A 6.5 g bullet leaves the muzzle of a rifle with
a speed of 347 m/s.
What total constant force is exerted on the
bullet while it is traveling down the 0.83 m
long barrel of the rifle?
a = (V^2-Vo^2) / 2d.
a = ((347)^2-0) / 1.66. = 72536 m/s^2.
F = ma = 0.0065kg * 72,536 = 472 N.
Well, you definitely don't want to mess with a bullet leaving the muzzle of a rifle. It's kind of like a really angry bee leaving its hive! Now, to calculate the total constant force exerted on the bullet while it's traveling down the barrel, we can use the equation:
Force = Mass x Acceleration
First, we need to figure out the acceleration of the bullet. We can use the kinematic equation:
v^2 = u^2 + 2as
where v is the final velocity (which is 347 m/s in this case), u is the initial velocity (0 m/s since it starts from rest), a is the acceleration, and s is the distance traveled (0.83 m).
Plugging in the values, we can solve for acceleration:
347^2 = 0 + 2a(0.83)
a = 347^2 / (2 * 0.83)
a ≈ 71181 m/s^2
Next, we can calculate the force using the mass of the bullet, which is 6.5 grams (or 0.0065 kg):
Force = 0.0065 kg x 71181 m/s^2
Force ≈ 462.6 N
So, the total constant force exerted on the bullet while it's traveling down the 0.83 m long barrel of the rifle is approximately 462.6 Newtons. That's no joke!
To find the total constant force exerted on the bullet, we need to make use of Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. In this case, we need to calculate the acceleration of the bullet.
Given:
Mass of the bullet (m) = 6.5 g = 0.0065 kg
Speed of the bullet (v) = 347 m/s
Distance traveled by the bullet inside the barrel (s) = 0.83 m
We know that acceleration (a) is given by the equation:
v^2 = u^2 + 2as
Where:
u = initial velocity (which is 0 in this case, as the bullet starts from rest within the barrel).
Rearranging the equation to solve for acceleration (a):
a = (v^2 - u^2) / (2s)
a = (v^2 - 0) / (2s)
a = v^2 / (2s)
Substituting the given values:
a = (347 m/s)^2 / (2 * 0.83 m)
a = 60209 / 1.66
a = 36286.75 m/s^2
Now, we can use Newton's second law to find the force (F):
F = m * a
F = 0.0065 kg * 36286.75 m/s^2
F = 235.7 N
Therefore, the total constant force exerted on the bullet while it is traveling down the 0.83 m long barrel of the rifle is approximately 235.7 newtons.
To find the total constant force exerted on the bullet while it is traveling down the barrel of the rifle, we can use Newton's second law of motion, which states that force (F) is equal to the mass (m) of an object multiplied by its acceleration (a).
First, we need to find the acceleration of the bullet while traveling down the barrel. We can use the kinematic 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 displacement.
Given:
Initial velocity, u = 0 m/s (as the bullet starts from rest inside the barrel)
Final velocity, v = 347 m/s
Displacement, s = 0.83 m
Rearranging the equation and substituting the given values, we have:
(347 m/s)^2 = (0 m/s)^2 + 2a(0.83 m)
120,409 m^2/s^2 = 0 + 1.66a
120,409 m^2/s^2 = 1.66a
Now, to find the acceleration (a), we rearrange the equation:
a = 120,409 m^2/s^2 / 1.66
a ≈ 72,815 m^2/s^2
Now that we have the acceleration, we can find the force (F) using Newton's second law:
F = m * a
Given the mass of the bullet, m = 6.5 g = 0.0065 kg
F = 0.0065 kg * 72,815 m^2/s^2
F ≈ 474 N
Therefore, the total constant force exerted on the bullet while it is traveling down the 0.83 m long barrel of the rifle is approximately 474 Newtons.