a spring with a force of 2.9 n / m has a relaxed 0.85 m distance, when a 0.5 kg mass is attached it stretches to 1.65 m. what is the elastic potential energy?

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09-12-2022
Physics

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a spring with a force of 2.9 n / m has a relaxed 0.85 m distance, when a 0.5 kg mass is attached it stretches to 1.65 m. what is the elastic potential energy?

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rekharaypriti8910 rekharaypriti8910 · Answer 1 · 2022-12-15T22:13:31-05:00

The calculated answer is 1.96kg * m \s2\s /2. The energy that is stored when a force is used to deform an elastic object is known as elastic potential energy.

Until the force is released and the object springs back to its original shape, doing labor in the process, the energy is retained. The object may be compressed, stretched, or twisted during the deformation. Since you are aware of the spring constant k, which is 144 N/m, and the spring stretch from equilibrium position x, which is 16.5 cm, or 0.165, you can use the equation

PE = 2 1 kx 2

= 2 1144(.165) 2

to determine the potential energy of the spring.

This, if you prefer SI terms, is kg * m 2 /s2, or 1.96 Joules.

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