# Homework Help: Not a clue Springs

1. Apr 8, 2009

### joemama69

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

Ok sorry for the horrible drawing i freehanded it on paint.

The 5lb collar is released from rest at A and travels along the smooth guide. Determine the speed of the collar just before it strikes the stop at B. The spring has an unstretched length of 12in. k=2 lb/in

please note that from point C to point A it is a 1/4 circle of radius 12in
From C to B, it is horizontal.

2. Relevant equations

3. The attempt at a solution

I could do this problem if it was traveling from just C to B but the curve has me lost.

I attempted to find my Y and i got y=(144-x^2)^1/2+10
I solved that for x and got x=(144-(y-10)^2)^1/2
I did this to find the hypotenus or the s length of the spring. would it be correct for my s to be (x^2+y^2)^1/2. i feel like im way off hlep

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Last edited: Apr 8, 2009
2. Apr 8, 2009

### Staff: Mentor

Not sure what you are trying to do. What are y, x, and s?

Hint: Use energy conservation. Find the initial and final stretch of the spring.

3. Apr 8, 2009

### joemama69

s is the length of spring. the initial stretch is 22in, and the final is 12in, but how does that help

4. Apr 8, 2009

### Staff: Mentor

It will allow you to figure out the elastic potential energy stored in the spring. Be sure to take the unstretched length of the spring into account.

5. Apr 9, 2009

### joemama69

I attempted to find my Y and i got y=(144-x^2)^1/2+10
I solved that for x and got x=(144-(y-10)^2)^1/2

am i correct is saying that my s would the hypotenus using the x and y above.

I would then plug it into V(e) = .5ks^2 for elastic potential energy

I believe I also have to use V(g) = Wy

My main problem is that my y value is a peacewise function. From point B-C it is a different function than from C-A

Note the Picture attached above

6. Apr 9, 2009

### Staff: Mentor

Sorry, but I still have no idea what you are trying to calculate here.

No. All you need to find is the initial and final stretch of the spring from its unstretched length--the curved section is irrelevant. Your previous post had all the data needed to find the spring PE.

Yes, once you find the amount that the spring is stretched, that's how you find the elastic PE.

If you mean gravitational PE, then yes, you definitely need that.

So what? Energy is conserved. All you care about it is the total energy (elastic, gravitational, and kinetic) at points A and B.

7. Apr 9, 2009

### joemama69

Ok I think i got it, i thought it would be much more complicated, my book is confusing

ok heres what i did,

V(A) = V(e) - V(g) = .5ks^2 - Wy = -3292 k=2lb/in s=10in W=5lb*32.2 ft/s^2 <-- should this be in/s^2 since k is in lb/in...... & y=22in

T(A) = .5mv^2 = (5/2)v^2

And V(C) & T(C) = 0 because the spring is no longer stretcher and the y = o

so i got v = 36.29 in/s

Are my units correct.

8. Apr 9, 2009

### frankiben123

Last edited by a moderator: May 4, 2017
9. Apr 9, 2009

### Staff: Mentor

Why the minus sign?
I would put all quantities into standard units, distances in feet. (k = 2lb/in = 24lb/ft.) Note that 5 lbs is the object's weight, not its mass, so no need to multiply by g when finding the gravitational PE. (What's the object's mass?)

The collar starts from rest at point A.

The elastic and gravitational PEs are both zero at point B (not point C, at least according to your diagram). The KE is not zero.

10. Apr 10, 2009

### joemama69

I subtracted Wy because of a similar example in my book but i guess i read into it wrong

ok so my new mass is 2.26 slug

ok so my T(a) = 0, but my T(B) is now 1.134v^2

plug and chug and i got a velocity of 11.98 ft/sec, that just seems wrong, too fast

note the new pic

#### Attached Files:

• ###### Spring Problem.bmp
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258.8 KB
Views:
133
11. Apr 10, 2009

### Staff: Mentor

How did you get that?

As a sanity check for the speed, how fast would something be moving if it were simply dropped from that same height? (That speed must be less than the speed of the collar in this problem, since there's no spring pulling it down.)

12. Apr 11, 2009

### joemama69

1 pound = 0.0310809502 slug

5lb = .1554045... oops calc error

ok so i got at point A
V(g)=9.157366, V(e)=9.96, T=0

Point B
V(g)=0, V(e)=0, T=.77v^2

v=4.96 ft/sec I hope

13. Apr 12, 2009

### Staff: Mentor

That's better.

Your V(g) is close to my value: mgh = (5)*(22/12) = 9.167
Show your calculation for V(e).

Show your calculation for T.

Did you do the sanity check I suggested?

14. Apr 12, 2009

### joemama69

Point A

Vg = .554*32.2*1.83 = 9.157366<-- Different than urs Doc
Ve = .5*24*.83 = 9.96

Point B

T = .5*1.554*v^2 = .777v^2

15. Apr 12, 2009

### joemama69

oops Ve = .5 *24*.83^2 = 8.2668

v = 4.74

16. Apr 13, 2009

### Staff: Mentor

You must have a typo in here, since the left and right sides don't match.
Since mg is given as 5 lb, mgh = 5*(22/12) = 9.167
Since mg = 5; m = 5/g = 5/32.2 = 0.1553
The 0.83 (really 0.8333...) should be squared.

Wrong mass.

17. Apr 13, 2009

### joemama69

Sorry for the typo, i meant 1.554

I thought that the mass was 1.554 slug, and 5lb was the force

18. Apr 13, 2009

### Staff: Mentor

You're off by a factor of ten. The mass is 0.1554 slugs. (See my previous post.)

19. Apr 13, 2009

### joemama69

man i keep making stupid mistakes

V(g) = .1554*32.2*1.83333333333 = 9.1738
V(e) = .5*24*.8333333333333333^2 = 10
T(B) = .5*.1554*v^2 = .0777v^2

Therefor v = 15.71 ft/sec

20. Apr 13, 2009

### Staff: Mentor

Redo this one.