Question

A force of 570 N keeps a certain ideal spring stretched a distance of 0.900 m.
Part A:
What is the potential energy of the spring when it is stretched 0.900 mm?
Express your answer with the appropriate units.
Part B:
What is its potential energy when it is compressed 9.00 cm?
Express your answer with the appropriate units.

Answer 1

Hooke's law states that the elongation of a material is proportional to the applied stress within the elastic limits of the material.

Using Hooke's law

F = kx

570 = k * 0.9

k = 712.5 N/m

(a) energy at x = 0.8 m

U = 0.5 k x^2 = 0.5* 712.8* 0.8^2

U = 228 J

(b) U = 0.5 * 712.8* 0.09^2

U = 2.887 J

Elastic potential energy is the potential energy released as a result of the deformation of an elastic body. B. Spring extension is stored. This is equal to the work done to stretch the spring, which depends on the spring constant k and the stretched distance. The force exerted by the spring is the restoring force that helps return the spring to its equilibrium length.

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