# Net work done on the backpack

1. Nov 13, 2007

### MIA6

1. The problem statement, all variables and given/known data
a) Determine the work a hiker must do on a 15.0-kg backpack to carry it up a hill of height h=10.0 m b) determine the work done by gravity on the backpack, and c)the net work done on the backpack. Assume the motion is at constant velocity.

I don't understand why the question asks you the work done by hiker and done by gravity. What's the difference? Is that the work done by hiker is in the direction of the motion, i mean the slope. However, the work done by gravity is in the direction of the altitude of the hill, height? On the book, the diagram is like a backpack, and then the force which points upward is the force by Hiker, the force points downward is the gravity. Hope you can clarify, thanks.

2. Nov 13, 2007

### XJellieBX

Gravity doesn't actually do any work on the backpack because it is not moving downwards and you can't have a negative value for work. Work is only done when the object moves in the direction of the force being applied to it. Hope that helps.

3. Nov 13, 2007

### MIA6

On my book, it finds the component of gravity which acts in the direction of motion, so it's not direct gravity, but the component.

4. Nov 13, 2007

### PhanthomJay

No, that's not right, the gravity force does do work, and since its direction is downward, and the backpack is moving upward, it does negative work. The hiker does positive work, since she must apply an upward lifting force on the backpack in the same direction as its vertical motion. Since the movement is at constant speed, what does this say about the relationship betwen the 2 forces and work done by each?

5. Nov 13, 2007

### MIA6

If I need to find the work done by hiker, W=Fd, so force is the upward force, but I think this force should be the component of the force exerted by the hiker, since the hiker climbs a slope which is some degree angle inclined to the ground; she is not climbing vertically. How about the force exerted by the gravity? Should I find the component of it, too? Why it is not in the direction of the height which is vertical but also that incline?

6. Nov 13, 2007

### PhanthomJay

remember that work is not Fd it is Fdcostheta, that is, work is force times distance in the direction of the force. The gravity force is a conservative force such that the work done by gravity is independent of the path taken, so you just need to know the height in both cases.

7. Nov 13, 2007

### MIA6

I know that a conservative force is independent of path it takes, but it starts down the ramp, and finishes up the ramp, so it's not the vertical distance, but between these two points?

8. Nov 14, 2007

### PhanthomJay

Lets assume the ramp is sloped at 30 degrees with the horizontal (60 degrees with the vertical). The box, which weighs 150N, is raised to a height of 10m. This means that along the direction of the incline, it travels 20m, using basic trig. Since work is Fdcostheta, the work done by gravity is (-150)(20)(cos60) = -1500J. OR more simply, the work by gravity is just weight times vertical distance = (-150)(10) = -1500J.
Similarly, the work done by the person against gravity can be calculated the same way. Note that since this is a constant velocity problem, the net work done by both must be zero, per the work energy theorem.