An 11.8 g bullet leaves the muzzle of a rifle
with a speed of 578.5 m/s.
What constant force is exerted on the bullet
while it is traveling down the 0.6 m length of
the barrel of the rifle?
Answer in units of N
To find the constant force exerted on the bullet while it is traveling down the length of the barrel, we can use Newton's second law of motion, which states that force (F) is equal to mass (m) multiplied by acceleration (a). In this case, the bullet's mass is given as 11.8 g, which is equivalent to 0.0118 kg.
First, let's find the acceleration of the bullet. We can use the equation for acceleration, which is a = (v - u) / t, where v is the final velocity, u is the initial velocity, and t is the time taken. Given that the bullet leaves the muzzle of the rifle with a speed of 578.5 m/s and travels a distance of 0.6 m, we can assume that it does so in a very short time (t ≈ 0) since it's moving very fast. Therefore, the initial velocity (u) is zero, and the final velocity (v) is 578.5 m/s. Plugging these values into the acceleration equation, we get:
a = (578.5 - 0) / 0
a = 578.5 / 0
However, dividing by zero is undefined, so this approach doesn't work. Instead, let's use the equation: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled.
Since the bullet starts from rest, the initial velocity (u) is zero. We can rearrange the equation to solve for acceleration (a):
a = (v^2 - u^2) / (2s)
a = (578.5^2 - 0^2) / (2 * 0.6)
Simplifying this expression, we have:
a = (334,255.225 - 0) / 1.2
a = 278,546.0208 m/s^2
Now that we have the acceleration, we can calculate the force exerted on the bullet using Newton's second law. The mass of the bullet is 0.0118 kg, and the acceleration is 278,546.0208 m/s^2. Plugging these values into the equation F = ma:
F = 0.0118 kg * 278,546.0208 m/s^2
F ≈ 3,284.6824 N
Therefore, the constant force exerted on the bullet while it is traveling down the length of the barrel of the rifle is approximately 3,284.6824 N.