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Elastic potential energy stored in stretched springs

Given that the work done in stretching a spring to extension x is =
(lambda)x^2/2L

Could anyone clarify the following for me please?

I understood that the work-energy principle is:
"Work done by external resultant force on a particle = change in mechanical energy of particle"

So if we stretch a light spring horizontally, with no particle/mass attached, then clearly the total mech. energy has increased (elastic potential gone from 0 -> y Joules). Then from above principle, the work done by resultant force causing the spring to extend = this increase in EPE.

This is where I get confused:


the textbook says that the gain in EPE = work done by spring's tension


Now I believe that increase in EPE = work done resultant force, and although the displacment is the same, the value of the resultant force does not necessarily equal the tension at any point. So why are they able to equate resultant and tension, to say that EPE gained = work doen to stretch spring??


As usual I fear that the authors are being slightly sloppy with their definitions etc.

Thank you.
Original post by Choochoo_baloo


This is where I get confused:

the textbook says that the gain in EPE = work done by spring's tension



I would say that the gain in EPE= the work done against the spring's tension/compression. I.e the work done by the applied force.


Now I believe that increase in EPE = work done resultant force, and although the displacment is the same, the value of the resultant force does not necessarily equal the tension at any point.


The value of the applied force will equal the tension/compression at any given time.

Newton: Action and reaction are equal and opposite.

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