# A bullet of mass 15g leaves the barrel of a rifle with a speed of 800 m/s. If the length of the barrel is 75 cm, what is the the average force that accelerates the bullet?

Jul 4, 2015

The bullet is accelerated by a force of 6400 N.

#### Explanation:

In order to be able to determine the force that accelerates the bullet, you first need to determine the bullet's acceleration.

You know what the length of the gun barrel and what the bullet's exit velocity are, which means that you can write

v_"exit"^2 = underbrace(v_0^2)_(color(blue)("=0")) + 2 * a * l

Solve the above equation for $a$ to get

$a = {v}_{\text{exit}}^{2} / \left(2 \cdot l\right)$

a = (800""^2"m"^(cancel(2))/"s"^2)/(2 * underbrace(0.75cancel("m"))_(color(blue)("=75 cm"))) = "426,666.7 m/s""^2

This means that the force that's accelerating the bullet is equal to

$F = m \cdot a$

$F = \underbrace{\text{0.015 kg")_(color(blue)("=15 g")) * 426,666.7"m"/"s"^2 = 6400underbrace(("kg" * "m")/"s"^2)_(color(blue)("=Newton")) = color(green)("6400 N}}$