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Math Help - spring force

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    spring force

    A 135 g ball is dropped from a height of 62 cm above a spring of negligible mass. The ball compresses the spring to a maximim displacement of 4.45327 cm. The acceleration of gravity is 0.8 m/s^2.

    Calculate the spring force constant k. (units are N/m>
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    Quote Originally Posted by DINOCALC09 View Post
    A 135 g ball is dropped from a height of 62 cm above a spring of negligible mass. The ball compresses the spring to a maximim displacement of 4.45327 cm. The acceleration of gravity is 0.8 m/s^2.

    Calculate the spring force constant k. (units are N/m>
    I think you meant "9.8 m/s^2" yes?

    For simplicity take this problem in two parts. (You can do this in one step, but it's messy.)

    First part: The ball falls to the level of the top of the spring (ie. falls 62 cm).

    How fast is it moving just before it hits the spring? I've got a coordinate system at the point where the ball is dropped and I'm calling +y upward.
    v^2 = v_0^2 + 2a(y - y_0)

    v^2 = 2(-9.8)(-0.62)

    v = -3.48597~m/s
    (Negative since it's falling downward.)

    Second part: The ball compresses the spring.

    This is a work - energy problem. I'm defining the 0 level for the gravitational potential energy to be at the top of the uncompressed spring, and the 0 level for the spring potential energy to be at the top of the uncompressed spring, and in both cases I'm taking +y upward. There are no non-conservative forces at work here, so:
    W_{nc} = \Delta E

    \left ( \frac{1}{2}mv^2 - \frac{1}{2}mv_0^2 \right ) + (mgy - mgy_0) + \left ( \frac{1}{2}ky^2 - \frac{1}{2}ky_0^2 \right ) = 0

    The ball starts at the top of the spring ( y_0 = 0~m) with the speed from the last part of the problem, and ends at y = -0.0445327~m with a speed of 0. So:
    -\frac{1}{2}mv_0^2 + mgy + \frac{1}{2}ky^2 = 0

    k = \frac{\frac{1}{2}mv_0^2 - mgy}{\frac{1}{2}y^2}

    k = 1421.39~N/m = 14.2139~N/cm

    -Dan
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