# A bullet of mass 0.002 kg moving at 421 m/s

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A bullet of mass 0.002 kg moving at 421 m/s
impacts a large fixed block of wood and travels
6.9 cm before coming to rest.
Assuming that the deceleration of the bullet
is constant, find the force exerted by the wood
on the bullet.
In this case, the bullet is applied with a force by the block in the direction opposite to its velocity, which is the net force and creates the acceleration. Then the key is to find the acceleration from the settings. A simplified physics version of the story is that the bullet has an initial velocity of 421 m/s and a final velocity of 0 m/s. The distance traveled is 0.069 m. Then using the equation below, we can find its acceleration.  $$V_f^2-V_i^2=2a\Delta x$$.
Once $a$ is found, the force can be found with $ma$.