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.
A). 24 J
B). 3.8 J
C). 5.2 J
D). 2.6 J

Answer 1

Hint: The definition of displacement, velocity, acceleration and work done with their mathematical expressions can be used to solve these types of questions. Students also need a good understanding of differential calculus as well as integral calculus to solve for the solution.

Formulas used:
\eqalign{ & {\text{Velocity, }}v{\text{ }} = {\text{ }}\dfrac{{ds}}{{dt}} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left( 1 \right) \cr & \Rightarrow ds = vdt{\text{ }}\left[ {by{\text{ rearranging equation}}\left( 1 \right)} \right] \cr & \cr & {\text{Acceleration, }}a = \dfrac{{dv}}{{dt}} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \left( 2 \right) \cr & {\text{Work Done,}}W{\text{ = Force }} \times {\text{ Displacement}} \cr & {\text{Also, Force = Mass }} \times {\text{ Acceleration}} \cr & \Rightarrow {\text{Work Done = Mass }} \times {\text{ Acceleration }} \times {\text{ Displacement}} \cr & \Rightarrow W = \int {mavdt} \ldots \ldots \ldots \ldots \ldots \ldots \left( 3 \right) \cr}

Complete solution Step-by-Step:
In order to solve this question first we need to find the force, F acting on the body of mass 3kgs. And then the distance travelled by the body in the given time period of 2 seconds. And thus finally the work done by the force in 2 seconds.
\eqalign{ & {\text{Given:}} \cr & {\text{ }}M = {\text{ }}3kg{\text{ }} \cr & {\text{and }}s = \dfrac{{{t^3}}}{3} \cr}

Differentiating s with respect to time t, we get:
\eqalign{ & \dfrac{{ds}}{{dt}} = \dfrac{{d\left( {\dfrac{{{t^3}}}{3}} \right)}}{{dt}} \cr & \Rightarrow \dfrac{{ds}}{{dt}} = \dfrac{{3{t^2}}}{3} \cr & \Rightarrow \dfrac{{ds}}{{dt}} = {t^2} \cr & \Rightarrow v = {t^2}{\text{ }}\left[ {\because \dfrac{{ds}}{{dt}} = v} \right] \cr}
Again differentiating the above equation with respect to time t, we get:
\eqalign{ & \dfrac{{dv}}{{dt}} = \dfrac{{d{t^2}}}{{dt}} \cr & \Rightarrow \dfrac{{dv}}{{dt}} = 2t \cr & \Rightarrow a = 2t{\text{ }}\left[ {\because \dfrac{{dv}}{{dt}} = a} \right] \cr}
Now,
\eqalign{ & {\text{Work Done, }}W = \int {mavdt} \cr & \Rightarrow W = \int\limits_0^2 {m.2t.{t^2}} dt \cr & \Rightarrow W = 2m\int\limits_0^2 {{t^3}} dt \cr & \Rightarrow W = 2m\left| {\dfrac{{{t^4}}}{4}} \right|_0^2 \cr & \Rightarrow W = 2 \times 3 \times \dfrac{{2 \times 2 \times 2 \times 2}}{4} \cr & \Rightarrow W = 24J \cr}
Hence the correct answer is A., i.e., 24 joules.

Note: Students make a lot of errors while applying the formulas of integral and differential calculus. For the above solution the formulas applicable are:
\eqalign{ & \dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} \cr & \int {{x^n}} dx = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C,{\text{where n}} \ne {\text{1}} \cr}
Be sure to apply the formulas correctly and avoid silly calculation mistakes.