Resisting Force and Stopping Time of a Bullet in Wood

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The discussion focuses on calculating the resistive force exerted by wood on a bullet and the time it takes for the bullet to come to rest after penetrating 15 cm. The bullet, weighing 8 g and traveling at 390 m/s, experiences deceleration upon impact with the wood. To determine the resistive force, one can use the kinematic equations provided, specifically relating to the bullet's initial and final velocities, distance, and acceleration. Additionally, the time taken for the bullet to stop can be derived from the same equations. Understanding these concepts is crucial for solving the problem effectively.
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Homework Statement


A rifle with a barrel length of 55 cm fires a 8 g bullet with a horizontal speed of 390 m/s. The bullet strikes a block of wood and penetrates to a depth of 15 cm.

A) What resistive force (assumed to be constant) does the wood exert on the bullet?
B) How long does it take the bullet to come to rest?

Homework Equations


Vf = Vo + (a*t)
Xf = Xi + (Vo*t) + (1/2)a(t^2)
Vf^2 = Vo^2 + 2*a*d
F of friction = u*N

The Attempt at a Solution



(I'm really at a lost and I'm not understanding how to approach this problem.)
 
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The bullet has some initial velocity when it hits the block of wood. After 15cm into the wooden block, the bullet has been stopped. This means that some force acted to decelerate the bullet. 'A' is asking you to find the magnitude of that force. Based on the information you have, 'B' is asking how long (in time, instead of distance) does it take for the bullet to stop.
 
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