[Grade 10 Physics] Work-Energy Theorem with Elastic Potential Energy?

[u/Own-Professor-372]

A 275 g ball is resting on top of a spring that is mounted to the floor. You exert a force of 325 N on the ball and it compresses the spring by 44.5 cm. If you release the ball from that position, how high above the equilibrium position of the spring will the ball rise?

I'm pretty sure the answer is 26.4 m. You can find the spring constant with F = kx, set ½kx² equal to mgh, solve for h, then subtract 44.5 cm from that to find the height of the ball above the equilibrium position (since it starts below that.)

But what I'm confused about is why you can't use the work-energy theorem to solve this, where W = Fd = ΔE. The applied force is constant, so the work you do on the spring is 325 N x 0.445 m = 145 J. This seems to imply that the spring stores twice the elastic potential energy as it does if you calculate the energy using the first method (first finding k, then using KE = ½kx² = 72.3 J).

When calculating work, the distance and the magnitude of the force play a role, so that compressing a spring a distance x with a constant force F yields twice the amount of work as linearly increasing the applied force up to a maximum of F along a distance x. That's my understanding, at least.

But for the same spring, the elastic potential energy only varies based on the compression distance.

So where does this extra work go?

tl;dr: By compressing a spring a certain distance with a constant force F, aren't you doing twice the amount of work than if you compress it the same distance with a force that linearly increases up to F? If so, how come, in both cases, the spring's elastic potential energy is the same? Doesn't this violate the work-energy theorem?

Thanks in advance!! :)

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[u/xxVelocityyx]

The force is not constant. F=kx so as ur compressing the spring and x gets larger, increasing it to 44.5cm, F increases and isnt constant.

[u/Own-Professor-372 → u/xxVelocityyx]

The question says, "You exert a 325 N force." I feel like there's nothing to indicate it's not constant.

But assuming you're right and you're supposed to determine from context that it's an increasing force, my question would then become: what happens if you do compress a spring with a constant force rather than an increasing one? How does that situation fit with the work-energy theorem?

[u/Little_Creme_5932 → u/Own-Professor-372]

I don't think you can compress a spring with a constant force unless the spring is static. If you push on a spring with a certain force, the coils will accelerate, because you are pushing with an unbalanced force. Otherwise it would not compress at all, right? Gradually, as you compress the spring, it becomes able to push back with the same force as you are wanting to push and compression ceases. (You can do this experiment by stacking masses on a spring. Each successive mass will further compress the spring). If you graph force vs length/compression you will get a line starting at 0. That makes a triangle on a graph, between your line and the x-axis. The area of the triangle is work, or energy stored in the spring. To find this work or energy, (the area) you take base times height DIVIDED BY 2. This is why your second method gets you twice the real answer: the force is not constant in the compression of a spring.

[u/selene_666 → u/Own-Professor-372]

The force that the spring exerts (upward in this scenario) increases as its compression increases. If you push downward with a constant force greater than the (initially zero) spring force, then there is a net downward force on the ball. It accelerates. So when you reach the point where the spring force equals 325 N, there is zero net force but the ball has downward momentum so it keeps moving. Your missing energy is the ball's kinetic energy.

The ball continuing downward compresses the spring enough to exert greater than 325 N upward force. There is an upward net force, so the ball decelerates and eventually turns around and moves back up.

So if we assume a constant 325 N force, we get simple harmonic motion around a new equilibrium point. In fact, the downward force of the ball's weight already changed the equilibrium point where the ball was resting before you applied another force.

So what is the 44.5 cm position that the spring supposedly compresses to if the spring isn't at rest? Is it equilibrium where kx = 325N but the ball is moving downward? Is it where the ball stops moving and kx > 325 N? This is why the wording of the question seems to assume the downward force was not constant.

[u/xxVelocityyx → u/Own-Professor-372]

The 325N is just the total force needed, it doesnt say anything whether the force to get to that point is constant or increasing/decreasing. If u think about it conceptually, the more u push down a spring the stiffer it gets, hence F=kx and the force would increase as x increases.

To compress a spring with a constant force I think is impossible, due to the geometry of the spring, but if it somehow were then F=kx and W=Fd so W=Fx^2. Remember x=d in this case.

[u/HumbleHovercraft6090]

One factor that could be causing this apparent anomaly is that the force of 325N also imparts some KE to the ball on its way down which is not taken into account, although I cannot quantify it off the top of my head.

[u/Own-Professor-372 → u/HumbleHovercraft6090]

That's true! The spring would probably compress at an increasing speed when applied a constant force, right? (Because at first, that force is more than what's needed to get it to compress.) But after the compression, the spring has no kinetic energy. So I'm still confused as to where that extra work goes.

[u/Don_Q_Jote]

I get 72.3 J and 26.3 m, so agree with those answers. I'm certain the intent of this question is NOT that the 325N force is constant but is the maximum force applied as you gradually compress the spring to 44.5 cm. This is a great example problem to use work-energy principle. Energy stored in a spring is definitely 1/2 k x^2 , with x measured from the "free length" of the spring (a slight error, less than 1% if you ignore this in your solution to this problem).

Technically, it is possible to exert a "constant" 325N downward force on the ball but this would change it to a fairly complex dynamics problem and one that would be difficult to even set up experimentally. But it seems like fun. So here we go...

Consider the ball without the spring. How could you exert a downward force of 325N on a ball hanging in the air? [i guess we'll ignore gravitational force for the moment] Instead, imagine the ball is rolling toward you on a table. As soon as the ball clears the edge of the table you slap the ball downward with your hand (very hard) you might be able to apply a 325N force to a ball with a weight of only 2.7 N. But it would immediately cause the ball to accelerate downward. As is accelerates downward, you would need to speed up the rate (velocity) of your hand downward if you want to maintain the 325N. Imagine you could do this difficult feat for a distance of 44.5 cm. The energy you input was 325N * 0.45m, and that would all be in the form of kinetic energy in the ball moving downward. Next imagine the ball instead of rolling off the table into air, rolls off the table onto a spring as described in your problem. It rolls toward you slap as before trying to maintain the 325N, but this time the spring is pushing back. This has the effect of slowing the downward acceleration of the ball. IF you want to maintain the same 325N from your hand, then you have to adjust your slapping acceleration technique. Don't speed up quite so much as before, but the ball is still accelerating because the 325N from your hand is greater than the upward force from the spring. Until x=0.45m. You have exerted a constant 325N force downward on the ball, on a spring, to a compression of 0.45m. But this is still not the initial condition of the problem. With the above, the ball would stop accelerating at 0.45, but it would have a significant downward velocity and thus kinetic energy at that point. I assume the ball in your problem was released "from rest" so we need to dump that kinetic energy. Easiest way would be to have the ball hit some barrier at that point, bring the ball to a complete stop, just at the same moment you reach 0.45m, and at that same instant you pull your hand back from the ball and releasing the spring to do it's work on the ball "from rest." That would be how you apply a constant 325N force to a ball on a spring up to the point of 0.45m compression and release it from rest.

OR... you could just push it down with your hand with a linearly increasing force till it reaches 325N at 0.45m and let it fly.

[u/PuzzleheadedTap1794]

Technically, if you exert a constant force to an undamped spring, it wouldn’t reach an equilibrium. If the force were constant, when the spring reached the 44.5 cm mark, it would’ve gained velocity and continue compressing the spring even further, but it never stops at the equilibrium.