# Conceptual "Work" Question

1. Oct 20, 2014

### Agent 337737

My question isn't a homework question, it's a general conceptual question. We were going over a "work" problem in class and our professor told us that the work required to lift an object up to a certain height was equal to "mgh". However, is this only the case when we assume that acceleration is zero? Work is equal to force times displacement, and the net force = ma.
Therefore, if we say that the force we apply on an object to lift it up is equal to "mg", that means our net force = mg - mg= 0, which can only happen when acceleration equals 0, which is how we obtain W= (mg)(h). Am I right?

Last edited by a moderator: Oct 20, 2014
2. Oct 20, 2014

### Staff: Mentor

The work is $mgh$ even if the acceleration is not zero.

In a real problem where we're lifting a real object that starts and ends at rest, we have to apply a force slightly greater than $mg$ at the beginning to get the object moving but we also have to apply a force slightly less than $mg$ at the end so that the object decelerates and comes to rest at the end of the lift. The extra force at the beginning means that we do a bit a more work at the beginning, and that is exactly cancelled out by the bit less work that we do at the end.

In an idealized problem, which is what you'll find in most physics textbooks, we assume that both the acceleration from rest to the constant lifting speed at the bottom and the deceleration from the constant lifting speed to zero at the top happen so quickly that negligible distance is covered during these periods so the work done during them is negligible. This is a simplifying assumption that saves us the trouble of having to calculate two quantities that will just cancel each other out anyway.

Last edited: Oct 20, 2014
3. Oct 20, 2014

### Agent 337737

Oh, I see what you're saying! Out of curiosity, how would I go about proving mathematically that this slightly larger initial force and lower final force cancel each other out?

4. Oct 20, 2014

### Agent 337737

Not that I dont believe you, of course. It's just satisfying to be able to prove it to myself.

5. Oct 20, 2014

### Staff: Mentor

You can get there by considering the kinetic energy as well as the potential energy. While the object is moving at a constant speed $v$ during most of the lift, its kinetic energy is also constant and equal to $mv^2/2$. At the top and the bottom of the lift, the speed and hence kinetic energy has to be zero. The "extra" work at the beginning is what is required to increase the object's kinetic energy by that amount, and that's exactly the amount of energy we need to shed at the end of the lift to get the speed back to zero.

6. Oct 20, 2014

### Agent 337737

Thank you so much for your help! It makes sense now!

7. Oct 20, 2014

### robphy

You might consider drawing the force-vs-position graph.