Calculating Block Positions: Spring Question Solution | Edugen Course 1650

[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!

 

[thepanda]

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

 

[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

?

 

[AnnieD]

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

 

[thepanda]

Yeah you shouldn't 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.

 

[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.

 

[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.

 

[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??

 

[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.

 

[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.

 

[thepanda]

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

 

[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.

 

[thepanda]

It's 800 N*cm.

 

[AnnieD]

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

Thanks for your help! :)

 

[thepanda]

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

 

[AnnieD]

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