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8 December, 03:30

The massless spring of a spring gun has a force constant k=12~/text{N/cm}k=12 N/cm. When the gun is aimed vertically, a 15-g projectile is shot to a height of 5.0 m above the end of the expanded spring. (See below.) How much was the spring compressed initially?

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  1. 8 December, 03:58
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    0.011 m.

    Explanation:

    Energy stored in the spring = Energy of the projectile.

    1/2ke² = mgh ... Equation 1

    Where k = spring constant, e = extension or compression, m = mass of the projectile, g = acceleration due to gravity, h = height.

    make e the subject of the equation

    e = √ (2mgh/k) ... Equation 2

    Given: k = 12 N/cm = 1200 N/m, m = 15 g = 0.015 kg, h = 5.0 m

    Constant: g = 9.8 m/s²

    Substitute into equation 2

    e = √ (2*0.015*5/1200)

    e = √ (0.15/1200)

    e = √ (0.000125)

    e = 0.011 m.
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