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physics homework help?

A bullet of mass 0.002 kg initially moving at 493 m/s embeds itself in a large fixed piece of wood and travels 0.65 m in the wood before coming to rest. The acceleration of the bullet is constant.

what force is exerted by the wood on the bullet?

1 Answer

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  • 1 decade ago
    Favorite Answer

    By the work energy theorem:-

    =>W = ∆KE

    =>F x s = 1/2mv^2

    =>F = mv^2/2s

    =>F = [0.002 x (493)^2]/[2 x 0.65]

    =>F = 373.92 N

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