Question

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

Answer 1

We can find the answer by using the equation for work which is force multiplied by displacement in the direction of force. $w = \int\limits_{}^{} {F \cdot ds}$
Where, F is the force and ds is the small displacement
From newton's second law we know that force is mass multiplied by acceleration that is
$F = ma$
Where m is the mass and a is the acceleration.
On substituting this in the equation for work and integrating we can reach the final answer.

Complete step-by-step solution:
We know that velocity is rate of change of displacement
$v = \dfrac{{ds}}{{dt}}$
Where s denotes the displacement and t denotes the time taken . Displacement is the shortest distance between the initial and final position.
It is given that displacement S is related to t as $S = \dfrac{{{t^2}}}{4}$
Where t is the time.
Now let us substitute this in the equation for velocity. Then we get,
$v = \dfrac{{d\left( {\dfrac{{{t^2}}}{4}} \right)}}{{dt}}$
$\Rightarrow v = \dfrac{1}{4}\dfrac{{d\left( {{t^2}} \right)}}{{dt}}$
$\Rightarrow v = \dfrac{1}{4} \times 2t = \dfrac{t}{2}$
Since acceleration is the change in velocity divided by time taken we can write
$a = \dfrac{{dv}}{{dt}}$
Let us substitute $v = \dfrac{t}{2}$ in this equation .
Then we get
$a = \dfrac{{d\left( {\dfrac{t}{2}} \right)}}{{dt}} = \dfrac{1}{2}$
We need to find the work done by force in 2 seconds.
Let us find that.
We know work is force multiplied by displacement in the direction of force. In equation form we can write the work done as
$w = \int\limits_{}^{} {F \cdot ds}$
Where, F is the force and ds is the small displacement
From newton's second law we know that force is mass multiplied by acceleration that is
$F = ma$
Where m is the mass and a is the acceleration.
On substituting this in equation for work we get,
$w = \int {mads}$
Now substitute the given values.
Mass is given as $6\,kg$ .
$ds = \dfrac{{2t}}{4}dt$
$\Rightarrow w = \int\limits_0^2 {6 \times \dfrac{1}{2} \times \dfrac{{2t}}{4}dt}$
$\Rightarrow w = \dfrac{6}{4}{\left[ {\dfrac{{{t^2}}}{2}} \right]_0}^2$
$\therefore w = 3J$
So, the correct answer is option D.

Note:- Remember to substitute the derivative of the given displacement in the equation for work. Don’t directly substitute the given relation $S = \dfrac{{{t^2}}}{4}$ in place of ds. Take the derivative of this equation on both sides to get the value of ds as
$ds = \dfrac{{2t}}{4}dt$.