Hookes Law

[lucy1368]

Hi,

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? 

( 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. 

 

http://www.dpcdsb.org/NR/rdonlyres/70ACCCCE-D6A6-44CA-982F-8E319A8ADE82/89746/momentumenergyquestions.pdf (the solution is posted here).

 

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. 

[FizziksGuy]

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...