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an athlete stretches a spring an extra 40.0 cm. how much energy has he transferred to the spring, if the spring constant is 5286.0 N/m?

Us= 1/2kx^2

=1/2(5286.0)(.40)=1057.2 J

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Answered by HELP
Fixed Answer

Us= 1/2kx^2

=1/2(5286.0)(.40)^2=422.88 J
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