A 4.2 g bullet leaves the muzzle of a rifle with

a speed of 372.4 m/s.
What constant force is exerted on the bullet
while it is traveling down the 0.5 m length of
the barrel of the rifle?
Answer in units of N

V^2 = Vo^2 + 2a*d

a = (V^2-Vo^2)/2d
a = (372.4^2-0)/1 = 138,682 m/s^2

F = m * a = 0.0042 * 138,682 = 582.5 N.

To find the constant force exerted on the bullet while it is traveling down the length of the rifle barrel, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a).

In this case, we need to find the acceleration of the bullet. We can use the formula for acceleration:

a = (final velocity - initial velocity) / time

Since the bullet is just leaving the muzzle of the rifle, we can assume that the initial velocity is zero. Also, we are not provided with the time it takes for the bullet to travel down the length of the barrel, but we can find the time using the given information.

The formula to find time is:

t = distance / velocity

Given:
Mass of the bullet (m) = 4.2 g = 0.0042 kg
Initial velocity (u) = 0 m/s
Final velocity (v) = 372.4 m/s
Distance (d) = 0.5 m

First, let's find the time it takes for the bullet to travel down the length of the barrel:

t = d / v
t = 0.5 m / 372.4 m/s
t ≈ 0.001343 s

Now that we have the time, we can find the acceleration:

a = (v - u) / t
a = (372.4 m/s - 0 m/s) / 0.001343 s
a ≈ 277581.55 m/s²

Now, we can substitute the values of mass (m) and acceleration (a) into the formula for force (F):

F = m * a
F = 0.0042 kg * 277581.55 m/s²
F ≈ 1166.47 N

Therefore, the constant force exerted on the bullet while it is traveling down the length of the barrel is approximately 1166.47 N.