# Calculating work done - Need help with this problem

1. Jun 21, 2009

### ncondon

1. The problem statement, all variables and given/known data
A 12kg box is pulled 7m along a 35 degree incline by a person pulling a rope that makes a 20 degree angle with the incline. The tension in the rope is 100N and the coefficient of kinetic friction between the box and incline is 0.3

Calculate the work done by:
(i) Tension in rope
(ii) Gravity
(iii) Friction
(iv) normal force

2. Relevant equations
Wd = Fcos(angle in degrees) * distance

3. The attempt at a solution
Wd Tension = Tcos(angle) * distance - I'm not sure which angle to use for this one
= 100cos(20) * 7m
Wd gravity = -mgcos(angle) * distance
= - 12kg * 9.8 *cos(35) * 7
wd Friction = coeffiicientKE *mg* cos(angle) * distance
= 0.3 * 12 * 9.8 cos(35) * 7
wd Normal Force = mg cos(angle) * distance
= 12 * 9.8 cos 35 * 7

Is this right. Can anyone verify that it is or make the appropriate corrections please

2. Jun 21, 2009

### dx

First one is correct, the rest are wrong.

The angle θ in the formula W = F⋅cos(θ)⋅d is the angle between the force and the displacement. For example, for the second one, if you draw a picture it's pretty easy to see that the angle between the gravity force and the displacement is 90 + 35, so W = F⋅cos (90 + 35)⋅d = mg⋅(-sin(35))⋅d.

3. Jun 21, 2009

### ncondon

ok thanks for that.

For the WD due to normal force i have worked out as it is negative to the work done due to gravity. Is that correct?

I'm not too sure how to do work done due to friction. I'm pretty sure it's not simply WDfriction = uk * mg * distance or would it be WDfriction = uk * mg* sin 45.

4. Jun 21, 2009

### dx

No. The normal force is perpendicular to the displacement, so θ = 90 and cos(θ) = 0.

What is the direction of the friction force? Why did you choose θ to be 45?

5. Jun 21, 2009

### ncondon

so the work done due to normal foce is 0 then since cos(90) =0?

come to thnk of it the work done due to friction should have been WDfriction = uk * mg * cos(0) therefore it should just simply be WDfriction = uk * mg?

6. Jun 21, 2009

### dx

Yes.

No. What is the magnitude of the normal force? (It's not mg.)

7. Jun 21, 2009

### ncondon

well it wouldn't be the 100N from the tension force. other than that I'm not too sure what the normal force would be since i thought it would be mg. since I can't seem to find other information in the question. would the angle have been 0 but

8. Jun 21, 2009

### ncondon

or is is -mg?

9. Jun 21, 2009

### dx

The normal force is the normal (perpendicular) force exerted by the inclined plane on the block, and is perpendicular to the inclined plane. Find the total force on the block due to gravity and tension, and find it's component perpendicular to the inclined plane. Since the block stays on the inclined plane, the normal force must balance this component, and this condition determines the normal force.

Last edited: Jun 21, 2009
10. Jun 21, 2009

### ncondon

so to balance the forces out we do the equation:

Fgsin(-35) + Tcos(20) + n = 0

11. Jun 21, 2009

### dx

No, how did you get that? The component of the tension force perpendicular to the incline is Tsin(20), and the component of the weight perpendicular to the incline is -mgcos(35).

12. Jun 21, 2009

### ncondon

well i have worked n = mgcos(35) - Tsin(20). so is the frictional force * -1.

13. Jun 21, 2009

### ncondon

oops i meant to say *0.3 then the distance

14. Jun 21, 2009

### dx

Frictional force is normal force times the coefficient of kinetic friction, pointing opposite to the direction of motion.

15. Jun 21, 2009

### ncondon

Frictonal force = 0.3(mgcos(35) - Tsin(20)) * -1

WdFriction = Frictional force * 7m

Is that right

16. Jun 21, 2009

### dx

Well, the magnitude of the frictional force is just 0.3(mgcos(35) - Tsin(20)) without with -1, but you have to multiply by -1 when you calculate the work because it is pointing opposite to the displacement (cos(180) = -1).

17. Jun 21, 2009

### ncondon

so eventually i have come up with the right calculations

18. Jun 21, 2009

Yes.