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2 years ago

11

A spring whose stiffness is 980 N/m has a relaxed length of 0. 50 m. If the length of the spring changes from 0. 25 m to 0. 81 m

, what is the change in the potential energy of the spring

1 answer:

Gelneren [198K]2 years ago

8 0

151.9j

Explanation:

PE=1/2kx^2

PE=1/2(980)(.50)= 245j

PE=(1/2)(980)(.81)= 396.9j

396.9- 245= 151.9j

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