# Energy, effort, and mv

1. Jun 30, 2007

### asamaid1

Hello,

I have question regarding this. suppose , a rest object of mass m1is exerted
a force of F for time t1 to give a final velocity vf1, and traveled distance is x1. So kinetic energy would be Ek = W = F * x= (m * V^2)/2.

see the formula for kinetic energy is based on how much work is done on it to get the kinetic energy, and it is formulated as w = F *x = Ek= (m * V^2)/2. Now if the mass increases one can still achieve the same distance travaled by keeping the F same, but now time has to be increased, u have to force it longer than before. Ok if i this way it will be esier to understand:

for m2>m1,f1= f2, x1=x2

W1= W2 but t2 > t1.

MY QUESTION IS IF WE MEASURING THE EFFORT WHEN WE ARE MEASURING ENERGY, THEN IN THE CASE OF m2 WE ARE OBVIOUSLY MISSING SOME EFFORT, BECAUSE WE FORCED A BIGGER MASS LONGER TIME , AND W2 DOES NOT TAKE ANYACOUNT OF THAT, I MEAN HOW IT IS POSSIBLE THAT AFTER SMALLER MASS BEEN FORCED FOR SHORTER TIME GAIN THE SAME ENERGY AS THE BIGGER MASS WHICH WAS FORCED TO THE SAME BUT FOR LONGER TIME, even if the distance travelled is same in the both cases.

But If we consider mv to be the measure of the efort then it is protionate to the time, So mv increases whenever the time increases, so isnt it very logical to use mv to measure the effort? Please could anyone clarify pls .thanx.

2. Jun 30, 2007

### Staff: Mentor

OK. You do some work on an object of fixed mass and thus increase it's KE.

Now you push an object of larger mass with the same force through the same distance. You do the same work and thus achieve the same increase in KE. Sure it takes you longer, since the acceleration is less.

Not sure what your question is or what you mean by "missing some effort". If you are talking in human terms, then you have to push for a longer time--sounds harder to me. But that's biology, not simply a question of the change in energy of the object. After all, pushing a wall requires effort even though you are doing zero work on the wall. To account for the "effort" you need to consider all the biological factors involved--including the repeated contraction/relaxation of your muscle fibers.

Again, what if you push the wall? Still exerting effort, yet no change in momentum.

(I will move this into a separate thread.)

3. Jul 1, 2007

### asamaid1

hello,

R u trying to say that if u put same engine in a car and in a car carrying Lorry and put same amount of fuel, and let the both vehicle go at their full speed they will travevel the same distance before they run out of the fuel, regardless whatever time they take?

4. Jul 1, 2007

### mahoutekiyo

it is very difficult to understand what you are saying. But I think the answer to your car question is, "no". If they could continuously accelerate, then it might be true assuming the drag forces are equal, but it takes more power to sustain velocity of a vehicle with more mass than it does a vehicle with less mass (and therefore more fuel consumption is required by the car with more mass). But cars cannot continuously accelerate at full throttle indefinitely.

I never really understood the relationship between work and kinetic energy.

KE = w
1/2*m*v^2 = F*d
1/2*m*(d/t)^2 = m*a*d
1/2*m*(d^2/t^2) = m*(d/t^2)*d
1/2*m*(d^2/t^2) = m*(d^2/t^2)
everything matches accept for that "1/2"... anyone care to explain?

5. Jul 1, 2007

### Staff: Mentor

You made several errors in your analysis. I assume you are looking at the simple case of an object starting from rest and being uniformly accelerated for a distance "d" by a constant force "F":
(1) The v in the KE formula is the final speed; but d/t is the average speed, which is half the final speed. So: v = 2d/t.
(2) The acceleration equals v^2/(2d), not v^2/d.​
Correct those errors and you'll see that both sides match.

Last edited by a moderator: Apr 22, 2017
6. Jul 1, 2007

### asamaid1

DOC AL Wrote

[[Now you push an object of larger mass with the same force through the same distance. You do the same work and thus achieve the same increase in KE. Sure it takes you longer, since the acceleration is less.]]

Now i m reffering to my first post that :

[[suppose , a rest object of mass m1is exerted a force of F for time t1 to give a final velocity vf1, and traveled distance is x1. So kinetic energy would be Ek = W = F * x= (m * V^2)/2.

the formula for kinetic energy is based on how much work is done on it to get the kinetic energy, and it is formulated as w = F *x = Ek= (m * V^2)/2. Now if the mass increases one can still achieve the same distance travaled by keeping the F same, but now time has to be increased, u have to force it longer than before. Ok if i this way it will be esier to understand:

for m2>m1,f1= f2, x1=x2

W1= W2 but t2 > t1. ]]

Now Here the scenario was that two different mas m1 and m2 (where m2>m1)was forced the same, with a force of F, and the both mass traveled the same distance, but m1 took time t1 and m2 took time t2(t2>t1), now as the force and the distance travelled are same for the both object the kinetic energy for both is same.

So we can say that same work has been done on a small mass and a bigger mass and the energy is converted into kinetic enrgy and they travelled the same distance. So is it not the same scenario as the car scenario? same fuel= same work done on the masses, same distance by the cares = same distance by the car, same force by the car(as the engines are same)= same force on the masses. But we disagree with the later scenario. So should disagre with the former as well. Am i right or wrong?

7. Jul 1, 2007

### Staff: Mentor

Right.

If you make your car scenario identical to the first, then of course they'll agree. If the force exerted on both cars is the same, if the only horizontal force is the accelerating force (no drag), thus if the cars uniformly accelerate throughout the motion, if the engines are equally efficient at their different speeds... lots of ifs here--then of course the fuel consumption will be the same. Why complicate things by introducing the car scenario?

8. Jul 1, 2007

### asamaid1

Because in that case it means that with the same amount of energy u can move a pen one inch and a table one inch as well. With the same amount of energy u can move anything one inch no problem. Sorry , let me ask a very simple question, how this W= E= F.d came about?

And E= 1/2(mv^2) what we can understand from this quantity? suppose from mv we know that any object having mv will be able to bring about same mv amount ofmovement before it stop! but does E say how much movement we can get fropm it?

9. Jul 1, 2007

### asamaid1

Because in that case it means that with the same amount of energy u can move a pen one inch and a table one inch as well. With the same amount of energy u can move anything one inch no problem. Sorry , let me ask a very simple question, how this W= E= F.d came about?

And E= 1/2(mv^2) what we can understand from this quantity? suppose from mv we know that any object having mv will be able to bring about same mv amount ofmovement before it stop! but does E say how much movement we can get fropm it?

10. Jul 1, 2007

### Staff: Mentor

Sure, why not? Imagine a mouse and an elephant both on a frictionless surface. If you exert the same force on each for the same distance they will end up with the same KE. So what? If you exert an unbalanced force on something, of course it will accelerate.

Note that we are merely describing the mechanical work done on the object, not the complex chemical/biological work done in the human body to create the needed force. Pushing that mouse a certain distance will undoubtedly requre less human "effort" and fuel than pushing the elephant through the same distance, since you need to maintain the force for a longer period in the case of the elephant.

Check out the thread I linked in post #5 for various derivations.

Not quite clear what you are asking here, but momentum (mv) and energy (KE) are very different quantities. Both (and more!) are needed to fully describe an interaction. Note that momentum is a vector quantity, but KE is not. Note that for a simple "point" object, knowing one of those quantities allows me to figure out the other--they are not independent.

11. Jul 1, 2007

### asamaid1

Hello,

could u pls prove the conservation of kinetic energy in a colision of two elastic object!

12. Jul 1, 2007

### Staff: Mentor

Not sure what you mean by "prove". The definition of an elastic collision is that KE is conserved.

13. Jul 1, 2007

### asamaid1

But u have to prove first that Ek is conserved!

14. Jul 1, 2007

### Staff: Mentor

Sorry, but we are going in circles. By definition, an elastic collision is one in which KE is conserved. Nothing to prove!

15. Jul 1, 2007

### asamaid1

If 1/2(mv^2) is conserved in a colisionthen we can understand that it can be a candidate for the measure of E. Beacuase Energy must be conserved.

Now Refering to my previous post,Now i want to know what 1/2(mv^2) refers to because we need to measure the energy because we want to know how much movement we can get out of a certain moving object, or how much heat will give me how much movement, so does 1/2(mv^2 ) tells us these things? if not why we are using it for? on the contrary mv always says how much movent u will get from a certain moving object!

16. Jul 1, 2007

### asamaid1

refering to my previous post, why u think the work done should be measured by force muliplied by distance? What is the reason for this?what is the logic behind it?

17. Jul 1, 2007

### Staff: Mentor

Total energy must be conserved, not mechanical energy. Take two lumps of clay moving towards each other at a speed v (with respect to the ground). They collide and stick together. Mechanical energy is not conserved. The combined macroscopic translational KE of both objects goes from mv^2 to zero.
Again, energy is a different concept from momentum. Not sure what you are getting at. Give a specific example of what you are concerned with.

Last edited: Jul 1, 2007
18. Jul 1, 2007

### Staff: Mentor

That's the definition of work! Why the net work done on a point mass equals the KE is a simple derivation--see that other post for details. (That's the work-energy theorem.)

19. Jul 1, 2007

### asamaid1

Well in an elastic colision kinetic energy should be almost conserved.

Now let me tell u what is my concern, if energy it proportional to force than if u eexert force for longer time the object should gain more kinetic energy!

20. Jul 1, 2007

### Staff: Mentor

That's true--by definition.

Energy equals the work done, which is force X distance (assume constant force for simplicity). KE is not proportional to the force unless the distance is held constant. Of course, the longer you keep pushing the same object, the more KE it will have.

Realize that the same force exerted on a more massive object gives you a smaller acceleration: it takes longer to move that same distance. So?