# Incline plane problem

1. Apr 25, 2012

### albert12345

I've tried a few times on this problem but I can't get the right answer...

A box with mass m=1,05 kg slides down a ramp with angle=42.39°. Kinetic friction μ=0,1646 between the ramp and the box. It starts from rest. g=9,806m/s². The box hit's a spring with: k=792 N/m after sliding the distance: s= 2,875m, What is the maximum compression of the spring (x)?

I've tried the following solution based on the law of conservative energy:
(m*g*(s+x)*sin(42,39)) - (μ*m*g*(s+x)*cos(42.39)) = 0.5*k*x²

What am I doing wrong?

Thanks,

2. Apr 25, 2012

### Staff: Mentor

Your solution looks OK to me.

3. Apr 25, 2012

### albert12345

I agree. I came up with the answer: x ≈ 0,203m

4. Apr 25, 2012

### Staff: Mentor

What textbook is this problem from? (Just in case I have it.)

(My answer is close to yours.)

5. Apr 25, 2012

### albert12345

It's from an assignment. I really belive that the answer is wrong, but it has never happend before so...I must have done something wrong.

6. Apr 25, 2012

### Staff: Mentor

I see the problem. Is the spring on the incline? Or on the horizontal section after the incline?

Assume the latter and you'll get the given answer.

7. Apr 25, 2012

### tms

So is mine, if that means anything: 0.211 m.

Last edited: Apr 25, 2012
8. Apr 25, 2012

### albert12345

There is no pic. for the problem, but I get your point. Does it really make that big differernce? And We have never had such problems before, so i think the spring is aligned with the plane

9. Apr 25, 2012

### Staff: Mentor

Try it and see for yourself!

10. Apr 25, 2012

### albert12345

I really don't think it's right. It's going to be the same since Ug2=the height + the compression of the spring.

11. Apr 25, 2012

### Staff: Mentor

Assume the spring is horizontally oriented after the mass slides down the incline. You'll get an answer that matches the one you say is correct.

12. Apr 25, 2012

### albert12345

Ok! But I don't understand how I should calculate it :/

13. Apr 25, 2012

### Staff: Mentor

Similar to what you did before: you'll use energy conservation. Only this time it's even easier to solve. (Only the incline has friction--assume the flat part after the incline has no friction.)

14. Apr 25, 2012

### albert12345

I just tried it! It's wierd cause I got the same answer again :/

k1 = (1/2)mv^2 equals U2= (1/2)kx^2

15. Apr 25, 2012

### Staff: Mentor

Try again. Show me an equation similar to what you had in your first post.

16. Apr 25, 2012

### albert12345

I calculated the velocity when the box has slided down the plane:
v= 5,581963937m/s (The speed is just before the box hit's the spring

Then I put in the velocity in the ecuation I wrote earlier, together with the spring constant and then I solved for x

17. Apr 25, 2012

### albert12345

I've tried the same formula I used when I started this topic on another assignment of the same type. It shows that the solution in the assignment was wrong!!!!!

IDIOTS! We were right al along :D Thanks for your help guys!

18. Apr 25, 2012

### Staff: Mentor

It all depends on whether the spring is on the incline or after the incline. You need a diagram.

Edit: Oops... I think I made an error when I solved the problem the second way (with the spring horizontal). It doesn't give the 'correct' answer. So your suspicion is correct: the answer is wrong no matter how you slice it. Your instructor probably made the same error that I made! That's why I got the 'correct' answer that way. :uhh:

Last edited: Apr 25, 2012