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

\frac{M V^{2}}{T}

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\frac{1}{2} \frac{M V^{2}}{T^{2}}

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\frac{M V^{2}}{T^{2}}

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\frac{1}{2} \frac{M V^{2}}{T}

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Solution

The correct option is D $\frac{1}{2} \frac{M V^{2}}{T}$
Power delivered in time T is
$P = \to F \cdot \to V = M a V$
or $P = M V \frac{d V}{d T}$ $\Rightarrow$ $P d T = M V d V$
$\Rightarrow P T = \frac{M V^{2}}{2}$ or $P = \frac{1}{2} \frac{M V^{2}}{T}$

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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

The correct option is D 12MV2TPower delivered in time T is P=→F⋅→V=MaVor P=MVdVdT ⇒ PdT=MVdV⇒PT=MV22 or P=12MV2T

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

The correct option is D $\frac{1}{2} \frac{M V^{2}}{T}$
Power delivered in time T is
$P = \to F \cdot \to V = M a V$
or $P = M V \frac{d V}{d T}$ $\Rightarrow$ $P d T = M V d V$
$\Rightarrow P T = \frac{M V^{2}}{2}$ or $P = \frac{1}{2} \frac{M V^{2}}{T}$