stretching s spring

Calculus Level 3

A spring with natural length of 10 10 cm. is streched by 1 2 \dfrac{1}{2} cm. by a 12 12 newton force. Find the work done in stretching the spring from 10 10 cm. to 18 18 cm. Express your answer in Joules.


The answer is 7.68.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

Force constant of the spring is 12 0.005 = 2400 \dfrac{12}{0.005}=2400 N/m. Stretch of the spring is 0.18 − 0.10 = 0.08 0.18-0.10=0.08 m. Therefore the work required to stretch the spring by this amount is 1 2 × 2400 × ( 0.08 ) 2 = 7.68 \dfrac{1}{2}\times 2400\times {(0.08)^2}=\boxed {7.68} Joules.

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Thanks for posting a solution.

A Former Brilliant Member - 1 year, 3 months ago

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