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9 October, 19:48

A vertical spring with a spring constant of 430 N/m is mounted on the floor. From directly above the spring, which is unstrained, a 0.30 kg block is dropped from rest. It collides with and sticks to the spring, which is compressed by 2.2 cm in bringing the block to a momentary halt. Assuming air resistance is negligible, from what height (in cm) above the compressed spring was the block dropped

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  1. 9 October, 20:12
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    3.54 cm

    Explanation:

    From the question,

    The potential energy of the block = energy needed to compress the spring

    mgh = (ke²) / 2 ... Equation 1

    Where m = mass of the block, h = height from which the block was dropped, g = acceleration due to gravity, k = spring constant of the spring e = compression.

    Firstly, we make h the subject of the equation

    h = ke²/2mg ... Equation 2

    Given: k = 430 N/m, e = 2.2 cm = 0.022 m, m = 0.3 kg, g = 9.8 m/s²

    Substitute these values into equation 2

    h = 430 (0.022²) / 2*0.3*9.8

    h = 0.20812/5.88

    h = 0.0354 m

    h = 3.54 cm
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