# Confusion about the workdone in lifting an object in the constant gravitational field of the Earth

If we lift an object of mass $$m$$ from the ground at $$z=0$$ to height $$z=h$$ without acceleration, the lifting force must equal and opposite to the downward force of gravity. The work done by the lifting force is $$\vec{F}_{\rm lift}=mg\hat{z}$$ is $$W=\int\limits_{\ell=0}^{\ell=h}\vec{F}_{\rm lift}\cdot\vec{d\ell} =mgh$$ using $$\vec{d\ell}=d\ell\hat{z}$$. But if we lift it with a force $$\vec{F}'_{\rm lift}$$ which is greater in magnitude than the downward force of gravity $$-mg\hat{z}$$, will the workdone still be the same? If not, what happens to the "extra work" when the body is brought to rest at $$z=h$$?

You can apply a non-constant force to lift the object, but if you lift it in such a way that it is at rest when $$z=h$$ then the work done against gravity is always $$mgh$$ - there is no “extra work”.

You can see this intuitively by imaging you lift the object with a force greater than $$mg$$ for the first part of the lift. Then if the object comes to rest at $$z=h$$ the force during the second part of the lift must be less than $$mg$$.

The reason for this is that gravity is a conservative force - the work done by or against gravity depends only on the initial and final configuration of a system (where configuration includes the positions and moments of all objects) and not on the route taken from start to finish.

Of course, you could lift the object with a force that was greater than $$mg$$ throughout the lift - but in that case it would not be at rest at $$z=h$$, and the extra work done over and above $$mgh$$ is accounted for in the non-zero final kinetic energy of the object.

• I don't need a nonconstant force. I am applying a constant lifting force that is greater than the downward force of gravity. Aug 21, 2020 at 14:19
• @mithusengupta123 In that case the object is accelerating throughout the lift and so it cannot be at rest at $z=h$. Aug 21, 2020 at 14:21
• I have a problem with this. What stops you to bring your hand instantaneously to rest? Aug 21, 2020 at 14:24
• @mithusengupta123 When you suddenly bring your moving hand to rest, you apply force on your hands to be able to do so. Since the object you are lifting and your hand are connected bodies, this force also gets applied to the object, doing negative work on it and bringing it to rest Aug 21, 2020 at 14:26
• @mithusengupta123 If you are holding onto the object then you have exerted a large downwards force on it and this creates a negative term in your work integral. If you are not holding onto the object then it does not come to rest because of its inertia. Aug 21, 2020 at 14:28

The work done is the line integral of the sum of the forces acting on the object. $$W=\int\limits_{\ell=0}^{\ell=h}(\vec{F}_{\rm lift}+\vec{F}_g)\cdot d\vec{l} =mgh,$$ where $$\vec{F}_g$$ is in the opposite direction to $$\vec{F}_{\rm lift}$$.

The only way to can get the object to move upwards is if the magnitudes $$F_{\rm lift} > F_g$$.

However, if you want it to stop at $$h$$, then at some point in the second phase of the motion then the magnitudes $$F_{\rm lift} < F_g$$.

If you apply a constant lifting force, then of course the body will not be at rest when it reaches $$h$$. You will have done more work and the extra work will be accounted for by the kinetic energy of the body.

I do not understand what you mean by "bring it to rest". That implies exerting a force in the opposite direction to gravity. i.e. reducing the magnitude of $$F_{\rm lift}$$ to a negative value, which will jusr reduce the time duration of the necessary second phase beyond that if the decelerating force is provided by gravity alone.

• I can hold the object in my hand and no matter how fast I lift it, I think I can stop my hand instantaneously at $z=h$? Aug 21, 2020 at 14:11
• @mithusengupta123 In which case you have done exactly the same amount of work. You said the body is "brought to rest". If you are just using gravity to do that, well gravity is a conservative force; the path taken to get from A to B does not matter. Aug 21, 2020 at 14:12
• There is a clash with path-independence and that's exactly what I am hinting at. What stops you to lift the body with a force greater than the downward force of gravity and bring it to rest at $z=h$? At one hand, it seems we are doing a work greater than $mgh$ on the other hand path independence implies the work done must be $mgh$. Aug 21, 2020 at 14:18
• @mithusengupta123 Note that the path in “path independence” is the path in configuration space, not in physical space. Aug 21, 2020 at 14:24