# Why is Work equal to Force * distance? [duplicate]

I totally get the mathematical part, but I cannot imagine how this works. I apply a force to a ball. Why does the distance over which it moves matter to me? Sure, if I calculate the kinetic Energy of the ball after applying a force of 1N over the distance of 1m, it all works out to an energy increase of 1J, no matter how fast or heavy the ball is.

But if I imagine applying the same amount of force to a light ball that already rolled in quickly over this distance, and compare it with accelerating a heavy ball with the same force over the same distance - I can't imagine that it's the same amout of work. After all, the first thing took me seconds because the ball is so light and already had some momentum. But the heavy ball? It took me minutes to get it rolling this 1 meter because it's so heavy. I applied the force over the same distance, and added the Energy of 1J to each ball, but I cannot imagine that the Work I have done is the same, even though one time the force lasted so much longer.

So yeah, basically: Why is the Energy transfered by a force dependant on length over which it was applied, and not the time?

## marked as duplicate by knzhou, Jon Custer, Martin, Yashas, GiorgioPMar 21 at 18:27

• If the same force is applied it will. The velocity of the heavier ball will be much smaller but then, if some arrangement allows to lift a weight both heights should be equal...why? because that's what is observed... – santimirandarp Oct 18 '18 at 18:31
• – user190081 Oct 18 '18 at 18:33
• Don't "imagine" what the answer "ought to be". Do the math, and believe what it tells you! – alephzero Oct 18 '18 at 19:14
• – knzhou Oct 18 '18 at 20:06

So yeah, basically: Why is the Energy transfered by a force dependant on length over which it was applied, and not the time?

But the distance and time are not independent. For a simple example, consider a constant net force $$\vec F = F\hat{\mathbf{x}}$$ applied to an object with mass $$m$$ that initially has zero velocity and is located at the origin.

The position of the object, as a function of time is just

$$x(t) = \frac{F}{2m}t^2$$

and so the work done by the force, as a function of time, is just

$$W(t) = Fx(t) = \frac{(Ft)^2}{2m}$$

thus, the work done does depend on the time.

Now, since the initial velocity is zero, the work done equals the kinetic energy of the object

$$KE(t) = \frac{p^2_x(t)}{2m} = W(t) = \frac{(Ft)^2}{2m}$$

where $$p_x$$ is the x momentum of the object. Then you can see that, in this simple case, the product of the (constant) force and the time over which it is applied is just the momentum of the object.

It does depend on the amount of time that the force is applied to the ball. In your scenario you can check that if you want to apply a force of $$1N$$ over a distance of $$1$$ meter to two balls of different mass, then the amount of time that you will need to apply this force to each ball is dependent on their masses. This follows directly from Newton's law of motion:

$$F=ma \Rightarrow a=\frac F m$$ Starting with no velocity you have $$a=\frac {2d}{t^2}$$, then:

$$\frac F m=\frac {2d}{t^2} \Rightarrow t=\sqrt{\frac {2md}{F}}$$

For $$F=1$$ and $$d=1$$:

$$t=\sqrt{2m}$$

So the larger the mass, the longer it will take for it to cover the same distance. which is intuitive, since the heavier the object, the slower it moves under the same force. The lighter ball will cover this distance much faster than the heavier one.

Assume you believe, already in the definition of Kinetic energy:

$$KE = \frac{1}{2}mv^{2}$$

OK, now, I want to calculate the rate of change of kinetic energy:

\begin{align} {\dot KE} &= m {\vec v}\cdot {\dot {\vec v}}\\ &= m {\vec v} \cdot {\vec a}\\ \Delta KE &= \int dt\, {\vec F} \cdot {\vec v} \\ &= \int d{\vec x} \cdot{\vec F} \end{align}

Where in the last step, we used the change of variables formula in calculus to change the integral from one over t to one over x, and used the rule $$dt = (dt/dx)dx = dx\frac{1}{v}$$ in each of the three components of $$\vec v$$.

But if I imagine applying the same amount of force to a light ball that already rolled in quickly over this distance, and compare it with accelerating a heavy ball with the same force over the same distance - I can't imagine that it's the same amount of work.

I have emboldened the word imagine in the sentence above because I think that you are surmising as to what might happen if you actually tried to perform the "experiment".

The work that your body is doing is not the same as the work which is done on the ball(s). You would expend more energy applying a force of $$1\,\rm N$$ for one minute than for one second. Think of holding up a weight at a "constant" height above the ground where it would appear that you need to do no work but if fact your body must be doing work as you do get fatigued as time progresses. This is probably the most important factor which distorts your imagination of what happens.

So yeah, basically: Why is the Energy transferred by a force dependent on length over which it was applied, and not the time?

Time does come into it as the time the force is applied depends on the speed of the ball.
There is a practical problem of applying a force of $$1\,\rm N$$ over a distance of $$1 \,\rm m$$ when the ball is moving. For the large ball starting from rest you can imagine that this can be done relatively easily because the time over which the force need to be applied will be several seconds but what about a ball which is initially moving at speed of $$10 \, \rm m \, s^{-1}$$? Now the force would have to be applied in under a $$\frac {1}{10}^{\rm th}$$ of a second. (Almost) impossible to do (a push vs a kick) and hence difficult to imagine doing it under controlled conditions?

Now what might you actually observe?
With the large ball, mass $$1\,\rm kg$$, which lets suppose started from rest you would easily observe an increase in its speed.
work done = change in kinetic energy $$\Rightarrow 1 \times 1 = \frac 12 \times 1 \times \left (v^2_{\rm final} - 0^2 \right) \Rightarrow v_{\rm final} \sim 1.4 \,\rm m\, s^{-1}$$

What about the lighter ball, mass $$0.1\,\rm kg$$, which is moving at $$10 \, \rm m \, s^{-1}$$ before the force is applied?
work done = change in kinetic energy $$\Rightarrow 1 \times 1 = \frac 12 \times 0.1 \times \left (v^2_{\rm final} - 10^2 \right) \Rightarrow v_{\rm final} \sim 11 \,\rm m\, s^{-1}$$
Is this an easily observable change in the speed of the ball?
Granted that if the mass of the ball was even small, say $$0.01 \,\rm kg$$ then the final speed would be nearly double, $$17 \,\rm m\, s^{-1}$$, and you would probably notice the difference in speed.