# Homework Help: Using force equations and kinematic equations

1. Sep 29, 2007

### aligass2004

1. The problem statement, all variables and given/known data

A 1.0kg wood block is launched up a wooden ramp that is inclined at a 38 degree angle. The block's initial speed is 14 m/s. (Use uk = .20 for the coefficient of kinetic friction for wood on wood.) a.) What vertical height does the block reach above its starting point? b.) What speed does it have when it slides back down to its starting point?

2. Relevant equations

F = ma
Frictional force = uk(n)
kinematic equations

3. The attempt at a solution
I broke the weight into its components. Then I found the forces for each direction. I then solved for the normal force, and got 59.249N. I then substituted that in for the equation in the x-direction to find the acceleration. I got -67.179 m/s^2. I then tried using Vxf^2 = Vxi^2 + 2(ax)(delta x).

2. Sep 29, 2007

### rootX

I cannot see friction in your equations(or method), and normal force cannot be 59 N. It is much smaller

3. Sep 29, 2007

### aligass2004

Normal - Wy = m(ay) or Normal - mgcos (theta) = m(ay) right?

4. Sep 29, 2007

### rootX

m(ay) is 0
so
N = mgcos theta

N = 1 * g * cos (t)

and how u got >10?

5. Sep 29, 2007

### aligass2004

The normal force is 7.73N. So now I substitute the normal force into -fricitional force-Wx = m(ax).

6. Sep 29, 2007

### learningphysics

Are you allowed to use work/energy for this problem?

7. Sep 29, 2007

### aligass2004

I don't think so. We haven't gone over work or energy yet.

8. Sep 29, 2007

### learningphysics

yup. this looks right.

9. Sep 29, 2007

### aligass2004

For the acceleration, I got -7.586 m/s^2. I tried using Vxf^2 = Vxi^2 + 2(ax)(delta x), but it didn't work.

10. Sep 29, 2007

### lightgrav

what did you use for the final velocity?
... by the way, this distance is NOT the height!

11. Sep 30, 2007

### aligass2004

Very good point. I hadn't realized that at first. So how do I go about finding the vertical height?

12. Sep 30, 2007

### learningphysics

use trig. When you get the distance along the ramp... the height is just dsin(38).

13. Sep 30, 2007

### aligass2004

How do I find the distance? By using Vxf^2 = Vxi^2 + 2(ax)(delta x)?

14. Oct 1, 2007

### aligass2004

15. Oct 2, 2007

### learningphysics

Yes. What did you get for distance?

16. Oct 2, 2007

### aligass2004

I got 1.459m for the distance

17. Oct 2, 2007

### learningphysics

How did you get that? Can you show your calculations?

18. Oct 2, 2007

### aligass2004

Well, I assumed that Vxf would be zero since the block would stop on the ramp. Then I used 14 for Vxi and -67.179 for the acceleration.

-196 = 2(-67.179)deltax
-196 = -134.358deltax
delta x = 1.459

19. Oct 2, 2007

### learningphysics

-67 isn't the right acceleration. how did you get that?

20. Oct 2, 2007

### aligass2004

-(ukn)-Wx = m(ax)

21. Oct 2, 2007

### aligass2004

I've got equations everywhere. I think I may have used the wrong acceleration. The acceleration is -7.586. So, is the distance 12.919m?

22. Oct 2, 2007

### learningphysics

yeah, that looks better. I'm getting acceleration as -[gsin(38) + uk*gcos(38)] = -7.578 m/s^2.

so height is distance*sin(38)

23. Oct 2, 2007

### aligass2004

I got the height to be 7.954 m. So how do I start part b?

24. Oct 2, 2007

### learningphysics

find the acceleration. then use that to find the velocity.

25. Oct 2, 2007

### aligass2004

How do I find the acceleration?