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# Hookes Law

## Question

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.

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.

## Recommended Posts

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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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