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Hookes Law



For the following question:

A 1.37-kg fish is hung from a vertical spring scale with a force constant of 5.20x10² N/m. The spring obeys 
Hooke’s Law. 
(a) By how much does the spring stretch if it stretches slowly to a new equilibrium position? 
(B) If the fish is attached to the un-stretched spring scale and allowed to fall, what is the net force on the 
fish when it has fallen 1.59 cm? 
© Determine the acceleration of the fish after it has fallen 2.05 cm. 
I was wondering why you have to use forces to solve(kx) rather than the formula for elastic potential energy(kx/2). When I tried using energy, it did not give me the same answer. Why would energies not work?
Thank you in advance for taking the time to answer my question. 

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Hi Lucy.  I bet an energy approach would work if applied properly, but in part a, that's a basic application of Hooke's Law, so that would be much simpler, and in part b, since you're looking for a force, I'd go with Hooke's Law again (then Newton's 2nd).  Just more straightforward...

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