What is the force exerted on a bullet leaving a rifle?

[physicsquest]

Homework Statement

A 5.0g bullet leaves the muzzle of a rifle with a speed of 320 m/s. What force (assumed constant) is exerted on the bullet while it is traveling down the 0.82 m long barrel of the rifle?

Homework Equations

$$\Sigma$$F=ma

The Attempt at a Solution

 

[Donaldos]

It might be simpler to use the work-energy theorem here:

$$W=\Delta E_k$$

 

[tomcue]

To determine the force exerted on the bullet while traveling down the barrel of the rifle, we can use the equation \SigmaF=ma, where \SigmaF represents the net force, m represents the mass of the bullet, and a represents the acceleration of the bullet. We know that the mass of the bullet is 5.0g, or 0.005kg, and the acceleration of the bullet can be calculated using the equation v^2=u^2+2as, where v is the final velocity (320 m/s), u is the initial velocity (0 m/s), and s is the distance traveled (0.82 m). Plugging in these values, we get a = (320^2)/2(0.82) = 391707.32 m/s^2. Therefore, using \SigmaF=ma, the force exerted on the bullet while traveling down the barrel of the rifle is 0.005kg x 391707.32 m/s^2 = 1958.54N. This is the force that propels the bullet out of the rifle and allows it to travel at a speed of 320 m/s. It is important to note that this is an assumed constant force, as there may be other factors such as air resistance that can affect the actual force exerted on the bullet.