Question

A body of mass $3kg$ is under a force, which causes a displacement in it given by $S = {\dfrac{t}{3}^3}$ (in $'m'$ ). Find the work done by the force in the first $2$ seconds.

Answer 1

Hint : The basic premise of the question is that we are given the displacement in the form of a function in $t$ which is given by:
$S = {\dfrac{t}{3}^3}$
Then since we have to find the work done first we will find the velocity by differentiating the function and then we will differentiate the velocity to find the acceleration then we will find the force by multiplying the calculated value by mass given. And at last we will find the work done by using the formula W = \mathop \smallint \nolimits_{{t_2}}^{{t_1}} F.dS = \mathop \smallint \nolimits_{{t_2}}^{{t_1}} ma.dS

Complete Step By Step Answer:
First we will find the value of velocity by differentiation of displacement.
$\dfrac{{ds}}{{dt}} = {t^2}$
Then upon further differentiating we will get our value of acceleration in terms of $t$
The value of acceleration is given by:
$\dfrac{{{d^2}s}}{{d{t^2}}} = 2t$
Then we will find the value of work done using the formula
W = \mathop \smallint \nolimits_{{t_2}}^{{t_1}} F.dS = \mathop \smallint \nolimits_{{t_2}}^{{t_1}} ma.dS
Mass here is given as $3$ kg. and the ${t_1} = 2{\text{ and }}{t_2} = 0$
The expression will be written as:
\mathop \smallint \nolimits_0^2 3 \times 2t \times {t^2}dt
Upon solving we get,
\mathop \smallint \nolimits_0^2 6{t^3}dt and
Upon integrating we get:ss
$\dfrac{3}{2}\left[ {{t^4}} \right]_0^2$
And after putting the values we get our answer as:
$24J$ .

Note :
The rate of change of displacement is called velocity so to calculate velocity always differentiate the displacement, similarly rate of change of velocity is called acceleration to calculate acceleration we double differentiate displacement or differentiate the velocity. The rate of change of acceleration is called a jerk in scientific terms.