# Work of the same force acting once on a moving object and once on a motionless one

If the same force $$F$$ acted once on an already moving object and once on a motionless object. would it give more energy to the moving object than to the motionless one?. Note that both objects have the same mass and are similar.

according to the work formula $$W = Fd$$ (in this question work and displacement are in the same direction), the force will give more energy to the moving body because its displacement will be larger due to the fact that it already had some speed before the force acted on it. So is it true that a moving object gains more energy from the same force?

Note that in both cases the force acts for the same time.

Yes, the work will be larger if the object is already moving in the direction of the force.

The mechanical power (work per time) is given as:

$$P = \frac{\Delta W}{\Delta t} = \frac{Fd}{\Delta t} = F v$$

where $$v = \frac{d}{\Delta t}$$ is the velocity.

So, yes, it costs extra energy to apply the same force to an already moving object than if it was at rest!

That might be unintuitive at first. It can help to remember that kinetic energy is $$E=\frac{1}{2} mv^2$$. Because velocity is squared, increasing the speed from, say, 0 to 1 m/s costs significantly less energy than increasing it from 100 m/s to 101 m/s.

If you confused by this, you are not alone. For instance, rocket engines puzzled engineers & physicists for a long time. While it is burning, the engine produces a roughly constant thrust. According to the equation above, the power $$P=Fv$$ delivered by the engine will increase as the rocket accelerates. Eventually, the power will exceed the chemical energy released by burning the fuel. (I am not going to spoil the fun by posting the solution to that paradoxon right here!)

Note that in both cases the force acts for the same time.

Per the work energy theorem: the net work done on an object equals its change in kinetic energy, or for a constant force acting through a distance $$d$$

$$W_{net}=F_{net}d=\frac{1}{2}mv_{f}^{2}-\frac{1}{2}mv_{i}^{2}$$

So if you want to compare the work done on the two objects, you must base it on an equal distance through which the force acts on each, not the same amount of time the force acts. Since

$$d=\frac{1}{2}at^{2}+v_{i}t$$

Clearly, for the same amount of time, the distance traveled will be greater for the object having an initial velocity making the work done greater.

Hope this helps.