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27 April, 10:56

9. Santa brings a special baby a bouncy swing. The child bounces in a harness, with a spring constant k, and is suspended off of the ground. a. If the spring stretches x1 = 0.27 m from equilibrium while supporting an 7.15-kg child, what is its spring constant, in newtons per meter?

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  1. 27 April, 11:11
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    259.52N

    Explanation:

    Force initiated on a spring that causes an extension is related by the expression below;

    F = K*e

    Where F is the Force

    K is the spring constant

    e is the extension caused by the spring.

    By change of subject formula for K;

    K = F/e

    Now F is the same as weight and is given by mass * acceleration. In this case acceleration is g=9.8m/s2; this is because the child's mass is going to be under the influence of gravity as it swings up the harness}

    Hence W = 7.15*9.8=70.07N

    Hence K = 70.07/0.27 = 259.5185N/m

    =259.52N/m to 2 decimal place.
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