Question

A body of mass ‘m’ accelerates uniformly from rest to velocity ${{V}_{1}}$ in time ‘T’. Calculate the instantaneous power delivered to the body as a function of time.

Answer 1

Power delivered to the body of mass ‘m’ depends on the acceleration of the body and the instantaneous velocity at time t. First we will determine all the quantities related to the power. Further we will substitute those in the expression of power, to determine the instantaneous power.
Formula used:
$V=U+at$
$P=\dfrac{W}{t}$
$F=ma$

Complete answer:
Power of a body is defined as the rate at which work (W) is done by the body in time t. mathematically this can be written as, $P=\dfrac{W}{t}$.
Work done (W) is defined as the product of force times the displacement (d). And from Newton’s second law Force F is defined as the product of mass times the acceleration. Therefore work done can be mathematically written as,
$\begin{aligned} & W=Fd \\\ & \Rightarrow W=mad \\\ \end{aligned}$
Substituting the work done in equation of power we get,
$\begin{aligned} & P=\dfrac{W}{t} \\\ & \Rightarrow P=\dfrac{mad}{t}\text{, }\because \dfrac{d}{t}=v \\\ & \Rightarrow P=mav \\\ \end{aligned}$
Where v is the instantaneous velocity.
Let us say that a body moves with acceleration ’a’. if the initial velocity of the body is ‘U’ than its velocity V at time ‘t’ by Newton’s first kinematic equation is given by,
$V=U+at$
It is given in the question that the body initially at rest accelerates with uniform acceleration (a), such that its velocity ${{V}_{1}}$ is at instant of time T. Hence from Newton’s first kinematic equation the acceleration a is equal to
$\begin{aligned} & {{V}_{1}}=aT \\\ & \Rightarrow a=\dfrac{{{V}_{1}}}{T} \\\ \end{aligned}$
At any instant of time ‘t’ velocity of the body V is given by,
$\begin{aligned} & v=U+at \\\ & \Rightarrow v=at \\\ & \Rightarrow v=\dfrac{{{V}_{1}}}{T}t \\\ \end{aligned}$
Hence substituting the instantaneous velocity of the body and acceleration in the equation of power we get,
$\begin{aligned} & P=mav \\\ & \Rightarrow P=m\dfrac{{{V}_{1}}}{T}\dfrac{{{V}_{1}}}{T}t \\\ & \Rightarrow P=\dfrac{m{{V}_{1}}^{2}}{{{T}^{2}}}t \\\ \end{aligned}$

Note:
It is given in the question that the body moves with uniform acceleration. Hence the above equation is valid. If the body is not under uniform acceleration then we can say the power delivered to the body also changes as a function of acceleration.