# Negative work on a moving body

1. Nov 5, 2014

### alba

Suppose the box (stone, bullet..., $$m =1$$) is not moving upward because of a lift, but has been shot and has $$v = 20m/s$$ and $$KE = 200 J$$

Gravity is doing negative work and subtracting energy:
$$F = -ma = 10 \cdot 1 = -10 N$$
We cannot use the palallelogram here as KE is not a force, how do you deal with this situation, how do you describe it mathematically?

It is trivial that the box will reach the height KE/ F = 20 m and that there KE will be 0 and PE = 200, and at any height h $KE_h = 200 -PE_h$, but how do you describe the relation, process, the action of negative work done on the body?
The same problem, I suppose, arise if we try to describe the flight of an arrow, on which drag does negative work.

Last edited by a moderator: Nov 5, 2014
2. Nov 5, 2014

### Staff: Mentor

You can describe the action of the forces (gravity in your example) either dynamically (using Newton's 2nd law) or energetically (using work done). You'll get the same answer either way.

Dynamically, since the only force acting is gravity you know the resulting acceleration is -g. Using work, F*d is negative since the force (down) and displacement (up) are in opposite directions.

3. Nov 5, 2014

### alba

Thank you, sir, but it is rather hard for me to understand that, in general terms. Can you refer to the concrete example above? In the case of forces we use the parallelogram and subtract ( if upward F were 15 N we would know = + 5N) in this case how do you descrbe mathematically this opposition? Can we only use the makeshift of mechanical energy?

4. Nov 5, 2014

### Staff: Mentor

By 'parallelogram', I assume you mean vector addition to find the net force. In your example you only have one force, gravity. F = -mg