Question

Power supplied to a particle of mass $2\text{kg}$ varies with time as $P=\frac{3t^2}{2}\text{W}$, where $t$ is in seconds. If velocity of particle at $t=0$ is $v=0$, the magnitude of velocity of particle at time $t=2\text{s}$ will be

• A. $1\text{m/s}$
• B. $4\text{m/s}$
• C. $2\text{m/s}$
• D. $2\sqrt{2}\text{m/s}$

Answer 1

The correct option is C $2\text{m/s}$

Total work $W=\int Pdt$
$W=\underset{0}{\overset{2}{\int }}\frac{3t^2}{2}dt$
$W={\left[\frac{t^3}{2}\right]}_0^2=4J$
From work-energy theorem,
$W_{net}=\Delta KE=K_f-K_i$
$\frac{1}{2}m(v_f^2-v_i^2)=W$
$\frac{1}{2}m(v^2-0)=4$
$\frac{1}{2}\times 2v^2=4\Rightarrow v=2\text{m/s}$