A 4.2-g bullet leaves the muzzle of a rifle with a speed of 338 m/s. What force (assumed constant) is exerted on the bullet while it is traveling down the 0.83-m-long barrel of the rifle

289.05N

Explanation:

The mass of the bullet is 4.2-g bullet, if we convert it to "kg" we have

4.2-g bullet/1000

= 0.0042kg

The speed of the bullet leaving the missle is 338 m/s.

The Lenght of the barrel= 0.83-m-long

From kinematics, v²= u² +2al

V= final speed when it leaves the barell=

338 m/s

U= intial speed=0

L= length of the barrel travelled by the bullet

V²= 2al( initial speed=0)

If we make "a" subject of the formula we have

a=v²/2l

a= (338)²/(2×0.83)

=68821.687 m/s²

We can now calculate the force acting on the body using the Expression below

F=ma

Where F= force acting on the body

m= mass

a= Acceleration

F= 0.0042kg×68821.687

= 289.05N

Therefore, the force exerted on the bullet while it is traveling down the 0.83-m-long barrel of the rifle is 289.05N