# Why is work done when lifting an object with a constant velocity = weight times height?

While calculating work done when lifting an object, we take acceleration = gravity. Why is that? Initially, the object is at rest. When we balance out the gravitational force for the object by applying an upward force, shouldn't the object still remain at rest because it takes no force to maintain the state of motion or rest. The way I interpret it, we should have to apply a force > weight of an object to lift it but then we can reduce it to become equal to the weight of the object so that we can maintain the upward velocity.

• It's the same reason why moving an object horizontally (without friction, from rest, to rest) is zero work. The acceleration and deceleration cancel each other out.
– SF.
Nov 8, 2021 at 13:09
• I assume you're asking about the work done by the force of the person lifting the object, not the net work done on it. Nov 8, 2021 at 13:26

You are correct.

$$W=mgh$$ is also correct, but it brushes something under the rug. It ignores the force required to accelerate an object from rest, and it ignores the opposite force that slows the object to a stop. In between speeding up and slowing down the velocity is constant which, as you point out, implies the net force is zero. The lifting force magnitude is the same as that of gravity, and it is clear that $$W=mgh$$ during that interval. So what about starting and stopping? The extra vertical work needed to accelerate the object is balanced by the reduced vertical work needed to bring it to a stop. So in the end, $$W=mgh$$.

• To stop the object you can remove the force you apply and let gravity bring the body to rest. Of course then you need to apply a force to maintain the body at rest. Nov 8, 2021 at 14:31

For a very small change in the kinetic energy of the body, the total work done (by you and by gravity) is zero, by the definition of work in classical mechanics (change in kinetic energy due to the net force acting over a distance). The work done by gravity is typically considered as the negative of the change in gravitational potential energy, $$- mgh$$, and the average work done by you is $$Fh$$ where $$F$$ is the average force you exert. For very small change in kinetic energy, $$F \approx mg$$ and $$Fh -mgh \approx 0$$. So the total work done is $$0$$ and the work done by you is $$mgh$$. See the response by @garyp for how the force you exert first causes an increase in kinetic energy of the body and later a decrease. Most discussions in basic physics texts assume a very slow movement of the body due to you exerting a constant force infinitesimally greater than the force of gravity.

Regarding your question, we do not take acceleration = gravity. For slow movement the acceleration is $$0$$ since the force you apply essentially equals the negative of the force of gravity. If you accelerate then later decelerate the body the acceleration changes from positive to negative.

In reality you are right but in theoretical physics we do not account time taken in the displacement of the object. So if you want to lift an object you apply force little greater than the weight but we neglect the very small extra force and just write the force equal to the weight. And that's the reason why work done by you when lifting an object is weight times height.
Feel free to ask for any further information or doubt.
Hope it helps. :)

Weight times height is the work done by gravity (with the proper sign). The work done by any other force can be anything, depending on how that force act and what is the motion of the object.