# Work and Frames of Reference

1. Jun 8, 2014

If I stand on the surface of the Earth holding an object from my frame of reference I do no work because the object is at rest and the displacement is zero. However from the frame of reference of a second object moving toward the earth the first object would be moving up and I would be doing work. So am I doing work here?

2. Jun 8, 2014

### A.T.

Yes.

3. Jun 8, 2014

But the amount of work I do should correspond to a certain amount of expended chemical energy. So would the amount of chemical energy expended depend on the frame of reference? In other words if I held a five kilogram mass on the surface of the earth for an hour I would expend 4.9 X 10^9 Joules if there was an object moving at one hundred thousand kilometers per hour towards the earth?

4. Jun 8, 2014

### jbriggs444

If you adopt a frame of reference in which you are moving vertically at thousands of kilometers per hour, you would indeed be doing a great deal of work supporting the object. But at the same time the surface of the earth would be doing a great deal of work supporting you. The net work that you need to supply is zero regardless of the reference frame that you choose.

5. Jun 8, 2014

### Andrew Mason

This is a very good question. Start with the definition of work done on a body A in moving from position 1 to position 2 against a force $\vec{F}$ supplied by body B:

$$W_{1→2}= \int_1^2 \vec{F}\cdot d\vec{s}$$

Your question, essentially, is: what is the reference frame in which $\vec{s}$ (the displacement vector) is defined? This is often glossed over in texts.

The short answer to your question is that the displacement vector is measured in the reference frame of the body that is supplying the force, i.e. body B.

AM

6. Jun 8, 2014

### Staff: Mentor

This is not correct. The displacement vector can be measured in any inertial frame. Newtonian mechanics is frame invariant.

7. Jun 8, 2014

### Staff: Mentor

There are multiple definitions of work. The standard one is the F.d definition that Andrew Mason mentioned earlier. A better one (IMO) is that work is the transfer of energy by means other than heat. These two definitions are equivalent for the most part, but I find the second one easier to use.

In your example there are three objects: you, the weight, and the earth. There are two forces acting on the weight, the contact force and gravity. There are three forces acting on you, the two contact forces and gravity. There are also three forces acting on the earth, two gravity forces and the contact force with you.

Using the F.d definition of work all of the upwards forces do positive work and all of the downwards forces do negative work in a reference frame where you are moving upwards. In each case the positive work on each object equals the negative work so there is no net energy transfer. So you can make a somewhat silly loop showing how the work that you do on the weight does work on the earth which does work on you and round and round.

Using the transfer of energy definition you simply note that no energy is changing so no work is being done and you don't need to worry about the loop.

A more interesting scenario is, for example, a spring pushing two masses apart (changing elastic PE to KE). There work is done by the second definition, and it is interesting to see how it works out in different frames.

8. Jun 8, 2014

DaleSpam, just to clarify, what your saying is that I don't expend any chemical energy because the net work on all three objects is zero, and therefore I don't do any work regardless of the frame of reference?

But lets say (relative to the Earth's surface) I lifted the weight at a constant velocity. Since the acceleration is zero the net force has to be zero. Since the net force is zero the net work has to be zero, but the weight is gaining gravitational potential energy so wouldn't the net work then not be zero?

9. Jun 8, 2014

### Staff: Mentor

Yes. Using the "transfer of energy" definition of work it is clear that no energy is transferred and therefore no work is done.

Using the "F.d" definition of work is more complicated, but comes to essentially the same conclusion.

Yes. The work is not zero. Again, using the "transfer of energy" definition it is easy since it is clear that energy is transferred from you to the weight.

You can work it out using the "F.d" definition also, but it is more cumbersome. Just note that the d moved by your hands is greater than the d moved by your feet, and the d moved by your center of mass (gravity) is the same as the d moved by your feet. When you do that you get that there is net work being done by you.

10. Jun 8, 2014

### Andrew Mason

I agree that the laws of motion are frame invariant ie. mass, time, force and acceleration. Since displacement, velocity, work and energy are not, when one is talking about work and energy one has to define the reference frame. By writing W = ∫ Fds one is implicitly assuming a frame of reference. Otherwise W has no meaning. Work is always done by some body on another body so the displacement has to do with the change in relative positions of the two bodies. No?

Anyhow, I did say it was the short answer.

AM

Last edited: Jun 8, 2014
11. Jun 8, 2014

### Staff: Mentor

Sure. But that frame does not have to be "the reference frame of the body that is supplying the force". Any inertial frame will do.

12. Jun 9, 2014

### mattt

$$\int_{t_0}^{t_1}\vec{F}(t)\cdot\vec{v}(t)dt$$ obviously depends on the reference frame.

I guess you mean the particle is at rest at the beginning, say $$\vec{v}(t_0) = 0$$, then I lift it some meters, and at the end I hold it again, at rest $$\vec{v}(t_1)=0$$ but higher than at the beginning.

If we analyze this system with respect to a reference frame bound to the Earth surface AND we suppose it is an inertial frame, then we have:

$$\int_{t_0}^{t_1}\vec{F}(t)\cdot\vec{v}(t)dt = \frac{1}{2}m v^2(t_1) - \frac{1}{2}m v^2(t_0) = 0$$

(where $$\vec{F}(t)$$ is the total force upon the particle)

That is, the total work is cero. What are the forces on the particle?

One is the weight force $$m\vec{g}$$ on the particle, that is constant, downwards, and conservative. The other force is the force $$\vec{F}_1(t)$$ you apply upon the particle, and (we can assume to simplify things) this force vertical, upwards, non-constant, and it is not conservative.

At the beginning, $$F_1(t_0)>mg$$ that is why the particle starts to move upwards.

Nearing the end, that is, for $$t$$ close to $$t_1$$ we know that

$$F_1(t) < mg$$, that is why the particle is decreasing its upward speed nearly the end and at the vey end, at $$t_1$$, it is at rest again.

We have:

$$0 = \int_{t_0}^{t_1}\vec{F}(t)\cdot\vec{v}(t)dt = \int_{t_0}^{t_1}(\vec{F_1}(t) + m\vec{g})\cdot\vec{v}(t)dt =$$

$$= \int_{t_0}^{t_1}\vec{F_1}(t)\cdot\vec{v}(t)dt + \int_{t_0}^{t_1}m \vec{g}(t)\cdot\vec{v}(t)dt$$

You have to undertand that $$(\vec{F_1}(t) + m\vec{g})\cdot\vec{v}(t) > 0$$ at the beginning, and $$(\vec{F_1}(t) + m\vec{g})\cdot\vec{v}(t) < 0$$ nearing the end.

Also you have to understand that $$\left(\int_{t_0}^{t_1}\vec{F_1}(t)\cdot\vec{v}(t)dt\right) > 0$$

that $$\left(\int_{t_0}^{t_1}m \vec{g}(t)\cdot\vec{v}(t)dt\right) < 0$$ and that

$$\left|\int_{t_0}^{t_1}\vec{F_1}(t)\cdot\vec{v}(t)dt\right| = \left|\int_{t_0}^{t_1}m \vec{g}(t)\cdot\vec{v}(t)dt\right|$$

In fact we have that:

$$\int_{t_0}^{t_1}m \vec{g}(t)\cdot\vec{v}(t)dt = m.g.y(t_0) - m.g.y(t_1) = m g (y(t_0) - y(t_1))$$

And so we know that:

$$\int_{t_0}^{t_1}\vec{F_1}(t)\cdot\vec{v}(t)dt = - \int_{t_0}^{t_1}m \vec{g}(t)\cdot\vec{v}(t)dt = - mg(y(t_0) - y(t_1))$$

So the total work is zero, the Kinetic Energy Increment is zero (the particle is at rest at the beginning and also at rest at the end). The particle has more Gravitational Potential Energy at the end than at the beginning, and that is achieved because of the work done by the (non-conservative) force F_1 you apply upon the particle.

I hope you now understand this important example much better.

13. Jun 9, 2014

### nasu

The work of the NET force is equal to the change in kinetic energy. So zero net force means zero change in kinetic energy.
The change in potential energy is given by the work done by the field whose potential energy you consider, in this case gravitational. Not by the net force. This was discussed before on the forum.

And regarding the first "paradox", as it was mentioned the work and energy are not frame invariant. However, the chemical energy spent is a different story. This is not given by the amount of macroscopic work done to hold something (which may be zero).

The processes in the muscles need energy even to hold the arm in a fixed position, without any extra weight and without apparent motion. Things are moving inside the muscles, in order to keep your arm "steady". And this internal work (and the energy used) is independent of the reference frame. You will feel tired in the frame attached to your body or in any other frame.

14. Jun 9, 2014

### mattt

I guess you meant "against the field" (in this example).

15. Jun 9, 2014

In reference to the above scenario I'm still a little confused.

One definition for the net work is the change in kinetic energy (Work-Energy Theorem). Since the kinetic energy is the same at t0 and t1 the net work done on the weight equals zero.

Another definition for net work is the transfer of energy in a form other than heat. Since the object gained gravitational potential energy a positive net work was done on it.

So I don't understand if net work was done on the weight and how we would determine this.

16. Jun 9, 2014

### mattt

That is not a definition. That is a theorem.

The mathematical definition is just:

$$\int_{t_0}^{t_1}\vec{F}(t)\cdot\vec{v}(t)dt$$

that is, the number that equals the average of $$\vec{F}(t)\cdot\vec{v}(t)$$ (along the curve = trajectory of the particle with respect to a given reference frame) times the length of the curve.

Yes, because of the Work-Energy THEOREM, we know that in the example of my previous post, the total work (that is, the work done by the net force uppon the body throughout its trajectory) is zero (because the particle starts at rest, and it ends at rest again).

No, that is not a correct definition for the total or net work in Newtonian Mechanics. That is a useful definition in Thermodynamics, where part of the total work on the system is called "heat" for concrete given reasons.

Have you read my previous post? (post # 12 of this thread).

The net (or total) work is zero. The work done by you (your hand) is equal (but opposite sign) to the work of the weight force. With words:

0 = total work = (your hand + weight) work = (your hand) work + (weight) work

So

(your hand) work = - (weight) work = - (PEi - PEf) = PEf - PEi.

Please, reread post # 12 or at least how I just put it into words. :-)

17. Jun 9, 2014

### nasu

Well, by definition the change in potential energy is the negative of the work done by the field.
A force acting against the field may be present but is not required.

But you are right, I was a little sloppy. By "given" I did not mean "equal". Just that the change in potential energy is related to the work of this force and not of the net force.

18. Jun 9, 2014

### mattt

Yes, it is just a different way of saying the same thing.

In reality all this thing is crystal clear when you see it mathematically, but for some reasons, the moment we start puting names to some terms, people (at the beginning) start to get so confused about all this.

19. Jun 9, 2014

### nasu

You persist in your confusion. The potential energy is not related to the net work but just to the work of the field.

If you move a body upwards with constant velocity (as in your example) you need to act with an upward force equal to the weight. The net force is zero so there is no net work so the change in KE is zero (constant speed). However the work of gravity is not zero. It has a finite negative value.
The potential energy increases (positive change) and you have ΔPE=-Wg
where Wg is the work of gravity.

You don't need to worry about what happens when the body is accelerated from rest or slowed down at the end of the motion. It's enough to apply the definitions and theorems properly, for the portion with uniform velocity.

20. Jun 9, 2014