A bullet weighing 10 g moves at a speed of 800 m / s, hits a shaft 8 cm thick, and at the same time its speed decreases

A bullet weighing 10 g moves at a speed of 800 m / s, hits a shaft 8 cm thick, and at the same time its speed decreases by 2 times. The average value of the resistance force acting on the bullet is ..

The decrease in the kinetic energy of the bullet before and after passing through the shaft is equal to the work of the resistance force of the board A:

A = A1 – A2 = m (V1 ^ 2 – V2 ^ 2) / 2,

where m is the mass of the bullet, m = 10 g = 0.01 kg;

V1, V2 – bullet speed, respectively, before and after the shaft; V1 = 800 m / s; V2 = 800/2 = 400 m / s.

On the other hand, work A is equal to the product of the average resistance force F by the shaft thickness s = 8 cm = 0.08 m:

A = Fs.

Force F:

F = A / s = m (V1 ^ 2 – V2 ^ 2) / (2 s) = 0.01 × (800 ^ 2 – 400 ^ 2) / (2 × 0.08) = 30,000 N = 30 kN.

Answer: 30 kN.