How can time taken to do Work be different?

Question

Reading about Power today, I stumbled upon this question in my mind-

"How can the time taken to do a piece of work be different in 2 scenarios?" - when I came across, the example of '2 different bodies(with same mass) being pushed with the same force in the same direction in different time intervals' which showed different amounts of power being used[assuming the surface is frictionless] despite the Work done being the same. Here's why I think this is not possible(I may be wrong) -

• When a certain force is applied on a body, it causes it to accelerate, which leads to displacement, ultimately leading to Work being in that scenario. Now, if the body is being displaced across the same displacement, in a different amount of time, the velocity of the body must be different. But since the body is accelerating due to application of force, the displacement is given by - s = ut + 0.5at^2 , assuming initial velocity to be zero, s = 0.5at^2. Since displacement was same, time was different, acceleration in both cases must be different, therefore, the force applied was different, and the Work done was different in both cases.

Summing up my question in a single sentence-

How is it possible for the same piece of Work with same force and displacement to be done in different intervals of time?

Answer 1

The key phrase in your argument is "assuming initial velocity to be zero".

If the same force $F$ is applied to two bodies with the same mass $m$ then they have the same acceleration $a = \frac F m$. If both bodies start from rest then it takes each body the same time $t = \sqrt{\frac {2d}{a}}$ to cover a distance $d$, and the work done on each body in this time interval is the same - it is $W = Fd$. And, of course, the kinetic energy gained by each body is

$\displaystyle E = \frac 1 2 mv^2 = \frac 1 2 m(at)^2 = \frac 1 2 m (2ad) = mad = Fd = W$

So in this case, if both bodies start from rest, then the work done on each body in a given time interval is the same.

However, if one body starts with an initial velocity $u > 0$ then it still has acceleration $a$, but now it covers the distance $d$ in a time $t' = \frac {\sqrt {u^2 + 2ad}-u} {a}$. So now the same work $W = Fd$ is done on the body in a shorter length of time, since $t' < t$. Note that the kinetic energy gained by the body is now

$\displaystyle E = \frac 1 2 m(v^2-u^2) = \frac 1 2 m((u+at')^2-u^2) = \frac 1 2 m (2ad) = mad = Fd = W$

as before.