# Homework Help: What is the net work done on the box?

1. Jun 30, 2010

### JWSiow

1. The problem statement, all variables and given/known data
A man picks up a 10kg box, and holds it at a height of 1m above the horizontal ground as he walks to a ramp, and then carries the box up the ramp to a height of 4m, and then down the ramp as shown, finally placing it on the ground at F.

Describe the work done at each stage of the motion. What is the net work done on the box?

2. Relevant equations
W=Fdcos$$\theta$$

3. The attempt at a solution
Not really sure with this one.

I said A-->B = no work done. B-->C = work done (don't think that's a good description though!). C-->D = no work done. D-->E = work done (would this be negative (or positive) work? E-->F = no work done.

So, the net work done on the box would = 0.

Last edited: Jun 30, 2010
2. Jun 30, 2010

### n.karthick

Re: Work

Do you mean to say that the box is not displaced at all (i.e., is it lying at the same position at the end)?
Obviously no!

3. Jun 30, 2010

### JWSiow

Re: Work

I don't know, I don't really get the question. In the textbook, it says that you carry something in your hands, you don't do any work on it since the force to carry it is perpendicular to the displacement.

4. Jun 30, 2010

### JWSiow

Re: Work

Anyone? :S I need to know by tomorrow! (have an exam....)

5. Jun 30, 2010

### JWSiow

Re: Work

But in the textbook, it said, "You also do no work on the bag of groceries if you carry it as you walk horizontally across the floor at constant velocity"... which is why I thought no work would be done from A to B, C to D and E to F.

How would I work out the net work done on the box?

6. Jun 30, 2010

### n.karthick

Re: Work

Yeah, your text book is correct. Even though the box is displaced from A to F, the force is at right angles to direction of displacement.
So the work done is zero as you mentioned.

7. Jun 30, 2010

### JWSiow

Re: Work

Ok, thanks! That makes me feel much more relaxed! (cos it seemed like something that could come up in the exam) :)

8. Jun 30, 2010

### RoyalCat

Re: Work

But since you've not changed the kinetic energy of the box, nor its potential energy, it makes no difference to the energy where in the plane of reference you've placed it.

9. Jun 30, 2010

### n.karthick

Re: Work

In real sense, however it is not possible to displace a box without doing work even at constant velocity because friction comes in to picture.