A spring with a stiffness of K = 2 * 10 ^ 5 N / m is lengthened by 3 cm under the influence

univerkov | January 10, 2021 | education

A spring with a stiffness of K = 2 * 10 ^ 5 N / m is lengthened by 3 cm under the influence of the elastic force. What is the potential energy of the stretched springs.

Initial data: k (spring rate) = 2 * 10 ^ 5 N / m; Δl (spring extension) = 3 cm.
SI system: Δl = 3 cm = 3/100 m = 0.03 m.
The potential energy of a stretched spring can be calculated by the formula: En = k * ∆l ^ 2/2.
Let’s perform the calculation: En = 2 * 10 ^ 5 * 0.03 ^ 2/2 = 90 J.
Answer: The potential energy of a stretched spring is 90 J.

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