# Work done in lifting a mass vs work done by mass as it falls to starting position

There are variations of this question on here but I was unable to find exactly what I was looking for so please bear with me if this is old hat.

If I start with a mass m at rest at height H1 and raise it through a height of h to finishing height H2 what force is required? I often see the simple answer that the upward force required to raise the mass m at constant speed is simply mg, and that the work done with be mgh - equal to the gain in gravitational potential energy of the mass in rising from H1 to H2. The problem I have with this is that it seems to ignore the fact that I have to accelerate the mass from rest - if I simply apply an upward force mg to a resting mass m will it not simply remain at rest support by that force?

It therefore seems I must apply a force somewhat greater than mg, call it (F+mg, to get the mass moving upward. This then seems to imply that the work done on the mass is (F+mg)h which is greater than the increase in gravitational potential energy mgh. What has happened to the extra energy transferred ie Fh?

If at height H2 the mass is placed on a plunger which then falls (with some resistance) under the weight of the mass m as it returns from height H2 to H1, what is the work done by the mass? It would seem to be mgh - ie the change in gravitational potential energy as it falls through height h (the extra energy Fh is still missing). However, given the plunger is offering some resistance to the fall of the mass it seems it must be applying an upward force to the mass which is something less than the weight mg of the mass (otherwise it wouldn't fall). But doesn't the principle of action and reaction then mean that the force applied to the plunger by the mass is similarly less than its weight mg? I guess not - otherwise the work done on the plunger by the mass as it falls would be less than mgh and I would have some more missing energy to deal with!

Could you (gently) show me where I am going wrong - or if by some miracle I am on the right track, help me find the "missing" energy?

• Work is concerned with net force, so when you apply F+mg, and gravity applies -mg, the net force moving the object is simply F, and energy is still conserved. Jan 15, 2019 at 19:25
• Also, work is done by forces, not masses, so when the mass falls the gravitational force is doing work, not the mass. Edit: Unless you're talking about the force on the plunger from the mass, in which case you're right, my mistake. Jan 15, 2019 at 19:27

The problem I have with this is that it seems to ignore the fact that I have to accelerate the mass from rest - if I simply apply an upward force mg to a resting mass m will it not simply remain at rest support by that force?

It therefore seems I must apply a force somewhat greater than mg, call it (F+mg, to get the mass moving upward. This then seems to imply that the work done on the mass is (F+mg)h which is greater than the increase in gravitational potential energy mgh.

While correct at one level, there is no particular speed necessary. In principle, we can make this speed (and the extra force) arbitrarily small. It can be made so close to zero that it can be ignored (just by moving it slowly).

What has happened to the extra energy transferred ie Fh?

In the normal course of events, nothing. It's really hard to get anything like a "constant force". What happens instead is that the force is slightly greater for a period, then reduced to exactly $$mg$$. As the forces are equal, the mass continues to raise at a constant speed. Only at the beginning was the force greater, and all that extra energy went into the kinetic energy of the object. At the end, the situation is reversed and the force is lowered so that the object comes to a stop. The energy is returned.

If the force were greater during the entire lift, the object would be accelerating the entire way. The extra energy in all cases goes into the kinetic energy of the object.

If I start with a mass m at rest at height H1 and raise it through a height of h to finishing height H2 what force is required? I often see the simple answer that the upward force required to raise the mass m at constant speed is simply mg, and that the work done with be mgh - equal to the gain in gravitational potential energy of the mass in rising from H1 to H2. The problem I have with this is that it seems to ignore the fact that I have to accelerate the mass from rest - if I simply apply an upward force mg to a resting mass m will it not simply remain at rest support by that force?

So you are mostly correct. If the mass starts at rest, then yes you will need to exert a force that is larger than $$mg$$ to get it moving. Then if you want it to then move at a constant speed, you will have to reduce your force to $$mg$$. So while the mass is moving at a constant speed after you accelerated it you have done more work than gravity has. This makes sense, since the net work done on the object is $$W_{net}=W_{me}-mgh=\Delta K>0$$ However, if the mass then comes to rest at $$H_2$$, then during the time of deceleration gravity does a larger magnitude of work than you do. The overall result is that you and gravity have done the same amount of work going from $$H_1$$ to $$H_2$$ $$W_{net}=W_{me}-mgh=\Delta K=0$$

It therefore seems I must apply a force somewhat greater than mg, call it (F+mg, to get the mass moving upward. This then seems to imply that the work done on the mass is (F+mg)h which is greater than the increase in gravitational potential energy mgh. What has happened to the extra energy transferred ie Fh?

As we have said your force analysis is mostly correct. The extra work you have done goes into the kinetic energy of the mass. But if you keep your force at $$F+mg$$ then the object will continue to accelerate as you do more work than gravity and add more and more to the mass's kinetic energy.

Hopefully this will help in your analysis of the second scenario.

• Thanks Aaron, very helpful. As I write this there is still something about the plunger bit that puzzles me - but armed with what I have learnt I am going to ponder on that some more to see if I can work it out for myself. Jan 16, 2019 at 21:36
• @iMatt66 Glad I could help! If the answer is good enough, please consider giving it an upvote. Also, there are some pretty good answers here. It would be helpful for you to select one of them as the accepted answer by clicking on the check mark next to the answer. That way future readers can know what is the "correct" answer. Jan 16, 2019 at 23:06

In oder to move the mass from point 1 to point 2 from rest, clearly the mass needs to accelerate, which then of course gives it kinetic energy in addition to potential energy. This "extra energy" is eliminated by causing the mass to decelerate prior to reaching point 2 so that it comes to rest at point 2.

To get things going I need to apply an external upward force, $$F_{ext}$$, slightly greater than $$mg$$, to give it small upward acceleration $$a$$. Let’s say I do this for a brief time $$dt$$ and therefore over a short distance $$dh$$. The mass thereby attains a small velocity $$v=adt$$ and a small increase in kinetic energy of $$½ m(adt)^2$$ and a small increase in the potential energy of the mass m of $$(mg) dh$$. So the differential amount of work done on the mass during the period $$dt$$ is

$$dW = ½ m(adt)^2 + (mg) dh$$

We now immediately reduce our upward force so that it equals the downward gravitational force. The mass is now rising at constant velocity v and so there is no subsequent change in KE, however its potential energy keeps increasing.

Prior to reaching point 2 we need to bring the mass to a stop. I reduce my external upward force, $$F_{ext}$$, to slightly less than $$mg$$, to give it small acceleration downward. I do this for sufficient time to bring the mass to rest at point 2. Gravity now does a small amount of work downward resulting in a negative change in kinetic energy negating the positive change in kinetic energy that was necessary to start raising the mass from point 1:

$$dW =-½ m (adt)^2$$

The end result in going from 1 to 2 is an increase in potential energy of

$$\int_1^2 (mg)dh = (mg)h$$.

Hope this helps.

• Very helpful Bob, thank you very much for taking the time to help me out. Jan 16, 2019 at 21:41
• @Bob D I know this is an old post, but it was very helpful, well-done mate. Jan 24, 2021 at 16:03

Work done in lifting an object is normally done via conservative forces, and so is path independent, and independent of any large or small accelerations required to get the object moving or slowed to a stop at the end of the movement. Of course, the rate that work is done, or the power required to move the given object through any given path, is dependent on the accelerations involved.

If you choose the mass as your system you should not think in gravitational potential energy terms.

The mass has a downward external force on it $$mg$$ due to the gravitational attraction of the Earth.
If the mass starts from rest then to get the mass moving upwards there must be an upward external force on the mass greater than $$mg$$ let’s call it $$mg + F$$.

The net upward external force on the mass is $$F$$.
If the mass rises a height $$h$$ the work done on the mass is $$Fh$$ and by the work-energy theorem this is the gain in kinetic energy of the mass ie at the end of its journey the mass will be moving upwards.

To get the mass to be at rest at the end of its journey proceed as follows (there are in fact an infinite number of ways of doing this).
Arrange for the net upward external force on the mass for half its journey to be $$F$$ and then for the other half of its journey arrange for the net downward force on the mass to be $$F$$.
The total work done on the mass is $$F\frac h2 -F \frac h2 =0$$ and so at the end of its journey rising a total distance $$h$$ there is no change in the kinetic energy of the mass - the mass starts and finishes at rest.

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Now I am struggling as to the purpose of the plunger but it might be a way of applying an upward force of $$mg +F$$ on the mass on its descent from $$H2$$ to $$H1$$?
Whilst not in contact with the plunger the mass overshoots height $$H2$$ reaches a maximum height when has no kinetic energy and the returns to height $$H2$$ having the same kinetic energy that it had before $$Fh$$ but now moving downward.

The plunger now exerts a upward force $$mg+F$$ on the mass so there is a net upward external force $$F$$ on the mass.
In falling a distance $$h$$ that external force does $$-Fh$$ amount of work and thus by the work-energy theorem the mass loses kinetic energy $$Fh$$ and finishes up at rest.

• Thank you very much for your help on this - I have found all the answers very illuminating. As to the plunger: Jan 16, 2019 at 21:26
• Thank you very much - as you say the plunger is there to apply some upward force to the mass. Armed with what I have learned from the answers to my question I should be able to work out exactly what is happening as the mass and plunger descend together - and will have fun doing so. Jan 16, 2019 at 21:39