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

Question

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 $z=h$?

Answer 1

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.

Answer 2

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.