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10 August, 06:28

You are launching a 2 kg potato out of a potato cannon. The cannon is 2.0 m long and is aimed 70 degrees above the horizontal. It exerts a 52 N force on the potato. What is the kinetic energy (in J) of the potato as it leaves the muzzle of the potato cannon?

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  1. 10 August, 06:46
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    The net force on the potatoes is given by:

    F = 52 - mgSintheta

    F = 52 - (2*9.8 * Sin70°)

    F = 52 - 18.4

    F = 33.58N

    Using Newton's 2nd law

    F = ma

    a=F/m = 33.58 / 2 = 16.79m/s^2

    Using the equation of motion:

    V^2 = u^2 + 2as

    V^2 = 0 + 2 * 16.79 x2

    V^2 = 67.16

    V=sqrt (68.16)

    V = 8.195m/s This is the exit velocity of the potatoes

    Kinetic energy, K. E = 1/2mv^2

    KE = 1/2 * 2 * 8.195^2

    KE = 67.16J
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