Elastic potential energy of a spring
A spring with a constant of 200 N/m is compressed by 0.1 m. How much elastic potential energy does it store?
Answer A: explanation for every choice
A
Correct. U = ½kx² = ½(200)(0.1)² = 1 J.
B
This leaves out the ½, using kx².
C
This uses x instead of x²: ½(200)(0.1).
D
20 N is the spring force kx, not the energy.
TakeawaySpring force: F = kx. Spring energy: U = ½kx².
Equation usedSprings (Hooke's law)
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