A bullet weighing 10 g, flying at a speed of 400, piercing a 5 cm thick board 5 cm thick, halved the speed.

A bullet weighing 10 g, flying at a speed of 400, piercing a 5 cm thick board 5 cm thick, halved the speed. Determine the resistance force of the board to the bullet movement.

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

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 board; V1 = 400 m / s; V2 = 200 m / s.

On the other hand, work A is equal to the product of the resistance force F and the thickness of the board s = 5 cm = 0.05 m:

A = Fs.

Resistance force F:

F = A / s = M (V1 ^ 2 – V2 ^ 2) / (2 s) = 0.01 × (400 ^ 2 – 200 ^ 2) / (2 × 0.05) = 12000 N = 12 kN.

Answer: 12 kN.