Answer:
The constant force exerted on the bullet is 1590.87 N.
Explanation:
It is given that,
Mass of the bullet, m = 8.6 g
Initial speed of the bullet, u = 0 Â Â Â Â Â
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,
[tex]v^2-u^2=2ad[/tex]
[tex]v^2=2ad[/tex]
[tex]a=\dfrac{v^2}{2d}[/tex]
[tex]a=\dfrac{(430.1)^2}{2\times 0.5}[/tex]
[tex]a=184986.01\ m/s^2[/tex]
Let F is the force exerted. It is given by :
[tex]F=ma[/tex]
[tex]F=8.6\times 10^{-3}\times 184986.01[/tex]
F = 1590.87 N
So, the constant force exerted on the bullet is 1590.87 N. Hence, this is the required solution. Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â
