Asked by rad

A spring with a spring constant k of 44 N/m is stretched a distance of 30 cm (0.3 m) from its original unstretched position. What is the increase in potential energy of the spring?
Answered by drwls
P.E. = (1/2) k X^2

Use X = 0.3 meters

The original value of X (deviation from the equilibrium position) was zero.
The answer will be in Joules
Answered by Anonymous
1.98
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