1. Dec 5, 2009

### Dark Visitor

There are multiple questions all related to the same problem. I want to make sure I get them all correct, so any help would be much appreciated. I need this by tonight.

A box slides at a constant speed down a rough inclined plane whose angle is 30 degrees above the horizontal. There are no ropes or outside agents pushing or pulling on it.

1. The work done by the normal force is:
* positive
* negative
* zero
* not enough information

2. The work done by friction is:
* positive
* negative
* zero
* not enough information

3. The work done by the weight is:
* positive
* negative
* zero
* not enough information

4. The total work (or net work) done on the box is:
* positive
* negative
* zero
* not enough information

My thoughts are that the work done by the normal force is positive, the work done by weight is negative, and the work done by friction is negative, which leaves total work to be negative. Please tell me if I am wrong, and where I am wrong.

2. Dec 5, 2009

### kuruman

What do you base your thoughts on? Remember that the work done by a constant force is the product of three things

W = F d cosθ

F is the magnitude of the force (always positive or zero)
d is the magnitude of the displacement (always positive or zero)
cosθ is the cosine of the angle between the force and displacement vectors (sometimes positive, sometimes negative, sometimes zero)

Moral: The algebraic sign of the work is the algebraic sign of the cosine.

3. Dec 5, 2009

### Dark Visitor

I drew a picture of a box, like the problem stated, with it on an incline, and drew my forces (weight, normal, friction) on it.

I know that W = F*d, and for this problem, W = F*d*cos30, but we don't have any numbers to apply here. So how would I apply that?

4. Dec 5, 2009

### kuruman

Look at your diagram and then at the first question. What is the cosine of the angle between the normal force and the displacement?

5. Dec 5, 2009

### Dark Visitor

30 degrees? Because the normal force is perpendicular to the plane the box is sliding on.

Or would it be 90 + 30 degrees = 120 degrees?

6. Dec 5, 2009

### kuruman

In what direction is the displacement vector?

7. Dec 5, 2009

### Dark Visitor

Perpendicular to the plane the box is on, which is up, which made me think positive.

8. Dec 5, 2009

### kuruman

Does the box move move in a direction perpendicular to the plane that it is on? Try again.

9. Dec 5, 2009

### Dark Visitor

No, the box moves straight down the plane. But normal always points perpendicular to the plane the object is on.

10. Dec 5, 2009

### kuruman

Yes, normal is perpendicular to the plane. My question is what is the angle between the normal force (that we know is perpendicular to the plane) and the displacement of the box?

Last edited: Dec 5, 2009
11. Dec 5, 2009

### Dark Visitor

If I am thinking this right, I say 90 degrees? But I feel confused...

12. Dec 5, 2009

### kuruman

It is 90 degrees, you are correct. So the direction of motion is perpendicular to the normal force. What is the cosine of 90 degrees?

13. Dec 5, 2009

### Dark Visitor

Zero. So the normal force does no work?

14. Dec 5, 2009

### kuruman

Yup. Whenever a force acts in a direction perpendicular to the displacement, the work done of the force is zero. Try to remember that.

One down three to go.

15. Dec 5, 2009

### Dark Visitor

Okay, I will try my best.

Well, I still think weight is negative, and friction is negative. Are either of those right?

16. Dec 5, 2009

### kuruman

Remember that the cosine of an angle is positive when the angle is less than 90 degrees, zero when the angle is 90 degrees and negative when the angle is greater than 90 degrees. Use this information and your diagram and you should be OK. I have to sign off for today. Good luck.