Question

A 5.8 g bullet leaves the muzzle of a rifle with a speed of 314 m/s. What force (assured constant) is exerted on the bullet while it is traveling down the 0.81 m long barrel of the rifle?

A) 3610 N

B) 3252 N

C) 2888 N

D) 2564 N

E) 2176 N

Answer 1

The force exerted on the bullet while it is traveling down the barrel is approximately 2250.77 N.

Step-by-step explanation:

The force exerted on the bullet while it is traveling down the barrel can be calculated using Newton's second law, which states that force is equal to mass times acceleration. The acceleration of the bullet can be found using the formula for average acceleration, which is given by the change in velocity divided by the time taken:

a = (vf - vi) / t

Where vf is the final velocity of the bullet, vi is the initial velocity of the bullet, and t is the time taken. In this case, the bullet starts from rest and reaches a velocity of 314 m/s, so the formula becomes:

a = 314 m/s / 0.81 m = 387.65 m/s^2

Using the formula for force:

F = m * a

Where F is the force, m is the mass of the bullet, and a is the acceleration, we can calculate the force:

F = 5.8 g * 387.65 m/s^2 = 2250.77 N

Therefore, the force exerted on the bullet while it is traveling down the barrel is approximately 2250.77 N.