A leaf spring in an off-road truck with a spring constant k of 87.6 kN/m (87,600 N/m) is compressed a distance of 6.2 cm (0.062 m) from its original unstretched position. What is the increase in potential energy of the spring (in kJ) ?

0.00367 kJ

Explanation:

Using,

ΔEp = 1/2ke² ... Equation 1

Where ΔEp = increase in potential energy, k = spring constant of spring, e = extension/compression

Given: k = 87600 N/m, e = 0.0062 m

Substitute into equation 1

ΔEP = 1/2 (87600) (0.0062²)

ΔEP = 3.367 J.

ΔEp = 0.00367 kJ

Hence the change in potential energy of the spring = 0.00367 kJ