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physics

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A spring (spring 1) with a spring constant of 520N/m is attached to a wall and connected to another weaker spring (spring 2) with a spring constant of 270N/m on a horizontal surface. Then an external force of 70N is applied to the end of the weaker spring (#2).
how much potential energy is stored in each spring

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    F = k1*x1
    70 = (520)*x1

    F = k2*x2
    70 = 270*x2

    Plug x's into potential energy for a spring equation U = 0.5kx^2 for respective springs

  • physics - ,

    Spring #1:
    70N * 1m/520N. = 0.1346 m.
    PE=0.5F * d = 0.5*70 * 0.1346 = 4.71 J.

    Spring #2:
    70N/270N * 1m. = 0.2593 m.
    PE=0.5F * d = 0.5*70 * 0.2593 = 9.07 J.

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