Hooke's Law Question?

When the block is pulled out to x = +4.0 cm, we must apply a force of magnitude 360 N to hold it there. We pull the block to x = 11.0 cm and then release it. How much work does the spring do on the block when the block moves from xi=+5.0 cm to (a)x= +4.0 cm, (b)x=-4.0 cm, (c)x=-5.0 cm, and (d)x=-10.0 cm?



I've tried working this and I get it wrong. Does stretching it to 11 and back to 5 change anything with the spring constant or does it just make 5 = equalibrium?Hooke's Law Question?
The equilibrium point is x = 0



The two equations are:



F = -kX

E = (1/2)k(X^2)



where:

F is the force

k is the spring constant

X is the displacement

E is the energy



You are given F for X = 4 so you can compute the spring constant.



Given the spring constant, you can compute the energy in the spring for each displacement.



If you assume no friction, any difference in the energy in the spring is either given to or taken from (the kinetic energy of) the block.



The only reason x = 11 is mentioned is to explain how the spring got to x = -10. You can ignore it (the answers would be the same whether the spring were pulled to 11, 12, 20 , etc.)



Response to addendum:

You can choose any units you want. If you want the spring constant in newtons/meter, then it would be 360/.04. If you want it in newtons/cm, then it would be 360/4. As long as your units of force, distance, and energy are consistent, it doesn't matter which set of units you use.



But as MKS is getting more popular and CGS less, I'd go for MKS all the way, which makes your spring constant 9000 newtons/meter.

http://www.unc.edu/~rowlett/units/cgsmks



Here's a site with a spring constant experiment. Look at their displacement, force, and spring constant values:

http://www.ap.smu.ca/demos/content/mecha

row cell

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