1

The work done by a body is equal to the product of the force applied and the displacement caused by it.
Work done,
$W = F \times s$
where, F = force in newtons(N) and s = displacement in metres (m) .
The SI unit of work is joules (J).

Complete step-by-step answer:
Step 1: Compute the displacement .
The displacement is given by the function, $s\left( t \right) = \dfrac{{{t^2}}}{4}$
At time, t = 2 sec
The displacement will be:
$s\left( 2 \right) = \dfrac{{{2^2}}}{4} = \dfrac{4}{4} = 1m$
Step 2: Compute the force
Given, mass of the body = 6 kg.
To obtain the acceleration , we have to differentiate the equation of displacement twice –
$a = \dfrac{{{d^2}s}}{{d{t^2}}}$
We have –
$s\left( t \right) = \dfrac{{{t^2}}}{4}$
Differentiating once,
$\dfrac{{ds}}{{dt}} = \dfrac{{2t}}{4} = \dfrac{t}{2}$
Differentiating again, we get –
$\dfrac{{{d^2}s}}{{d{t^2}}} = \dfrac{d}{{dt}}\left( {\dfrac{{ds}}{{dt}}} \right) = \dfrac{d}{{dt}}\left( {\dfrac{t}{2}} \right) = \dfrac{1}{2}$
Hence, the acceleration is a constant, which is equal to, $a = 0.5m{s^{ - 2}}$
Multiplying by mass, we can obtain the force.
$F = ma = 6 \times 0.5 = 3N$
Step 3: Obtain the work done
Work done, $W = F \times s$
Substituting, we get –
$W = 3 \times 1 = 3J$

Hence, the correct option is Option D.

Note: The work done is a scalar quantity, which only has magnitude and no direction. You may wonder as to how the product of two vector quantities such as force and the displacement can be a scalar.
Well, there are two types of multiplication of vectors : - Dot Product : Product is scalar - Cross Product : Product is vector
The work done is a dot product of two vectors, force and displacement.
In vector form,
$W = \overrightarrow F .\overrightarrow S = \left| {\overrightarrow F } \right|\overrightarrow {\left| S \right|} \cos \theta$