A(n) 4.8 g bullet leaves the muzzle of a rifle
with a speed of 345.1 m/s.
What constant force is exerted on the bullet
while it is traveling down the 0.5 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 0.5 m length of the barrel, we can use the equation:
Force = Mass x Acceleration
Given:
- Mass of the bullet (m) = 4.8 g = 0.0048 kg
- Speed of the bullet (v) = 345.1 m/s
- Length of the barrel (s) = 0.5 m
First, we need to find the acceleration of the bullet using the formula:
Acceleration = Change in Velocity / Time
The time can be calculated using the equation:
Time = Distance / Velocity
Since the bullet is traveling down the length of the barrel, the distance is given as 0.5 m, and the velocity is given as 345.1 m/s.
Plugging in the values:
Time = 0.5 m / 345.1 m/s
Time ≈ 0.0014513 s
Next, we can find the acceleration using the formula:
Acceleration = (Final Velocity - Initial Velocity) / Time
The initial velocity is 0 m/s (bullet starts from rest), and the final velocity is given as 345.1 m/s.
Plugging in the values:
Acceleration = (345.1 m/s - 0 m/s) / 0.0014513 s
Acceleration ≈ 237908.0727 m/s^2
Finally, we can find the force using the equation:
Force = Mass x Acceleration
Plugging in the values:
Force = 0.0048 kg x 237908.0727 m/s^2
Force ≈ 1143.48035 N
Therefore, the constant force exerted on the bullet while it is traveling down the 0.5 m length of the barrel is approximately 1143.48035 N.
To find the 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 the force acting on an object is equal to the product of its mass and acceleration.
The mass of the bullet is given as 4.8 g, which is equivalent to 0.0048 kg (since there are 1000 grams in a kilogram).
Next, we need to find the acceleration of the bullet. Since the bullet starts from rest at the muzzle of the rifle and achieves a final velocity of 345.1 m/s, we can use the equation of motion:
v^2 = u^2 + 2as
where:
- v is the final velocity (345.1 m/s),
- u is the initial velocity (0 m/s),
- a is the acceleration,
- s is the displacement (0.5 m).
Plugging in the values, we have:
(345.1)^2 = (0)^2 + 2a(0.5)
Simplifying the equation:
119014.01 = a
Therefore, the acceleration of the bullet is approximately 119014.01 m/s^2.
We can now substitute the values of mass (0.0048 kg) and acceleration (119014.01 m/s^2) into Newton's second law:
F = m * a
F = 0.0048 kg * 119014.01 m/s^2
Multiplying the values:
F ≈ 572.06712 N
Therefore, the constant force exerted on the bullet while it is traveling down the 0.5 m length of the barrel of the rifle is approximately 572.07 N.