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14 January, 23:57

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) ?

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  1. 15 January, 00:13
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    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
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