Answer: 1872 N
Explanation:
This problem can be solved by using one of the Kinematics equations and Newton's second law of motion:
[tex]V^{2}=V_{o}^{2} + 2ad[/tex] (1)
[tex]F=ma[/tex] (2)
Where:
[tex]V=611.9 m/s[/tex] is the bullet's final speed (when it leaves the muzzle)
[tex]V_{o}=0[/tex] is the bullet's initial speed (at rest)
[tex]a[/tex] is the bullet's acceleration
[tex]d=0.8 m[/tex] is the distance traveled by the bullet before leaving the muzzle
[tex]F[/tex] is the force
[tex]m=8 g \frac{1 kg}{1000 g}=0.008 kg[/tex] is the mass of the bullet
Knowing this, let's begin by isolating [tex]a[/tex] from (1):
[tex]a=\frac{V^{2}}{2d}[/tex] (3)
[tex]a=\frac{(611.9 m/s)^{2}}{2(0.8 m)}[/tex] (4)
[tex]a=234013.5063 m/s^{2} \approx 2.34(10)^{5} m/s^{2}[/tex] (5)
Substituting (5) in (2):
[tex]F=(0.008 kg)(2.34(10)^{5} m/s^{2})[/tex] (6)
Finally:
[tex]F=1872 N[/tex]
