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9 June, 07:44

A spring with a force constant of 5400 N/m and a rest length of 3.5 m is used in a catapult. When compressed to 1.0 m, it is used to launch a 48 kg rock. However, there is an error in the release mechanism, so the rock gets launched almost straight up. How high does it go (in m) ? (Assume the rock is launched from ground height.) m

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  1. 9 June, 11:20
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    5.51 m

    Explanation:

    From the question,

    The energy used to stretch the spring = the potential energy of the rock.

    (1/2) ke² = mgh ... Equation 1

    Where k = spring constant, e = extension/compression, m = mass of the rock, g = acceleration due to gravity, h = height of the rock above the ground

    make h the subject of the equation.

    h = ke²/2mg ... equation 2

    Given: k = 5400 N/m, e = 1 m, m = 48 kg.

    Constant: g = 9.8 m/s²

    Substitute into equation 2

    h = 5400 (1²) / (2*48*9.8)

    h = 5400/940.8

    h = 5.51 m.

    Hence the height of the rock = 5.51 m
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