As a small bullet leaves a gun, a net force is applied to shoot it through a barrel. If the bullet has a mass of 34.67 grams and leaves the barrel at a speed of 249.7 m/s. If the barrel of the gun is only 13 cm, what force is applied to the bullet as it leaves the gun?

F = 333.1 N

Explanation:

Force: This is defined as the product of mass of a body and its acceleration. The S. I unit of force is Newton (N).

From Newton's equation of motion.

v² = u² + 2as ... Equation 1

Where v = final velocity of the bullet, u = initial velocity of the bullet, s = length of the gun, a = acceleration of the bullet

Making a the subject of formula in the equation above

a = (v² - u²) / 2s ... Equation 2

Given: v = 249.7 m/s, u = 0 m/s (The bullet was at rest before the gun was shot), s = 13 cm = (13/100) m = 0.13 m.

Substituting these values into equation 2

a = (2497.7-0) / (2*0.13)

a = 2497.7/0.26

a = 9606.54 m/s²

Force = Mass * acceleration

F = M a ... Equation 3

Where M = mass of the bullet, F = force applied to the bullet

Given: m = 34.67 g = (34.67/1000) = 0.03467 kg, a = 9606.54 m/s²

Substituting these values into Equation 3

F = 0.03467 * 9606.54

F = 333.1 N