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Ok, taking the equation W=FD. Say a 30N force is acting on a 10kg object over 10s, causing it to move 150 metres over a frictionless surface. The work done by this force will be 30(150)J. However, if two 30N forces are placed on the same object, it will move 300 metres over the same time. But this indicates that the work done by each 30N force is 30(300)J, which is double if it was pushing alone. This doesn't make sense to me.

Furthermore, what is the displacement taken relative to? If we take the displacement relative to the CMB then across that 10 second interval the earth would have moved 6270000M and the work done will be 30(6270000)=a ridiculously high number

So the question boils down to what displacement value do I take for the equation W=fd?

The acceleration should be constant.

Qmechanic
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ReveaZapcre
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3 Answers3

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it will move 300 metres over the same time. But this indicates that the work done by each 30N force is 30(300)J, which is double if it was pushing alone. This doesn't make sense to me.

"Constant forces" are easy to calculate in physics classes, but they are unusual in real life, so intuition may not help as much with understanding. Imagine standing next to a merry-go-round that is stopped. Do you think you can push it with a 100N force? Probably without much difficulty. Now make it spin pretty fast. As the bars go by, do you think you can apply 100N to them? Probably not, and if you could it would take significantly more effort.

So it takes more power to apply a constant force to an object moving in the same direction, and over the same period of time, more energy is transferred.

If we take the displacement relative to the CMB then across that 10 second interval the earth would have moved 6270000M and the work done will be 30(6270000)=a ridiculously high number.

And yet, quite true. The kinetic energy of an object depends on the frame. Where did that energy come from? Whatever was pushing it must have been slowed down a bit and its KE in that frame dropped by almost the same amount.

Similar: Work done walking on moving train

BowlOfRed
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Displacement is of the object under force. If a 30N force acts on a 10Kg object for 10 sec displacement is 150 m. In case of 60N (or 2x30N) the net displacement is 300 m.

  • Now Force is a vector quantity so a 30N force does displacement of 150 m only, so if 2x the 30N force means double the displacement i.e. 300 m
  • You have to find the displacement of net force which can be done by 2 ways:

add up the displacement done by 2 force individually

or find net force and then find displacement using equation of motion

so

But this indicates that the work done by each 30N force is 30(300)J, which is double if it was pushing alone.

This is not true as 300 m is displacement of 60N force not a 30N one.

Universe A7
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what is the displacement taken relative to?

The displacement can be taken relative to any inertial frame. Work is a frame-dependent quantity.

It is important to know that although work is frame-dependent energy is conserved in all inertial frames.

But this indicates that the work done by each 30N force is 30(300)J, which is double if it was pushing alone. This doesn't make sense to me.

Your analysis is correct. It does indeed take more energy to exert the same force at a higher speed over the same time. If it were not so then energy would not be conserved.

Dale
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