Question

The bullet starts at rest in the gun. An 8.6 g bullet leaves the muzzle of a rifle with a speed of 430.1 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 1

The constant force exerted on the bullet is 1590.87 N.

Step-by-step explanation:

It is given that,

Mass of the bullet, m = 8.6 g

Initial speed of the bullet, u = 0

Final speed of the bullet, v = 430.1 m/s

We need to find the force exerted on the bullet while it is traveling down the 0.5 m length of the barrel of the rifle. Let a is the acceleration of the bullet. So,

`v^2-u^2=2ad`

`v^2=2ad`

`a=(v^2)/(2d)`

`a=((430.1)^2)/(2* 0.5)`

`a=184986.01\ m/s^2`

Let F is the force exerted. It is given by :

`F=ma`

`F=8.6* 10^(-3)* 184986.01`

F = 1590.87 N

So, the constant force exerted on the bullet is 1590.87 N. Hence, this is the required solution.