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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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- FiremanLv 71 decade agoFavorite 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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