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

The Correct Option is D

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 \(p=\frac{1}{2}\frac{MV^2}{T}\)