# Homework Help: Spring Question. Am I doing this right?

1. Oct 27, 2008

### AnnieD

In Fig.7-11
http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c07/pict_7_11.gif
we must apply a force of magnitude 82.0 N to hold the block stationary at x=-2.0 cm. From that position, we then slowly move the block so that our force does +8.0 J of work on the spring–block system; the block is then again stationary. What are the block's positions, in cm? ((a) positive and (b) negative)

This is what I did, but can anyone confirm if it's correct?
F = -kx
82 = -k(-2)
82 = 2k
41 = k

Convert 8J aka 8N/m to .08N/cm because the final answer has to be in cm.
U = 1/2kx^2
For positive:
82 + .08 = 1/2(41)x^2
solve for x .. 2.00cm
For negative:
82 - .08 = 1/2(41)x^2
solve for x.. 2.00cm
(has to have 3SF)

Something doesn't seem right.. maybe I messed up in converting?
Any help would be much appreciated.. thanks!

2. Oct 27, 2008

### thepanda

Yes... Joule is N*m, not N/m :p

3. Oct 27, 2008

### AnnieD

haha that would be a good start.

So does this make sense..

Convert 8J aka 8Nm to 800Ncm
U = 1/2kx^2
For positive:
82 + 800 = 1/2(41)x^2
solve for x .. 6.56cm
For negative:
800 - 82 = 1/2(41)x^2
solve for x.. 5.92cm

?

4. Oct 27, 2008

### AnnieD

No wait.. that doesn't make sense either.
Why would there be more N acting on a smaller area?
I'm lost!

5. Oct 27, 2008

### thepanda

Yeah you shouldnt be thinking about the 82 like that because it is a force, not energy, so it would not apply to that problem. You should think about the potential energy difference between the old x and the new x, that will equal the work you put in.

6. Oct 27, 2008

### AnnieD

PE = mgh
But there's no height.. or mass?

Is the solving for
F = -kx
82 = -k(-2)
82 = 2k
41 = k
at least correct?

I have 46min to solve this question.

7. Oct 27, 2008

### thepanda

ok U is the potential energy

So 1/2 k(x_new^2 - x_old^2) = 8J, you have x old so you can solve for x new.

That is for the positive work, if you have negative work the difference between the squares is flipped.

8. Oct 27, 2008

### AnnieD

okay.. so
8 = 1/2(41)[(x)^2 - (-2)^2)]
8 = 1/2(41)(x^2 - 4)
16 = 41x^2 - 164
180 = 41x^2
x = 2.10

And for negative:
8 = 1/2(41)[(-2)^2 - (x)^2)]
8 = 1/2(41)(4 - x^2)
16 = (41)(4-x^2)
16 = 164 - 41x^2
16 - 164 = - 41x^2
-148 = - 41x^2
x = 1.90cm

??

Last edited: Oct 27, 2008
9. Oct 27, 2008

### thepanda

Yes, but keep in mind that either the positive or the negative numbers will satisfy the problem. So since it is "slowly," it is probably the closest ones that work. So if your calcs are correct, the closest positive work on the system is at x = -2.10, and the closest negative work on the system is x = -1.90.

10. Oct 27, 2008

### AnnieD

I tried those answers, but they're incorrect. It's an online assignment so it tells us right away whether or not we have the right answers.

11. Oct 27, 2008

### thepanda

did you convert the 8 J into N*cm when you did the calculations?

12. Oct 27, 2008

### AnnieD

8J = .08Ncm?

This is the new question.. well really the same question, just different values.
we must apply a force of magnitude 81.0 N to hold the block stationary at x=-3.0 cm. From that position, we then slowly move the block so that our force does +7.0 J of work on the spring–block system; the block is then again stationary.

okay.. so
k = 27

.07 = 1/2(27)[(x)^2 - (-3)^2)]
.07 = 1/2(27)(x^2 - 9)
.14/27 - 9 = x^2
x = -3.00

And for negative:
.07 = 1/2(27)[(-3)^2 - (x)^2)]
.07 = 1/2(27)(9 - x^2)
.14/27 - 9 = -x^2
x = 3.00

That doesn't seem right either.
I think I'm almost ready to give up.

13. Oct 27, 2008

### thepanda

It's 800 N*cm.

14. Oct 27, 2008

### AnnieD

Nope, that didn't work either.
That's okay.. I give up.

15. Oct 27, 2008

### thepanda

Wait, what did oyu get for that second problem? I get -6.546 cm.

16. Oct 27, 2008

### AnnieD

Same thing.
The positive value was correct, but the negative position wasn't.
That's okay. :)