A 22.4-g bullet is fired from a rifle. It takes 3.37 × 10-3 s for the bullet to travel the length of the barrel, and it exits the barrel with a speed of 740 m/s. Assuming that the acceleration of the bullet is constant, find the average net force exerted on the bullet.
F=ma=m•v/t=...
Well, let's calculate that average net force, shall we?
First, we need to find the initial velocity of the bullet. Since we know the final velocity is 740 m/s and the time it took to reach that speed is 3.37 × 10-3 s, we can use the formula:
v = u + at
Rearranging the formula to solve for the initial velocity (u), we get:
u = v - at
Plugging in the values:
u = 740 m/s - (a)(3.37 × 10-3 s)
Now, we also know the initial velocity is 0 m/s because the bullet starts from rest, so we can set the formula to be:
0 = 740 m/s - (a)(3.37 × 10-3 s)
Let's solve for "a," the acceleration:
a = 740 m/s / (3.37 × 10-3 s)
a = 2.194 × 10^5 m/s^2
Now that we have the acceleration, we can find the net force using Newton's second law:
F = m × a
Plugging in the values:
F = (22.4 g)(2.194 × 10^5 m/s^2)
Converting the mass to kilograms:
F = (0.0224 kg)(2.194 × 10^5 m/s^2)
Finally, we get:
F ≈ 49,134 N
Looks like the average net force exerted on the bullet is approximately 49,134 N. I hope it didn't break a sweat!
To find the average net force exerted on the bullet, we can use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration.
Given that the mass of the bullet, m = 22.4 g = 0.0224 kg
The initial velocity, u = 0 m/s (since the bullet starts from rest at the beginning of the barrel)
The final velocity, v = 740 m/s
The time taken, t = 3.37 × 10^(-3) s
We can find the acceleration using the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
Rearranging the formula, we get a = (v - u) / t.
a = (740 m/s - 0 m/s) / (3.37 × 10^(-3) s)
a = 740 m/s / (3.37 × 10^(-3) s)
a ≈ 219525.47 m/s^2
Now, we can calculate the net force using the formula F = ma, where F is the net force, m is the mass, and a is the acceleration.
F = (0.0224 kg) × (219525.47 m/s^2)
F ≈ 4913.87 N
Therefore, the average net force exerted on the bullet is approximately 4913.87 N.
To find the average net force exerted on the bullet, we can use 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.
First, we need to determine the mass of the bullet. Given that its mass is 22.4 g, we need to convert it to kilograms (kg) because the SI unit for mass is kg. 1 g is equal to 0.001 kg, so the mass of the bullet is 22.4 g x 0.001 kg/g = 0.0224 kg.
Next, we need to calculate the acceleration of the bullet. We can use the equation for acceleration, which is given by a = (v - u) / t, where "v" is the final velocity, "u" is the initial velocity, and "t" is the time taken. In this case, the initial velocity is 0 since the bullet starts from rest in the barrel. So, we can rearrange the equation to solve for acceleration:
a = (v - u) / t
= (740 m/s - 0 m/s) / 3.37 × 10^(-3) s
= 740 m/s / 3.37 × 10^(-3) s
≈ 2.193 × 10^5 m/s^2
Now we have the mass of the bullet (0.0224 kg) and the acceleration (2.193 × 10^5 m/s^2). We can substitute these values into Newton's second law of motion:
F = m * a
= 0.0224 kg * 2.193 × 10^5 m/s^2
≈ 4.91 × 10^3 N
Therefore, the average net force exerted on the bullet is approximately 4,910 Newtons (N).