A 28.6-g bullet is fired from a rifle. It takes 2.36 × 10-3 s for the bullet to travel the length of the barrel, and it exits the barrel with a speed of 710 m/s. Assuming that the acceleration of the bullet is constant, find the average net force exerted on the bullet.

To find the average net force exerted on the bullet, we can use Newton's second law of motion, which states that the force acting on an object is equal to its mass multiplied by its acceleration:

F = m * a

In this case, we are given the mass of the bullet (m = 28.6 g = 0.0286 kg) and the time it takes to travel the length of the barrel (t = 2.36 × 10^-3 s).

To calculate the acceleration, we can use the formula for average acceleration:

a = Δv / Δt

where Δv is the change in velocity and Δt is the change in time. In this case, the bullet starts from rest (initial velocity, u = 0 m/s) and exits the barrel with a velocity of 710 m/s (final velocity, v = 710 m/s).

Δv = v - u = 710 m/s - 0 m/s = 710 m/s
Δt = t = 2.36 × 10^-3 s

Substituting these values into the formula, we can find the acceleration:

a = Δv / Δt = (710 m/s) / (2.36 × 10^-3 s) = 3.02 × 10^5 m/s^2

Now, we can calculate the average net force exerted on the bullet:

F = m * a = (0.0286 kg) * (3.02 × 10^5 m/s^2) = 8637.2 N

Therefore, the average net force exerted on the bullet is 8637.2 N.

The bullet goes from rest to a speed of 710 m/s in 2.36 * 10^-3 s

v = a*t

where v is the speed, a is the acceleration, and t is time

710 = a * 2.36*10^-3

Solve for a

Force = m*a

where m is mass