Question

An 8 g bullet leaves the muzzle of a rifle with

a speed of 611.9 m/s.
What constant force is exerted on the bullet
while it is traveling down the 0.8 m length of
the barrel of the rifle?
Answer in units of N.

Answer 1

Answer: 1872 N

Step-by-step explanation:

This problem can be solved by using one of the Kinematics equations and Newton's second law of motion:

`V^(2)=V_(o)^(2) + 2ad` (1)

`F=ma` (2)

Where:

`V=611.9 m/s` is the bullet's final speed (when it leaves the muzzle)

`V_(o)=0` is the bullet's initial speed (at rest)

`a` is the bullet's acceleration

`d=0.8 m` is the distance traveled by the bullet before leaving the muzzle

`F` is the force

`m=8 g (1 kg)/(1000 g)=0.008 kg` is the mass of the bullet

Knowing this, let's begin by isolating
`a` from (1):

`a=(V^(2))/(2d)` (3)

`a=((611.9 m/s)^(2))/(2(0.8 m))` (4)

`a=234013.5063 m/s^(2) \approx 2.34(10)^(5) m/s^(2)` (5)

Substituting (5) in (2):

`F=(0.008 kg)(2.34(10)^(5) m/s^(2))` (6)

Finally:

`F=1872 N`