0
$\begingroup$

The situation: I have an ideal spring, at equilibrium, sitting on a table. I then put a weight onto the spring which compresses the spring.

The spring has a height of $1\:\text{m}$ and spring constant $k$. The weight has a mass, $m$. The weight compresses the spring by $0.5\:\text{m}$. If we take the table height to be the zero of gravitational potential energy, then the work done on to the spring, by the weight, is:

$$W=-F\cdot\Delta h=mg(0.5\:\text{m}-1\:\text{m})=\frac12mg$$

My question is: What is the work done on to the weight by the spring?

If we keep the conventions the same (spring is system, weight is surroundings), then the spring should be doing negative resistive work on to the weight while it's lowering.

In other words, shouldn't the work actually be the NET force times distance? Net force is force of gravity - opposing spring force.

It's clear here that the net force must be smaller than the force just from gravity which is why I don't understand why the answer isn't the net force times distance and is instead just the gravitational force times distance.

$\endgroup$
  • $\begingroup$ I edited my answer to address your edits (I'm not sure if it notifies you when that happens). $\endgroup$ – JMac Mar 15 '17 at 17:28
1
$\begingroup$

The work done on the mass by the spring is equal to: $$W=\int_{X_1}^{X_2} \vec{F}_{\mathrm{on\ mass}}\cdot \mathrm{d}\vec{X},$$ where $X$ is the distortion of the spring. The force on the mass is $\vec{F}=-k\vec{X}$, and $\vec{X}\cdot\mathrm{d}\vec{X} = x\,\mathrm{d}X.$

For your problem, the initial distortion is zero ($X_1= 0$) and $X_2 = 0.5$ m. The work done is $$W=\int_0^{0.5}(-kX)\,\mathrm{d}X = \left.\frac{-kX^2}{2}\right|_0^{0.5}=\frac{-k}{8}.$$

The work by the spring on the mass is, indeed, negative.

Lagniappe: If the mass starts with zero velocity at the top and ends with zero velocity at the 0.5 m point, the net work should be zero, so we should get $$\frac{-k}{8}+\frac{1}{2}mg=0$$

$$\frac{1}{2}mg=\frac{k}{8}$$

The equilibrium position (the position at which the mass would continually rest on the spring) is going to be $$X_{\mathrm{equilibrium}}=\frac{mg}{k}= 0.25\ \mathrm{m},$$ halfway between the top point and the 0.5 m point in your problem. That's what we expect for a spring-mass-gravity oscillator.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Great explanation! I just have a few questions. Why do you say that "the net work should be zero." Also, what does it mean from a physical interpretation, that the work by the spring on the mass is negative? $\endgroup$ – Nova Mar 15 '17 at 18:13
  • 1
    $\begingroup$ @Nova When he says "net work should be zero" it's the same point as I made in my edited answer. Since the whole system is at rest and it started at rest, the net work of the system must be zero. There is still work done between the mass and spring though. The physical interpretation of negative work is that the spring gains potential energy, where positive work done creates a decrease in potential energy. $\endgroup$ – JMac Mar 15 '17 at 19:02
  • $\begingroup$ @JMac Is this reasoning correct: The spring gains potential energy because the weight does work on it. But at the same time it loses some potential energy because it's doing resistive work on the weight. $\endgroup$ – Nova Mar 15 '17 at 23:57
  • $\begingroup$ @JMac This will be a weird question. If this was a magical spring that was only capable of being compressed, but not resisting force. Would the total work done on the spring just be 0.5mg? Would that be a perpetual motion machine? $\endgroup$ – Nova Mar 15 '17 at 23:59
  • $\begingroup$ @Nova The potential energy stored in the spring is manifested by how the spring moves: more distortion means more potential energy because negative work is being done on the spring: work on spring = $\Delta U$. The total work done on the spring might come from several forces, so our accounting looks like this: several forces doing work result in a singular resulting change in potential energy. Potential energy change is a net effect. Your thought isn't totally wrong, but it could lead you down the wrong path later on. $\endgroup$ – Bill N Mar 16 '17 at 15:01
0
$\begingroup$

I'm not sure I follow where you decide to designate the zero of potential energy at the table height as this is where the potential energy of the spring is at its maximum as it is fully compressed. It seems counter-intuitive to me to want to set this as a zero (unless you're talking about a gravitational zero). Nonetheless, I shall try to answer.

There is a convention that work is gained from a loss of potential energy, in this case gravitational potential energy. You could say that the spring doesn't do any work on the weight because the spring's potential energy is increasing or you could say the spring is doing negative work. The weight's potential energy due to gravity is decreasing (its height with respect to the Earth/table decreases). Thus, according to this setup, gravity is doing positive work (via the weight) and the spring is doing negative work (via "resistive force"). The spring's elastic potential energy increases and the weight's gravitational potential energy decreases.

Work is indeed the net force times the displacement (dot product, actually) but whether or not you choose gravity or the spring or the weight depends on whose potential you care about and what sign/direction convention you adopt.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Yes, I meant gravitational potential energy. $\endgroup$ – Nova Mar 15 '17 at 17:09
  • $\begingroup$ I edited to explain how the net force should be different than the force just from gravity $\endgroup$ – Nova Mar 15 '17 at 17:11
  • $\begingroup$ Are you thinking about how the force of the spring changes as the spring compresses and why we're just taking E=mgh as a simple equation when the net force resisting the weight steadily increases ? $\endgroup$ – HsMjstyMstdn Mar 15 '17 at 17:20
0
$\begingroup$

The work done by the spring would be negative.

Initially you have a mass with a specific gravitational potential energy at it's height.

When you lower the mass, you are removing potential energy from it. At the same time, you're increasing the potential energy in the spring. Since the spring actually has more potential energy than when it started, the work done by the spring is negative. The mass now has less potential energy, so it did positive work.

As long as you're in a system where no energy is lost to dissipation effects (like friction) then when something does positive work onto something else, it is the same as that something else doing the same negative work. (This really all boils down to what you define as positive and negative in your sign convention)

To address your edit: When considering work between two objects you need to consider which object is applying what. The net force would be analyzing the system of both the spring and mass together.

When you do that, you get net force = 0, and therefore work done by the mass+spring system = 0. This is because they do no work on their surroundings. They only do work on each other. The mass can only do gravitational work (work due to it's weight) and the spring can only apply it's resistive force.

When you analyze what forces the block applies you'll see it's only $mg$, even though the net forces acting on it are both. I hope this clears up where you were confused.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

The work done by the spring is calculated using the force supplied by the spring. You do not use the net force on the weight, which will be zero if the weight is lowered with constant velocity.

For example, if a man pushes a block along the ground at constant velocity, the net force on the block is zero. The force supplied by the man is balanced by the friction force. But the work done on the block by the man is not zero, as he will tell you. It is the force he applies times the distance the block moves. This is also equal to the work done on the block by the friction force.

Both the man and friction do positive work on the block. The work done on the block by the man is removed by the work done by friction. You can see this more clearly if the man does the pushing on a frictionless surface, giving the block kinetic energy; the block then glides onto a rough surface, when friction reduces the KE to zero.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.