Question

1. -/1 points a 4.4 - g bullet leaves the muzzle of a rifle with a speed of 330 m/s. what force (assumed constant) is exerted on the bullet while it is traveling down the 0.77 - m - long barrel of the rifle? n resources read it

Explanation:

Step1: Convert mass to SI unit

The mass of the bullet $m = 4.4\ g=4.4\times10^{- 3}\ kg$.

Step2: Use the kinematic - equation $v^{2}=v_{0}^{2}+2ax$

The bullet starts from rest, so $v_{0} = 0\ m/s$, $v = 330\ m/s$ and $x = 0.77\ m$.
We can solve for the acceleration $a$:
\[

$$\begin{align*} v^{2}&=v_{0}^{2}+2ax\\ 330^{2}&=0 + 2a\times0.77\\ a&=\frac{330^{2}}{2\times0.77}\\ a&=\frac{108900}{1.54}\\ a& = 70714.29\ m/s^{2} \end{align*}$$

\]

Step3: Use Newton's second - law $F = ma$

\[

$$\begin{align*} F&=m\times a\\ F&=(4.4\times 10^{-3})\times70714.29\\ F&=311.14\ N \end{align*}$$

\]

Answer:

$311.14$