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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
$\frac{1}{2}\frac{MV^2}{T}$
Detailed Solution
By the work–energy theorem, work done on the particle = gain in kinetic energy: $W = \frac{1}{2}MV^2 - 0 = \frac{1}{2}MV^2$
This work is delivered in time T.
Power delivered (average) = $\frac{W}{T}$
$P = \frac{1}{2}\frac{MV^2}{T}$
This work is delivered in time T.
Power delivered (average) = $\frac{W}{T}$
$P = \frac{1}{2}\frac{MV^2}{T}$