The distance traveled by the wood after the bullet emerges is 0.16 m.
The given parameters;
- <em>mass of the bullet, m = 23 g = 0.023 g</em>
- <em>speed of the bullet, u = 230 m/s</em>
- <em>mass of the wood, m = 2 kg</em>
- <em>final speed of the bullet, v = 170 m/s</em>
- <em>coefficient of friction, μ = 0.15</em>
The final velocity of the wood after the bullet hits is calculated as follows;

The acceleration of the wood is calculated as follows;

The distance traveled by the wood after the bullet emerges is calculated as follows;

Thus, the distance traveled by the wood after the bullet emerges is 0.16 m.
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The answer would be point A.
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Answer:
SECOND LAW OF NEWTON
Explanation:
When the rocket fires the engines the gases leave at high speed and collide with the space station, transferring an impulse given by the expression
I = F t = Δp
As we can see this expression is a form of Newton's second law
F = m a
a = dv / dt
F = m dv / dt
F dt = m dv
p = mv
F dt = dp
Therefore the station moves through the SECOND LAW OF NEWTON