# Calculated by using W=Fs

1. Oct 17, 2008

### Mentallic

Work, calculated by using $$W=Fs$$ provides us with the energy required/used.

If a force is applied to a mass, but there is no displacement (due to friction/gravity/some other counteractive force) this means there is no work being done. My first issue with work is that energy is being used to create this force, but the 0 work means 0 energy. It doesn't take into account the energy required to exceed the countering forces.

If a force of 1N is applied to a mass of 1kg for 1 second, its displacement will be 50cm. This calculates to be 0.5 Joules of energy. If this same force is applied for the same amount of time to the same mass now travelling at 1ms-1, the displacement will now be 150cm, so the work done is equal to 1.5 Joules.
The same force was sustained on the same mass for the same time, yet in the 2nd situation there was 3x as much work done as the 1st situation.

I guess it's my limited understanding of work, but this concept seems to be getting the better of me and I'm struggling to come to terms with it. My teacher can't explain it to me either, so if anyone is able to explain what is happening here, it would be much appreciated.

2. Oct 17, 2008

### Tac-Tics

Re: Work

First, it doesn't take energy to create a force field. All objects are surrounded by a gravity field at all times, but it does not cost them any energy.

Work into or out of an object in this scenario really means a change in kinetic energy. The formula for kinetic energy is 1/2 m v^2. The key is that energy is not linear with time!

In the case where v0 = 0 and v1 = 1m/s, the initial (kinetic) energy E0 = 1/2 (1) (0)^2 = 0 and the final energy E1 = 1/2 (1) (1)^2 = 1/2. So the *change* in energy is E1 - E0 = 1 = 0 = 1/2.

In the later case with the moving object, v0 = 1m/s and v1 = 2m/s. So E0 = 1/2 (1) (1)^2 = 1/2 and E1 = 1/2 (1) (2)^2 = (1/2) 4 = 2. So E1 - E0 = 2 - 1/2 = 3/2.

So is it crazy that the latter requires three times the energy? Not at all! Because as we have just shown, the energy required to increase the velocity from 1m/s to 2m/s is three times as much as increasing the velocity from 0 to 1/ms.

I will admit, though, it confused me for a while. You just have to pick the interpretation that *does* make sense, and try to ignore any others =-P

Work is very useful and intuitive, but much moreso when you're working with potential energy in fields. In such instances, you don't need to worry about kinetic energy at all. The difference in energy potentials becomes a simple function of distance.

3. Oct 17, 2008

### Mentallic

Re: Work

Why is it that the work formula isn't used in this scenario? and how could this scenario be altered so as it is necessary to use the work formula, rather than the kinetic energy formula?

Yes the fact that gravity is a force, yet doesn't require energy whatsoever is a problem that I cannot seem to grasp, and don't understand how it is possible. But I'll ask this again on a later note so as to not drift this thread into 2 topics.

4. Oct 17, 2008

### Tac-Tics

Re: Work

This scenario *does* use the work formula (W = Fd). I was just showing that you could achieve the same results using the other formula, and that it is perfectly legitimate for a force to do more work on a faster moving object than a slower one, even though the distance through which it acts is the same. It's not clear why this can be true in the work equation, but it's not unreasonable when you look at it through the kinetic energy equation.

I don't think this is too far off topic, and it's something I was confused about for a long time, too.

When we think of energy and work in qualitative terms, we always think about how hard it would be for us, as humans, to do something. Things that make us tired, we assume, take more work than things that don't. That's just our intuition, but it's misleading.

Say you have a lighter friend. You hold her up in the air. It's hard, you get tired, you sweat, and you heat up. At some level, you believe that you need to eat your Wheaties for energy or you wouldn't be able to do such a thing.

But now put her down, lie on the floor, and ask her to lie on top of you (tell her it's for science!) Now her weight on top of you may make you feel squished, but it won't nearly as difficult. You could probably keep her above the ground as long as you wanted like this.

This is because of the physiology of your body. When she's laying on top of you, most of her weight is supported by your bones, which are rigid and hard. It doesn't take any energy at all to keep her suspended this way, just the same as it doesn't take any energy for her to lie on a table, just the same as it doesn't take any energy for the moon to stay in orbit.

But when you lift her with your arms, your muscles are not rigid in the same way, and your body is actually acting in a complicated manner. Your muscles are programmed to relax by themselves and your brain must continually correct them to contract. Your muscles heat up because they are contracting and expanding small amounts (a la forces through a distance), and *this* is why it's tiring.

The above might not be 100% biologically correct, but it's correct in principle. Your body is doing work that it doesn't need to do. Other kinds of animals have muscles that are able to hold up weights without effort, but not us.

5. Oct 17, 2008

### Mentallic

Re: Work

Aha! So it has something to do with the difficulty of a force being applied to a moving object rather than a still object. It's just that whenever I thought force, I thought of gravity for simplicity and this never occurred to me.

I completely tossed the idea of using energy to hold things in place since my teacher told me there is no work being done to hold something still, since there is no displacement of the object. The energy used to hold up the girl must be a different 'type' of energy to that of work?
That biological analogy was pretty good and I agree with that and the mass on the table, however, this is different to gravity as there is no movement being done in these cases. Gravity provides a constant force and somehow this is "free" energy, and a hell of a lot of free energy too!

6. Oct 17, 2008

### Staff: Mentor

Re: Work

There's no work being done on the girl. But internally, as Tac-Tics described, your muscles are doing work to maintain tension.

7. Oct 17, 2008

### Tac-Tics

Re: Work

This is correct.

This is not. No energy is used to hold up the girl. The energy goes from your wheaties to your muscles, but never to the girl. It is simply lost as heat.

In general, whenever it feels like something is taking up energy, but isn't, the answer is usually heat. When you run a mile, you start off at rest and you're at rest after the run. You're in the same place where you started. But it was tiring, you're sweating, and that energy from eating your wheaties turned into heat inside your body.

Whether it's the table under the girl or you, it doesn't matter. Whether it's the force of gravity, or you and hear are just wearing iron armor plating next to an enormous electromagnet, it doesn't matter. For matters we are talking about, all forces act the same. Gravity is exactly like any other imaginable force. No energy is coming to you for free. However, sometimes it seems you can "lose" it. But when you lose it, like I said above, it's just turning into heat and other forms of "useless" energy.

8. Oct 17, 2008

### Mentallic

Re: Work

Thankyou for directing me back on the right path. I guess I didn't understand the point of this the first time I read it.
As for the muscles, work is being done. Is it the force or rapid small fluctutations (resulting in a large distance) that is the major contributor to the work calculated?

Didn't your legs provide the force to make your body travel that mile? Isn't this also work?

Yes I see how much energy is inefficient and is turned into other forms besides the desired kinetic energy, but how is the gravitational energy being lost for a falling object? I am still failing to understand how the energy provided by gravity (since work is being done) can be maintained for an eternity, without consequences such as conversion of matter->energy.

9. Oct 18, 2008

### Tac-Tics

Re: Work

I'm not even sure of what your body does with the energy to begin with, but in the end, it just turns to heat.

But the distance traveled is at right angles the the field. The work equation (the calculus form) is a line integral involving the dot product of the path and the field. If the path is always at right angles to the field, their dot product will always be zero.

A falling object loses potential energy, but when it loses in potential, it gains in kinetic energy. Until it hits the ground, of course. Then it has lost that potential energy *and* its kinetic energy is now 0. Where did the missing work go? Heat.

It's very hard to get a grip on it when thinking of falling objects because of all the exceptional phenomena involved. But take any situation where the object that doesn't involve a person lifting weights and that doesn't involve the object hitting the ground, and it makes very good sense.