Question

A variable force acts on a $1\text{kg}$ particle such that acceleration as a function of time is given by $a=2t$. The particle is starting from rest. Find the work done by the force on particle in first $3$ seconds.

• A. $81\text{J}$
• B. $40.5\text{J}$
• C. $20\text{J}$
• D. $10.5\text{J}$

Answer 1

The correct option is B $40.5\text{J}$

Acceleration given, $a=2t$

Velocity obtained by the integrating on the both sides during time interval $t=0\text{sec}$ to $t=3\text{sec}$

${\int }_0^{v}dv={\int }_0^{3}2tdt$
$v={\begin{array}{c}{t}^2\end{array}|}_0^3=3^2-0^2=9$

$v_0=0\text{m/s}$, as the particle is starting from rest
$v=9\text{m/s}$

According to work energy theorem,
Net work done $=$ Change in kinetic energy
$=\frac{1}{2}\times m\times (v^2-v_0^2)$
$=\frac{1}{2}\times 1\times (9^2-0)$
$=40.5\text{J}$

Hence option B is the correct answer.