Question #147187
A spring has a spring constant of 12.3 N/m. If it is stretched a distance of 1.08 m, how much elastic potential energy is stored in the spring?
1
Expert's answer
2020-11-30T14:54:47-0500

W=kx2/2=12.3⋅1.082/2=7.2(J)W=kx^2/2=12.3\cdot 1.08^2/2=7.2(J)


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Good work..thanks to the expert
United Kingdom #333261 on Nov 2022