# Energy Question

boggleface
Why is work defined as Fd? If you apply the same force to 2 different masses over the same distance surely the larger mass has more energy. Why isn't work defined as Ft??

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Ambitwistor
Originally posted by boggleface
Why is work defined as Fd?

Why is a circle defined to be the set of points equidistant from a central point? It's just a definition. I don't understand the question. Do you just not like that the name for Fd is "work", as opposed to some other word? i.e., it doesn't correspond to what you intuitively think "work" ought to mean?

If you apply the same force to 2 different masses over the same distance surely the larger mass has more energy.

No. Look at a proof of the work-energy theorem: the change in kinetic energy of the body is equal to the work done, which is Fd.

Maybe you'd understand a special case more easily: consider a body being accelerated from rest through a distance 'd' by a constant acceleration 'a'. There is a kinematics formula that says the velocity at the end is v^2 = 2ad. So the kinetic energy at the end is,

1/2 mv^2 = mad = Fd

If you apply a fixed force to a body, the larger mass will accelerate less, so it will end up with a lower velocity. But also has a larger mass, so it ends up with the same kinetic energy.

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If you apply the same force for the same time to two different masses (say a truck and a bicycle) you don't get the same result do you?

It takes more energy to move a truck 100m than it does a bicycle, so more work is done pushing it the same distance.

You also say "surely the larger mass has more energy" What do you mean by this? If I push a truck along a flat road, I do work moving the truck, but don't increase it's energy state.

Ambitwistor
It takes more energy to move a truck 100m than it does a bicycle, so more work is done pushing it the same distance.

I think you're taking friction into account, since the friction increases with mass.

On a frictionless surface, if you push on two objects with the same force over the same distance, they will end up with the same (kinetic) energy.

You also say "surely the larger mass has more energy" What do you mean by this? If I push a truck along a flat road, I do work moving the truck, but don't increase it's energy state.

This is again an issue of friction: if you have to apply a constant force to counteract the constant force of friction, then you do work on the truck, but there is no NET work done on the truck, because friction is doing work opposite to you.

Yes, I was considering friction.

My reply to the first post was started before you answered Ambitwistor, but posted AFTER yours went up. I was just giving a simple illustration of where boggleface had gone a bit wrong.

(Although actually, if you consider Friction, the truck tyres will have heated up a bit, so it WILL be in a higher energy state!)

Homework Helper
Originally posted by boggleface
Why is work defined as Fd? If you apply the same force to 2 different masses over the same distance surely the larger mass has more energy. Why isn't work defined as Ft??

Fd is work, which is equal to the change in kinetic energy (assuming there are no other forces on it at the same time)

Ft is impulse, which is equal to the change in momentum (same assumption)

The confusion between momentum and kinetic energy is common, especially since they are both quantities of mass in motion. The only way out of the confusion is to stick to the definitiona of each, until you "grasp" it mentally.

boggleface
No. What i mean is that Fd does not sit with my intuition about energy. I imagine 2 different masses bieng accelerated over the same distance with the same constant force. My intuition tells me that surley the larger mass will have more energy. NOTHING TO DO WITH FRICTION! why isn't mv named energy?Is it just me, am i strange?

Staff Emeritus
Gold Member
You will find that physics frequently requires you to drop preconcieved notions in order to grasp the realities of the universe. Do not worry about how you think it ougth to be, concentrate on learning how it is.

boggleface
yeah I've heard this said many times about intuition and reality but never with regard to this

Ambitwistor
Originally posted by boggleface
No. What i mean is that Fd does not sit with my intuition about energy. I imagine 2 different masses bieng accelerated over the same distance with the same constant force. My intuition tells me that surley the larger mass will have more energy. NOTHING TO DO WITH FRICTION! why isn't mv named energy?Is it just me, am i strange?

mv isn't named energy because it's already named momentum. Whatever you intuitively feel "energy" is, isn't the same as what "energy" is actually defined to be in physics. Presumably, you think the words "energy" and "momentum" mean the same thing. I don't think it's too productive to argue over the English meanings of words vs. the physics meanings. They're just definitions.

If you apply the same force over the same distance to two objects of different masses, then the larger mass will hurt more if it hits you, because it will have a larger momentum. On the other hand, you will burn the same number of calories pushing either mass, so they have the same energy. Momentum and energy in physics are simply different concepts, regardless of what you think their English meanings are or ought to be.

boggleface
im arguing over my intuative concepts not words

boggleface
If I'm pushing two different masses over the same distance with the same constant force, imagining that no friction is present on the masses, i am pushing the larger mass for a greater amount of time. Surely i have used more calories pushing the greater mass.

Ambitwistor
Your intuition is based on the meanings of words. If the quantity 1/2 mv^2 were called "wakalixes", then you wouldn't have any problem believing that the two objects end up with the same wakalixes: you'd just look at the definition, look at the work-wakalixes theorem, and accept it. The problem is that you have some prior expectation about what the word "energy" means, and physics simply uses that word in way that is different from your expectation. It's not that you have a bad intuition, per se, it's just that the word isn't being used in the way you think it should.

Ambitwistor
Originally posted by boggleface If I'm pushing two different masses over the same distance with the same constant force, imagining that no friction is present on the masses, i am pushing the larger mass for a greater amount of time. Surely i have used more calories pushing the greater mass.

In fact, you have used more calories, but that's because of friction. If you had experience with frictionless surfaces, you'd find that you don't actually have to work harder to move it the same distance.

(Think about astronauts moving large objects in free fall: the objects move very slowly, but the astronauts don't have to exert themselves any harder. It might be hard to imagine, though, since you've never done it yourself.)

boggleface
well if I am in space, pushing 2 different masses with the same canstant force over the same distance, surley i feel that i have exerted myself more pushing the larger mass as i have applied the same force over a longer period of time to bring it to the same distance

Ambitwistor
Originally posted by boggleface well if I am in space, pushing 2 different masses with the same canstant force over the same distance, surley i feel that i have exerted myself more pushing the larger mass as i have applied the same force over a longer period of time to bring it to the same distance

The calories you burn are not proportional to force times time.

Originally posted by boggleface
well if I am in space, pushing 2 different masses with the same canstant force over the same distance, surley i feel that i have exerted myself more pushing the larger mass as i have applied the same force over a longer period of time to bring it to the same distance

"Exerted Myself" is not the same as "did Work", when work is defined as in physics. To make this clearer, let's try a different example. If you push something for a long period of time and it doesn't move, you feel as though you've expended energy. But with the physics definition, you've done no work. None. It may tire your muscles, you may have used up calories, but the work done on the object that did not move, and therefore its change in energy, is Zero. If you really want to account for the energy used by your body in the pushing, you can say that the efficiency of converting the food you ate into energy was Zero percent.

This is not just a thing about definitions. You are wrong if you think that it is force times time that matters in terms of bodily work done. Try pushing your wall with a 50 pound force for a minute. It's not "hard work". In fact you could rig up a stick to do the work for you, but that would be cheating. Now go out and push your car with a 50 lb force for a minute. Assuming the ground is level, it will start moving, and you'll have to maintain the same force in spite of doing it while walking/running. Check your pulse rate afterward. It will be much higher than when you pushed the stationary wall.

Other example: Take a multi-speed bicycle, and put it in a really tall gear. Apply as much force as you can to the pedals for one minute. Your muscles may cramp, but you won't be out of breath. Now choose the correct gear, apply the same amount of force for the same amount of time. By the end of this, you will be going much faster than in the tall gear case, and you will be way out of breath. These examples demonstrate that froce times time is not the relevant quantity to measure energy.

zoobyshoe
Since there is no friction once you get either mass into motion it will stay in motion until something other than friction stops it.

If the acceleration is the same, it will take a larger force to get the larger mass into motion than it will take to get the smaller mass into motion. If the acceleration is different the force will be different according to:

F=ma

Homework Helper
You burn calories for several reasons. The most relavent reason in this case is the energy required to cycle the states of the muscular fibers. For instance, you don't do physics work on a 100 lb. dumbell, by just holding it in your hand with your arm straight out from your body. However, this will certainly make you tired, and cause your muscles to fatigue. Even here, the fatigue is not even primarily caused by the lack of energy, but by a build-up of lactic acid. You will by the time you decide that you can't take any more and drop the dumbell, you have done precisely zero physics work on the dumbell, but, since your metabolism has presumably been functioning properly to keep your overall body alive, and since the muscle fibers have been expanding and contracting in your arm to hold the dumbell in place, you have used many Calories.