I recently encountered a question that made me think of this "paradox":

Imagine the following situation: enter image description here

There are two forces: Force 1 and Force 2. They accelerate a mass $m$ from the same initial velocity to the same final velocity in the same amount of time, but as you can see from the graph, when compared at the same instants in time, the two processes have different accelerations.

How does the work done by each process compare?

It is easy to see that the work done by both forces must be the same because the change in kinetic energy is equal, which is correct.

However, work can also be written as $Fd$. Now Newton's Second Law says $F=ma$, so $W=mad$.

From the graph, it seems that the average accelerations are equal for both forces, as they result in the same change in velocity over the same interval of time. But it's also clear that Force 1 results in greater displacement (area under a velocity-time graph). Therefore, shouldn't Force 1 do more work?

I have a vague sense as to why the first reasoning was correct and the second incorrect, which has to do with average accelerations vs. instantaneous accelerations, but can anybody give me a definite explanation as to why the work done for both processes are the same?

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    $\begingroup$ Have you learned calculus? $\endgroup$ – DanielSank Jan 23 '19 at 1:28
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    $\begingroup$ The work is not $F d$, it is $\int F \, dx$. You have shown the time averages of the two forces are equal, but what matters for work is the space average. $\endgroup$ – knzhou Jan 23 '19 at 1:28
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    $\begingroup$ @knzhou So the difference between $\int F(x) dx$ and $\int F(t) dt$? $\endgroup$ – RayDansh Jan 23 '19 at 1:32
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    $\begingroup$ Yup! In fact, the two forces have the same $\int F \, dx$ and the same $\int F \, dt$, even though they don't have the same $\int \, dx$. $\endgroup$ – knzhou Jan 23 '19 at 1:51
  • $\begingroup$ Okay I think I understand now, thanks @knzhou! (And yes I'm taking AP Calc right now) $\endgroup$ – RayDansh Jan 23 '19 at 2:10

The comments seem to qualitatively answer your question. Let's look at a specific example. There is probably a more elegant way to do this, but let's just use brute force.

Let's say $$v_2(t)=bt$$ and $$v_1(t)=bt(2-t)$$ so we have $v_1(0)=v_2(0)=0$ and $v_1(1)=v_2(1)=b$ similar to what you have in your picture.

Taking time derivatives and applying Newton's second law we arrive at $$F_2(t)=mb$$ and $$F_1(t)=2mb(1-t)$$

Now, the general definition of work in one dimension is given by $$W=\int F\ \text d x$$ but since we are given the force as a function of time we can perform a change of variables to arrive at $$W=\int Fv\ \text d t$$

Applying this we get $$W_2=\int_0^1mb^2t\ \text d t=\frac12mb^2$$ and $$W_1=\int_0^1(2mb(1-t))(bt(2-t))\ \text d t=2mb^2\int_0^1(t^3-3t^2+2t)\ \text d t=\frac12mb^2$$ so we see that $W_1=W_2$$^*$

From the graph, it seems that the average accelerations are equal for both forces, as they result in the same change in velocity over the same interval of time. But it's also clear that Force 1 results in greater displacement (area under a velocity-time graph). Therefore, shouldn't Force 1 do more work?

You are right in saying that the first object will have a larger displacement, but you also need to look at the forces, since $W=\int F\ \text d x$. $F_2$ is constant over the entire interval, whereas $F_1$ starts at $2F_1$ and ends at $0$. The work done does not just depend on the displacement, it also depends on the value of the force during this displacement.

$^*$This is the result of something more general: $$W=\int Fv\ \text d t=\int m\frac{\text d v}{\text d t}v\ \text d t=\int mv\ \text dv=\frac12m(v_f^2-v_0^2)=\Delta K$$

So we see that the work done just depends on the initial and final velocities, regardless of how long it takes for this progression to occur.


The problem with what you're doing is you're not accounting for the acceleration both positively and negatively that happens in case 1. To get to the peak of that curve it had to be accelerated more than the line, and subsequently accelerated the other direction to account for it. Yes, in both cases the average acceleration is the same, but the real acceleration to make the movement in figure 1 happen would be greater for sure.

Edit: Ahh the drawing was a bit funny, I see what you're getting at there though, the curve going up faster and then tapering off. The same is still true, and it's easy to get lost with equations and math so let's just talk reality. If two spaceships (cliche I know) were moving next to one another like you described, their velocities are constant unless something accelerates them. So if in the initial instance they are moving at the same velocity, and one of them is accelerated, it will continue to move faster until something accelerates it the other direction. This results in more work. This graph could be modified to show this by assuming one observer's reference frame. Meaning it's velocity is always zero, so the 0 line would actually just be the line you drew for it's increasing velocity. If you turn the picture and look at it that way, you can see clearly that the relative acceleration is indeed negative at one point.

I wish you broke it too, that would be more fun!

  • $\begingroup$ The acceleration is positive the entire time for both cases. $\endgroup$ – Aaron Stevens Jan 23 '19 at 1:56
  • $\begingroup$ I just updated my answer to be a bit more clear to account for that. $\endgroup$ – Seth Weinstein Jan 23 '19 at 2:07
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    $\begingroup$ You should edit your answer to be one clear answer. If something is wrong then just get rid of it. There is an edit history available for those who are interested. $\endgroup$ – Aaron Stevens Jan 23 '19 at 2:33

Another way to look at this. You are correct in saying that the work done by each process is equal, since the two processes produce the same change in final kinetic energy. You are also correct in the think that force required to move the particle over the curved path is larger than the force required to move the particle along the straight path. So why doesn't the curved path force do more work? It is because work is actually defined by

$W=\int \overset {\rightarrow } {F}\cdot \, d\overset {\rightarrow } {x} $

The dot product of the force and position element means that only the force parallel to the particle's motion contributes to work. On the curved path there is also a component of force perpendicular to the motion which causes the change in direction of the particle. This extra perpendicular component of the force does no work. By the way, you can get the two particles to the same spot in equal times, or with equal velocities, but not both.

  • $\begingroup$ This is a 1D problem. If you are referring to the curve is the graph, the graph is velocity v. time, not y v. x $\endgroup$ – Aaron Stevens Jan 23 '19 at 22:41
  • $\begingroup$ @AaronStevens: I see the graph, but nowhere in the question does it state that the two particles follow the same path. As I say, my answer is another way to look at it. $\endgroup$ – Bill Watts Jan 24 '19 at 0:51
  • $\begingroup$ The particles certainly don't follow the same path. $\endgroup$ – Aaron Stevens Jan 24 '19 at 2:02
  • $\begingroup$ It's just confusing because if we are talking about a non-1D system, then your talk about forces acting perpendicular to the path of the object does no work and does not change the magnitude of the velocity. Therefore, the graph only shows what happens due to force parallel to path, so it might as well be a 1D problem $\endgroup$ – Aaron Stevens Jan 24 '19 at 2:54
  • $\begingroup$ @AaronStevens: Well if we don't assume a 1-d system, and there was no reason to assume so, then everything I said is absolutely correct. It seemed to me that the answer to a 1-d system was too obvious to be what he meant, especially as he was talking about a larger force and was confused about that force not representing more work. But if he was talking about both forces following the same line, then I guess I answered a different problem. $\endgroup$ – Bill Watts Jan 24 '19 at 8:46

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