Asked by Brandon
A bullet of mass 0.0019 kg moving at 487 m/s embeds itself in a large fixed piece of wood and travels 0.55 m before coming to rest. Assume that the deceleration of the bullet is constant.
What force is exerted by the wood on the bullet?
What force is exerted by the wood on the bullet?
Answers
Answered by
drwls
Consider the work done by the bullet on the block. It gets turned into heat, and equals the initial kinetic energy of the bullet.
Force x distance = Initial Kinetic energy
F = (1/2)M V^2/0.55 = 410 N
Force x distance = Initial Kinetic energy
F = (1/2)M V^2/0.55 = 410 N
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