A 23 g bullet traveling at 230 m/s penetrates a 2.0 kg block of wood and emerges cleanly at 170 m/s. If the block is stationary on a surface with a coefficient of kinetic friction of 0.15 when hit, how far does it move after the bullet emerges?

The distance traveled by the wood after the bullet emerges is 0.16 m.

The given parameters;

mass of the bullet, m = 23 g = 0.023 gspeed of the bullet, u = 230 m/smass of the wood, m = 2 kgfinal speed of the bullet, v = 170 m/scoefficient of friction, μ = 0.15

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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