A spring that has a spring constant of 1400 cap n over m is stretched to a length of 2 point 5 m If the normal length of the spring is 1 point 0 m , how much elastic potential energy is stored in the spring?
Answer
A
1575 cap j
B
700 cap j
C
4375 cap j
D
1050 cap j
Explanation
The spring’s extension is 2.5 m − 1.0 m = 1.5 m. Elastic potential energy is E = ½kx², so E = ½(1400)(1.5)² = 1575 J. Therefore, the correct choice is 1575 J.
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