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# Energy, mass and velocity

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I'm struggling with the definition of kinetic energy as given by 1/2.m.v^2. I can get the derivation based on the definition of work, but it is this that I have the problem with.

I'm familiar with work defined as force applied over a distance, but find it really counterintuitive (why not force times time)?

For example, if I'm pushing an object, applying 100N of force to it, am I not expending the same amount of energy to apply that 100N each second regardless of how far that object travels?

From the standard formulae, if I push (in a vacuum, no gravity) an item of 100kg with a force of 100N for 10 seconds it will reach a final velocity of 10ms and cover a distance of 50m. Work [50m * 100N] = Energy [1/2.100Kg.10ms^2]=5,000j.

But if the object weighs 1000kg with a force of 100N for 10 seconds it will reach a final velocity of 1ms and cover a distance of 5m. Work [5m * 100N] = Energy [1/2.1000Kg.1ms^2]=500j.

So why is it that pushing the same object just as hard for the same time uses different amounts of energy?

As I say, it seems incredibly counterintuitive to me. Two thoughts I've had on why this might be the case are:

1) The same amount of energy is used, but different amounts are expended as work (so where does the rest go with my more massove object?).

2) The answer has something to do with applying a constant force to an accelerating body (as the lighter object is accelerating more quickly).

But I'm whistling in the dark here - I really don't get it.

Can anyone help?

Billy.

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A very interesting post Billy, one which I hope someone can answer!

As an extension of this, what happens if you exert a force against an object which won't move (say it needs more force than you can exert to overcome friction for instance). The distance is zero but work is still being done surely as the force is being applied for a period of time?

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10 minutes ago, Stu said:

A very interesting post Billy, one which I hope someone can answer!

As an extension of this, what happens if you exert a force against an object which won't move (say it needs more force than you can exert to overcome friction for instance). The distance is zero but work is still being done surely as the force is being applied for a period of time?

Yes but there will be no kinetic energy as the mass isn't moving.  The time is irrelevant if there is no velocity.

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Cancel that, re-reading it made no sense at all!!

This should answer it clearer.  The Kinetic energy is proportional to the square of its speed, so the faster mass has more kinetic energy.

This equation reveals that the kinetic energy of an object is directly proportional to the square of its speed. That means that for a twofold increase in speed, the kinetic energy will increase by a factor of four. For a threefold increase in speed, the kinetic energy will increase by a factor of nine. And for a fourfold increase in speed, the kinetic energy will increase by a factor of sixteen. The kinetic energy is dependent upon the square of the speed. As it is often said, an equation is not merely a recipe for algebraic problem solving, but also a guide to thinking about the relationship between quantities.

Edited by RayD

I think

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5 minutes ago, RayD said:

I think

Funny, you are where I am!

I'm guessing the energy gets spent within the object doing the pushing perhaps? For instance I get puffed out trying to push a car which goes nowhere. Where is the energy spent there?

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1 minute ago, Stu said:

Funny, you are where I am!

I'm guessing the energy gets spent within the object doing the pushing perhaps? For instance I get puffed out trying to push a car which goes nowhere. Where is the energy spent there?

Lots of grunting a groaning

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3 minutes ago, Stu said:

Funny, you are where I am!

I'm guessing the energy gets spent within the object doing the pushing perhaps? For instance I get puffed out trying to push a car which goes nowhere. Where is the energy spent there?

Quote

No work is done without motion.

It strikes most students as nonsensical when they are told that if they stand still and hold a heavy bag of cement, they are doing no work on the bag. Even if it makes sense mathematically that W=Fd gives zero when d is zero, it seems to violate common sense. You would certainly become tired! The solution is simple. Physicists have taken over the common word “work” and given it a new technical meaning, which is the transfer of energy. The energy of the bag of cement is not changing, and that is what the physicist means by saying no work is done on the bag.

There is a transformation of energy, but it is taking place entirely within your own muscles, which are converting chemical energy into heat. Physiologically, a human muscle is not like a tree limb, which can support a weight indefinitely without the expenditure of energy. Each muscle cell's contraction is generated by zillions of little molecular machines, which take turns supporting the tension. When a particular molecule goes on or off duty, it moves, and since it moves while exerting a force, it is doing work. There is work, but it is work done by one molecule in a muscle cell on another.

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If it were you pushing I suspect the energy would be used by you i.e. blood flow increase, sugar usage, muscle contract, heat etc.

It's a great question as kinetic energy of course is only present when there is velocity, until then it's just plain hard work!

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20 minutes ago, furrysocks2 said:

That all makes sense now!!

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Yes - just holding an object against gravity has you using energy - muscle flexing plus a load mysterious biological stuff.  (I only studied physics and chemistry.)

As for work done, it is indeed force times distance moved parallel to the force.  Time enters the situation when you consider RATE of working,  known as  power.  100N through 10m in 1 second is 1000 joules divided by 1s, i.e. 1 kilowatt.  If that work was done in 10s, the power dissipation would be just 100W.

The "parallel thing": consider a huge steel sphere on a flat smooth surface (no friction).  You could push it around with ease because you are moving it AT RIGHT ANGLES to the Earth's gravitational pull.  But could you raise it even 1mm??  Certainly not.

Doug.

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Ignore me.

Edited by furrysocks2

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This may be a case where non-precise everyday terms are at odds with the precise physics definitions.  The higher inertial mass resists acceleration and therefore has a lower kinetic energy than the light mass after the same force has been applied to both. As you point out, the definition of work done is force times displacement not force times time.

The energy expended by muscles sounds liike a red-herring in this case, though is absolutely true for holding a weight in the hand.

Edited by Bodkin

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The confusion arises because the definition of work done refers to the work done on the object.  Holding a 1kg mass at a constant height of 1m above the ground results in no work being done on the object but you would certainly eventually become tired .  Energy is of course being expended in the muscles in order to hold the mass but this is quite different from the measure of work done on the mass.  Work would be done on the mass if it were now to be raised vertically by say another 1m (force x distance).  Similarly, if the mass were held at 1m and moved horizontally by walking across the room, again no work would be done on the mass - the applied force must be in the same direction as the movement.  Pushing against an immovable object again results in the same conclusion - energy is expended in the body but no work is done on the object.  As Jonathan has suggested above, the confusion arises from the definition.  "Force x time" is a separate property known as Impulse which in turns equates to the change in momentum.

Jim

Edited by saac

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When you push on a wall (a solid one) it moves a very little bit. The constituent elements of the wall will gain a small amount of potential energy due to their new positions. This gain in potential energy is the work done on the wall. The tiredness that you eventually feel in your arm is the consequence of the wall pushing back at you. It is just like compressing a spring. There is no work being done when you and the wall are in equilibrium.

When you push an object that is free to move the force that you feel is inertial (the unwillingness of objects in nature to change their velocity) and the work done on the object manifests itself in the form of kinetic energy.

I hope this helps.

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Thanks for the info all - stillsomewhat confused, though my confusion has now moved on to how both energy and momentum are conserved. I appreciate kinetic energy isn't and that energy can end up as heat etc but I'm interested in how you put all that together to model the real consequences of a collision between two objects of known mass, velocity and (presumably) elasticity.

Can anyone advise on a good textbook covering this stuff that is not pitched at too high a level? I'd describe my maths as basically functional high school level, but maybe a bit rusty. Basic trig and geometry are fine, and simple calculus, but anything above that and I'm struggling.

I'll take a look at the light and matter link (looks good) but I p[refer to work from print if possible.

Cheers,

Billy.

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