Homework Help: Horizontal spring and wood block energy and equilibrium question

1. Jun 4, 2014

salmayoussef

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

A 200 g wood block is firmly attached to a horizontal spring. The block can slide along a table where the coefficient of friction is 0.40. A force of 10 N compresses the spring 18 cm. If the spring is released from this position, how far beyond its equilibrium position will it stretch on its first swing?

2. Relevant equations

(I've listed the equations and variables/given data in my attempt!)

3. The attempt at a solution

The following is my attempt at solving this problem, but my issue is that one of my friends got a different answer (0.17 m). I tried it another way and got his answer but I don't know which one is correct (there's no answer key).

The other equation I used was (F - μmg)/k = x which gave me 0.17 m. But I found this equation on Yahoo Answers and I'm not sure exactly how they got it.

Mine looks right, but is it? Thanks in advance!

2. Jun 4, 2014

Jilang

Try looking again at the forces involved at the starting postion.

3. Jun 4, 2014

dauto

Your solution looks right. I didn't double check the math, though. The other equation doesn't make any sense. You can't just pick some random equation from Yahoo and expect it to make sense.

4. Jun 4, 2014

salmayoussef

It wasn't just an equation. There was some rearranging involved to find it, but that's the part I don't really understand. I found it here: https://ca.answers.yahoo.com/question/index?qid=20070405203621AAhJVqX

I tried following through with it but ended up getting slightly different answers. What I wanted to know was which is the right way to do it?

5. Jun 4, 2014

dauto

You should have posted that link in the OP to begin with. The Yahoo answer is wrong.

6. Jun 4, 2014

Jilang

Are you sure that you have the right answer for k? I get something different.

7. Jun 4, 2014

salmayoussef

Calculating it again, I got the same answer. 55.56 N/m. Unless I'm using incorrect variables to find k, then it should be right! What are you doing differently?

8. Jun 4, 2014

Jilang

I am considering the friction at the beginning. I have k = 10.784/0.18.

9. Jun 4, 2014

dauto

I don't think you should consider friction at the beginning position. The problem says "A force of 10 N compresses the spring 18 cm" Nothing about friction here. In fact, if you include friction it becomes impossible to solve the problem since you can't calculate the friction (neither direction nor magnitude).

10. Jun 4, 2014

salmayoussef

I tried it and it gave me 0.17 m, as it should be, I'm guessing! Thanks for your help! :thumbs:

11. Jun 4, 2014

Jilang

You are most welcome.

12. Jun 4, 2014

dauto

As I said in post #11, Adding the friction to the force doesn't make sense. Why are you adding the maximum friction? Why not subtract it? After all, the friction might be in the other direction, right? Why use the maximum friction as opposed to - say - half the maximum friction? There are no good answers to those questions. The only sensible interpretation is that there is no friction force at the outset.

13. Jun 4, 2014

Jilang

I would suppose that if you were stretching or compressing the spring the friction would work in your favour, but I can see where you are coming from.

14. Jun 4, 2014

salmayoussef

I understand what you mean. To be safe, I will exclude it. Thank you also!

15. Jun 4, 2014

dauto

Why in the world would you suppose that? There is no reason to suppose that. If anything, the friction would actually work against you - as usual. But that's not a good assumption either. Based on the question text, the only sensible assumption is that there is no friction at the set up position.

16. Jun 4, 2014

Jilang

Ok, think about the force required to hold the spring in a compressed or extended position. Would friction make any difference? ( My car rolls a lot more easily than my last one because of a big effort to reduce friction these days to save energy. I always need to make sure my handbrake is on, whereas with my old car it mattered less.)

Last edited: Jun 4, 2014
17. Jun 5, 2014

dauto

No there is no friction. Keep in mind that the block is at rest.

18. Jun 5, 2014

Jilang

If I pull on a block resting on a table with a force that steadily increases from zero the block won't start moving straight away. It will start to move when the force I apply is large enough to overcome the friction. Before that the block is not moving but there is still friction. It is called static friction.
http://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html

Last edited: Jun 5, 2014
19. Jun 5, 2014

mafagafo

dauto just seems to be right all along. Got 1.5e-01 m here.

20. Jun 5, 2014

dauto

That's true but has nothing to do with the problem here.

21. Jun 5, 2014

dauto

Voc^e e' Brasileiro?

22. Jun 6, 2014

haruspex

Looks like they dropped a factor of 2:
Initial compression = d = 0.18m
Mass = M = 0.2kg
Initial force = F = 10N
Spring constant k = F/d
Initial energy = kd2/2
Distance travelled = d+x
Work done against friction = (d+x)μMg
Final potential energy kx2/2
(d+x)μMg = kd2/2 - kx2/2
μMg = kd/2 - kx/2

23. Jun 6, 2014

Nathanael

Shouldn't the "kd/2" term be "Fd/2" (or kd2/2 ... same thing) ?

EDIT:
Actually I don't even know what your equation means, so nevermind. I was just thinking of the equation I used which was:

$(d+x)mg\mu+\frac{Fx^2}{2d}=\frac{Fd}{2}$

Sorry, for some reason I thought that's the equation you were going for, but obviously not. (I don't know why I thought that)

Last edited: Jun 6, 2014
24. Jun 6, 2014

haruspex

No. I cancelled a factor (d+x) from each side of the preceding equation. This is one reason I made the post, to show that it is not necessary to solve a quadratic.