# Homework Help: Work done against gravity

1. Jun 15, 2007

### Mr Virtual

Hi all

The question I am going to ask is such that, had I asked it from my friends, they would have laughed at me. Instead I am asking you (atleast I won't have to see you laughing, whether you reply or not ).

We all learnt in junior classes that work done in lifting an object of mass m against gravity, to a height h = mgh
In higher standards, we learnt that opposite forces cancel each other.

Accordingly, if you have to lift an object of 80 N weight (mg =80 N) upto 1m height (h), you will have to apply a force of 80 N, thus doing 80 J work (=mgh).
But the thing I do not understand is, if 80 N applied by me is cancelled by
80 N weight of the object, how is it going to move. And if it is not going to move, how can we do 80 J work.

Where am I wrong? I know textbooks cannot be wrong. But where am I missing it?
So keep laughing, but do answer my question if it is worth it (or even if it is not !!).

Thanks
Mr V

2. Jun 15, 2007

### Chi Meson

THe key is understanding that the "net work" done over the move is zero.

When first starting to lift the object, you have to exert a force that is equal to the weight of the object PLUS a little more force to accelerate it slightly. Then, as you continue to lift the object at (more or less) a constant speed, the net force must be zero. Then as the object comes to a stop at the top, you decrease your force so that gravity is slightly more and the net force is now downward; again the net force causes an acceleration that decreases the speed, and the object stops.

Integrated over the entire move, the work done by you exactly equals the negative work done by gravity.

And sometimes textbooks are wrong. And this is a very good question, one that shows an acute appreciation of this concept. Look into the proper definition of the "Work-Energy Theorem" which says the net work done on an object equals the change in KE of the object.

3. Jun 15, 2007

### f(x)

In the case you mentioned, work done by gravity = -80 J
Work done by you = 80 J
If you consider both forces, consider the net work, which as expected, is 0

4. Jun 15, 2007

### Astronuc

Staff Emeritus
Last edited: Jun 15, 2007
5. Jun 15, 2007

### Mr Virtual

Thanks a lot to all of you for your helpful answers. So you didn't laugh after all. Thank god !!!

regards
Mr V

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