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A spring with spring constant of 30 N/m is stretched 0.19 m from its equilibrium position. How much work must be done to stretch it
an additional 0.077 m? Answer in units of J

  • physics -

    Using the equation for spring potential energy
    W=(1/2)kx²

    so
    ΔW=(1/2)k(x2²-x1²)
    =(1/2)(30 N/m)*(0.96²-0.19²)
    =13.3 J

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