A 21.3-g bullet is fired from a rifle. It takes 3.77 × 10-3 s for the bullet to travel the length of the barrel, and it exits the barrel with a speed of 779 m/s. Assuming that the acceleration of the bullet is constant, find the average net force exerted on the bullet.
Vf^2=Vi^2 + 2ad
but a= Force/mass
d=length of barrel, d=avgvelocity*time
d=1/2 Vf*timeinbarrel
solve for force.
Vf^2=2Force/mass*1/2 Vf*timeinbarrel
force=Vf*mass /timeinbarrel
To find the average net force exerted on the bullet, we can use Newton's second law of motion, which states that force is equal to mass multiplied by acceleration:
F = m * a
Given:
Mass of the bullet (m) = 21.3 g = 0.0213 kg
Time taken (t) = 3.77 × 10^(-3) s
Final velocity (v) = 779 m/s
First, let's calculate the acceleration (a):
Since we know that a = (v - u) / t, where u is the initial velocity (0 m/s in this case since the bullet starts from rest), we can substitute the given values to find the acceleration:
a = (779 m/s - 0 m/s) / (3.77 × 10^(-3) s)
a = 779 m/s / (3.77 × 10^(-3) s)
Now, we can substitute the calculated value of acceleration into Newton's second law equation to find the force (F):
F = m * a
F = 0.0213 kg * (779 m/s / (3.77 × 10^(-3) s))
Calculating this expression will give us the average net force exerted on the bullet.