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8 May, 22:54

A 30 g bullet moving a horizontal velocity of 500 m/s comes to a stop 12 cm within a solid wall. (a) what is the change in the bullet's mechanical energy? (b) what is the magnitude of the average force from the wall stopping it?

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  1. 8 May, 23:02
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    M = 30 g = 0.03 kg, the mass of the bullet

    v = 500 m/s, the velocity of the bullet

    By definition, the KE (kinetic energy) of the bullet is

    KE = (1/2) * m*v²

    = 0.5 * (0.03 kg) * (500 m/s) ² = 3750 J

    Because the bullet comes to rest, the change in mechanical energy is 3750 J.

    The work done by the wall to stop the bullet in 12 cm is

    W = (1/2) * (F N) * (0.12 m) = 0.06F J

    If energy losses in the form of heat or sound waves are ignored, then

    W = KE.

    That is,

    0.06F = 3750

    F = 62500 N = 62.5 kN

    Answer:

    (a) 3750 J

    (b) 62.5 kN
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