Calculate Force (N) on Bullet in 0.8m Rifle Barrel

[PepeF.]

a 5.3g bullet leaves the muzzle of a rifle with a speed of 618.9 m/s.

What constants force (N) is exerted on the bullet while it is traveling down the 0.8m lengh of the barrel of the riffle?

 

[Tzim]

What did you try to do?

 

[kerimek]

To calculate the force exerted on the bullet while it is traveling down the 0.8m length of the rifle barrel, we can use the equation F=ma, where F is the force, m is the mass of the bullet, and a is the acceleration of the bullet. In this case, the mass of the bullet is given as 5.3g, which we will need to convert to kilograms (1g = 0.001kg). This gives us a mass of 0.0053kg.

The acceleration of the bullet can be calculated using the formula v^2 = u^2 + 2as, where v is the final velocity (618.9 m/s), u is the initial velocity (0 m/s), and s is the distance traveled (0.8m). Rearranging this equation, we get a = (v^2 - u^2)/2s. Plugging in the given values, we get an acceleration of 190,830 m/s^2.

Now, we can plug these values into the equation F=ma to calculate the force exerted on the bullet. F = (0.0053kg)(190,830 m/s^2) = 1,012.47 N.

Therefore, while the bullet is traveling down the 0.8m length of the rifle barrel, it is experiencing a force of approximately 1,012.47 Newtons. This force is necessary in order to accelerate the bullet from rest to a speed of 618.9 m/s in a short distance of 0.8m. It is important to note that this calculation assumes ideal conditions and does not take into account factors such as air resistance or friction within the barrel.