Physics Asked on September 1, 2020
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 $vec{F}_{rm lift}=mghat{z}$ is $$W=intlimits_{ell=0}^{ell=h}vec{F}_{rm lift}cdotvec{dell} =mgh$$ using $vec{dell}=dellhat{z}$. But if we lift it with a force $vec{F}’_{rm lift}$ which is greater in magnitude than the downward force of gravity $-mghat{z}$, 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.
Correct answer by gandalf61 on September 1, 2020
The work done is the line integral of the sum of the forces acting on the object. $$W=intlimits_{ell=0}^{ell=h}(vec{F}_{rm lift}+vec{F}_g)cdot dvec{l} =mgh,$$ 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.
Answered by Rob Jeffries on September 1, 2020
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