A spring that has a spring constant of 1400 N/m is stretched to a length of 2.5 m. If the normal length of the spring is 1.0 m, how much elastic potential energy is stored in the spring?
Answer
A
700 J
B
1050 J
C
1575 J
D
4375 J
Explanation
The spring is stretched by 2.5 m − 1.0 m = 1.5 m. Elastic potential energy is E = ½kx², so E = ½(1400 N/m)(1.5 m)² = 1575 J. The correct choice is 1575 J.
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