Physics

A bullet of mass .03 kg moving with a speed of 400m/s penetrates 12cm into a fixed block of wood calculate the average force exerted by the wood on the bullet.

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  1. V^2 = Vo^2 + 2a*d = 0.
    400^2 + 2a*0.12 = 0,
    160,000 + 0.24a = 0, a = -666,667 m/s^2.

    F = M*a = 0.03 * (-666,667) = -20,000 N.

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  2. Can you please explain it in simple way .

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  3. K.E. of bullet = Workdone
    K.E. = 1/2 mv^2
    Workdone =Force* Distance
    1/2mv^2 = Force *0.12
    1/2*0.03*(400)^2 = Force *0.12
    Force = 20kn

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