A bullet weighing 7.9 g flies out under the action of powder gases from a bore 45 cm long at a speed of 54 km / h

A bullet weighing 7.9 g flies out under the action of powder gases from a bore 45 cm long at a speed of 54 km / h. Calculate the average force of the pressure of the propellant gases. Disregard the friction of the bullet against the barrel wall.

m = 7.9 g = 0.0079 kg.

V0 = 0 m / s.

V = 54 km / h = 15 m / s.

S = 45cm = 0.45m.

F -?

According to 2 Newton’s law, the force F, which acts on a body, is equal to the product of the body’s mass m by its acceleration a: F = m * a.

The acceleration of the bullet a in the barrel is expressed by the formula: a = (V ^ 2 – V0 ^ 2) / 2 * S, where V is the speed of the bullet when it leaves the barrel, V0 is the initial speed of the bullet, S is the movement of the bullet in the barrel, that is barrel length.

Since the bullet begins its movement from a state of rest V0 = 0 m / s, the formula will take the form: a = V ^ 2/2 * S.

The average force of gas pressure F is found by the formula: F = m * V ^ 2/2 * S.

F = 0.0079 kg * (15 m / s) ^ 2/2 * 0.45 m = 1.975 N.

Answer: the average force of gas pressure on a bullet is F = 1.975 N.