# Calculating work (positive or negative)

1. Mar 15, 2009

### okgo

1. The problem statement, all variables and given/known data

Estimate (to the nearest order of magnitude) how much work I must do to pick up a 12 kg box to a height of 1.3m. Does it mattert how I pick up the box (fast or slow, straight up or along a curvy path, etc.)?

2. Relevant equations

W=Fcos(x)S

3. The attempt at a solution

9.8(m/s^2)*12 kg = 117.6N
W=Fs
W=Fcos(angle)s
s=1.3m
If the box was lifted straight up, the angle is zero.
W=117.6cos(90)(1.3m)
W=0

It does not matter if you pick the box up slow or fast.

I'm confused at the curvy part of the question. If I pick up the box at an angle, is it always negative since I'm doing work against gravity?

2. Mar 15, 2009

### LowlyPion

If the angle is 0, why did you use 90?

3. Mar 15, 2009

### okgo

Sorry, previously, I did the problem thinking it was zero, but that's horizontal, not vertical. Since you pick up the box it should be some degree above zero. I did the calculation picking the box straight up.

The question ask what happens if you did a wavy motion which got me confused. ???

4. Mar 15, 2009

### LowlyPion

I think you better resolve the first issue first.

In the work equation, Force times distance, the Cosθ term is to account for the projection of the force in the direction of motion.

If you were calculating work in moving the box horizontally then true enough the work would be 0 if the force applied was perpendicular and none of it was involved in moving it horizontally.

But the question you are dealing with is that the direction of motion is UP. And that is against gravity which is down. And your force is UP. And the angle with respect to that movement is 0. And using θ of 0° would be correct.

5. Mar 15, 2009

### LowlyPion

6. Mar 15, 2009

### okgo

Okay so if I pick the box up at a curvy path, which might mean at an angle, then there would be work done.

But about the speed of the object. Does that mean if I pick up the box twice as fast I would have to do 2*N, therefore 2*W? Or does the equation only depend on the weight of the object?

7. Mar 15, 2009

### LowlyPion

There will be work going into accelerating it - that is true. But ... it comes to rest again. What it took accelerating, it gave back coming to rest didn't it?

That leaves just the vertical distance traveled then.

Now for your curved path consider that at any point if there is a slope that any movement horizontally is 0 work and you are then adding up only all the instantaneous incremental vertical displacements, just as if you lifted it straight up.