# Homework Help: Work energy theorem force diagram

1. Oct 29, 2006

### mawalker

I am working on the following problem.

Susan's 13.0 kg baby brother Paul sits on a mat. Susan pulls the mat across the floor using a rope that is angled 30 degrees above the floor. The tension is a constant 31.0 N and the coefficient of friction is 0.190 . Use work and energy to find Paul's speed after being pulled 2.60 m.

So far I have set up my force diagram with my x-axis along the ground. I have normal force pointing up. weight pointing down. tension to the left. and force susan to the right. I'm really not too sure where to go from here though. Any thoughts?

2. Oct 29, 2006

### Hootenanny

Staff Emeritus
Your tension should not be pointing left, it should be inclined at 30 degrees to the horizontal. Your next step should be to find the normal reaction force.

3. Oct 29, 2006

### mawalker

thanks hootenanny. i fixed my force diagram and i calculated the normal force to be 13kg * 9.8 ms = 127.4 N. Am I on the right track?

4. Oct 29, 2006

### Hootenanny

Staff Emeritus
Not quite. Look at all the vertical forces acting, you should have three;
1. Weight of the child
2. Normal Reaction Force
3. ?
What's the third?

5. Oct 29, 2006

### mawalker

the force of friction acting along the incline

6. Oct 30, 2006

### Hootenanny

Staff Emeritus
Not quite, there is no incline, the friction is acting horizontally. In addition to the two forces outline above, you will also have a component of the tension acting vertically (and a component acting horizontally). Can you see why?

7. Nov 1, 2006

### mawalker

yeah i think so, so would the force of tension vertically just be the 31 N * sin (30 degrees) = 15.5 N added to the 127.4 and i get 142.9 N total force. Is this correct?

8. Nov 1, 2006

### OlderDan

Take a careful look at the directions of the vertical forces.

9. Nov 1, 2006

### mawalker

right, the tension force would be pointing vertically downwards so the 15.5 would be subtracted from 127.4 and total force would be 111.9 N?

10. Nov 1, 2006

### OlderDan

The vertical component of the tension acting where the rope is attaced to the mat is upward, in the same direction as the normal. The normal force is upward, opposing the weight.

T_y + N = mg

So yes, the normal force is the weight minus the vertical tension component.

11. Nov 1, 2006

### mawalker

gotcha...i'm still kinda lost on where to go from this point now

12. Nov 1, 2006

### OlderDan

From the normal force you can find the friction, which opposes the horizontal component of the tension in the rope. The resultant of those two is a net horizontal force that accelertes Paul. You can find the acceleration, but the problem asks you to use work and energy to find the speed after a certain distance. What is the net work done by the force in the horizontal direction?

13. Nov 1, 2006

### mawalker

ok so to find the force of friction i took the coefficient of friction .190 times the normal force to give me 21.26 N. The tension in the horizontal direction would be 15.5 N, which would give me 36.7 N as the net work in the horizontal direction.

14. Nov 2, 2006

### OlderDan

That is not the horizontal component of the tension. That is the vertiacl compoonent. When you do find the horizontal component, remember that friction opposes the motion.