Question

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

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

Answer 1

Answer: (D)

$\frac{1}{2}\frac{MV^2}{T}$

Answer 2

The correct option is(D): $\frac{1}{2}\frac{MV^2}{T}$
Power delivered in time T is
$P = F\cdot V = MaV$
or P = MV $\frac{dv}{dt} \, \, \Rightarrow \, \, \, PdT = MVdV$
$\Rightarrow \, \, \, \, \, \, \, \, \, \, PT=\frac{MV^2}{2}$ or $p=\frac{1}{2}\frac{MV^2}{T}$