Aspring that is stretched 2.6 cm stores a potential energy of 0.053 j. what is the spring constant of this spring?

Recall the equation for the potential energy stored in a spring:

PE = 0.5kx²

PE is the potential energy.

k is the spring constant.

x is the length by which the spring is extended/compressed.

Given values:

PE = 0.053 J

x = 2.6 cm = 0.026 m

Substitute the terms in the equation with the given values and solve for k:

0.053 = 0.5×k×0.026²

k = 156.8 N/m

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