Work: Acceleration Doesn't Matter?

[justadream]

Mcat-review says

"Pushing an object at constant speed up a frictionless inclined plane involves the same amount of work as directly lifting the same object to the same height at constant speed."

If you pushed an object at non-constant speed (for example, use a rocket), shouldn't the amount of work be the same too?

The power would be different (P = W/t and the time would be less so the power would be greater).

Can someone confirm?

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[Cawolf]

Yes.

Work done against a conservative force (ie. gravity, spring, electric, magnetic) is path independent.

You are correct to state that the work done while accelerating would require more power, as the object is moved is less time.

 

[Chrisz]

Acceleration does make a difference about the amount of the work done. At a constant speed, your force is only necessary to counter the gravitational force. So the amount of work done is equal to increase in potential energy. If you applying a force that is more than necessary, your will accelerate the object and the object will gain KE at the same time it increases in potential energy.

Conservation of Energy.

In the first case (constant speed)
W=dPE
In the second case, (acceleration)
W=dPE+dKE

Power will be greater in the second case, since it will require more energy (PE+KE), assuming the hill is the same in both cases

 

[Cawolf]

Acceleration does make a difference about the amount of the work done. At a constant speed, your force is only necessary to counter the gravitational force. So the amount of work done is equal to increase in potential energy. If you applying a force that is more than necessary, your will accelerate the object and the object will gain KE at the same time it increases in potential energy.

Conservation of Energy.

In the first case (constant speed)
W=dPE
In the second case, (acceleration)
W=dPE+dKE

Power will be greater in the second case, since it will require more energy (PE+KE), assuming the hill is the same in both cases

I see what you are saying, but I think your formulas assume that the object has some kinetic energy at the "top of the hill". If we are saying the object is moved a fixed distance, then it will have no kinetic energy at the end of movement. So I don't think you can say that more work is done in the case the OP mentioned.

 

[Chrisz]

I think I should add one more thing to clarify @Cawolf 's misconception about path independence. The potential energy in a conservative system is independent, not that work done is independent, because you always can exert extra force to increase KE at the same time you increase PE.

 

[Chrisz]

I see what you are saying, but I think your formulas assume that the object has some kinetic energy at the "top of the hill". If we are saying the object is moved a fixed distance, then it will have no kinetic energy at the end of movement. So I don't think you can say that more work is done in the case the OP mentioned.

In this case, the final dKE has to be zero. merely saying moving along a fixed distance does not mean that the object is moving at the same speed at t=2, if you accelerate or decelerate at constant rate, the final velocity would be different, the work would be different. The only condition in a nonuniform acceleration pathway that the object moves along a hill is that the final velocity=initial velocity. This would mean that the object is accelerated at some instant and decelerated at some instant such that the final velocity is still equal to the initial velocity. If the final velocity is not defined, you cant make a definite conclusion.