A 5.4 g bullet leaves the muzzle of a rifle with a speed of 334 m/s. What force (assumed constant) is exerted on the bullet while it is traveling down the 0.95 m long barrel of the rifle?
___N
d = rt,
t = d/r = 0.95 m / 334 m/s = 0.0028 s,
V = at,
a = V/t = 334 m/s / 0.0028 s = 117606 m/s^2,
F = ma = 0.0054 kg * 117606 m/s^2 =
6351 N.
State the law of acceleration in its full form and give examples of how it works
To determine the force exerted on the bullet while it is traveling down the barrel of the rifle, we can use Newton's second law of motion. The formula is:
Force = Mass × Acceleration
First, we need to calculate the acceleration of the bullet as it travels down the barrel. We can use the formula:
Acceleration = (Final Velocity - Initial Velocity) / Time
Since the bullet leaves the muzzle of the rifle with a speed of 334 m/s and assuming we neglect air resistance, we can assume that the final velocity is 334 m/s. The initial velocity is 0 m/s because the bullet starts from rest. Therefore:
Acceleration = (334 m/s - 0 m/s) / Time
To find the time it takes for the bullet to travel down the barrel, we can use the formula:
Time = Distance / Speed
In this case, the distance is the length of the barrel, which is 0.95 m. The speed is 334 m/s. Therefore:
Time = 0.95 m / 334 m/s
Now, we can substitute the time back into the acceleration formula:
Acceleration = (334 m/s - 0 m/s) / (0.95 m / 334 m/s)
Next, we can calculate the mass of the bullet:
Mass = 5.4 g / 1000 (to convert grams to kilograms)
Finally, we can calculate the force exerted on the bullet:
Force = Mass × Acceleration
Substitute the values into the formula to find the force in Newtons.