# A bullet weighing 9 grams pierces a board 5 cm thick, while its speed decreases from 600 ms to 200 ms

**A bullet weighing 9 grams pierces a board 5 cm thick, while its speed decreases from 600 ms to 200 ms. Find the change in the kinetic energy, work and the value of the resistance, assuming it to be a constant value**

Data: m is the mass of the specified bullet (m = 9 g, in SI m = 0.009 g); s is the thickness of the punched board (s = 5 cm, in the SI system s = 0.05 m); Vp1 is the initial bullet velocity (Vp1 = 600 m / s); Vp2 – final speed (Vp2 = 200 m / s).

1) Calculate the change in kinetic energy: ΔEk = 0.5 * m * (Vp22 – Vp12) = 0.5 * 0.009 * (2002 – 6002) = -1440 J.

2) Let’s find a job: A = -ΔEк = – (- 1440) = 1440 J.

2) We calculate the value of the resistance force: A = Fcopr * s, from where we express: Fcopr = A / s = 1440 / 0.05 = 28.8 * 103 N.

Answer: The kinetic energy of the indicated bullet changed by -1440 J; the resistance force is 28.8 kN, its work is 1440 J.