Calculating work (positive or negative)

  • Thread starter okgo
  • Start date
  • #1
61
0

Homework Statement



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.)?

Homework Equations



W=Fcos(x)S

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?
 

Answers and Replies

  • #2
LowlyPion
Homework Helper
3,090
4
If the box was lifted straight up, the angle is zero.
W=117.6cos(90)(1.3m)
W=0
If the angle is 0, why did you use 90?
 
  • #3
61
0
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
LowlyPion
Homework Helper
3,090
4
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. ???
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.
 
  • #6
61
0
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
LowlyPion
Homework Helper
3,090
4
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?
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.
 

Related Threads on Calculating work (positive or negative)

  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
4
Views
8K
Replies
3
Views
7K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
4
Views
13K
Replies
1
Views
1K
  • Last Post
Replies
5
Views
22K
  • Last Post
Replies
2
Views
2K
Replies
6
Views
918
  • Last Post
Replies
5
Views
23K
Top