Question

A particle of mass $M$, starting from rest, undergoes uniform acceleration. If the speed acquired in time $T$ is $V$, the average power delivered to the particle is

• A. $\frac{M{V}^2}{T}$
• B. $\frac{1}{2}\frac{M{V}^2}{T^2}$
• C. $\frac{M{V}^2}{T^2}$
• D. $\frac{1}{2}\frac{M{V}^2}{T}$

Answer 1

The correct option is D $\frac{1}{2}\frac{M{V}^2}{T}$

Power delivered in time $T$ is
$P=\frac{W}{T}$
where $W$ is the work done in time $T$
According to the work energy theorem
$W=\Delta K$
$W=\frac{1}{2}M{V}^2-0$
where $V$ is the final velocity of the particle.
Hence $P=\frac{1}{2}\frac{M{V}^2}{T}$