January 19, 201412 yr 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.
January 19, 201412 yr 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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