An 18 g bullet is accelerated in a rifle barrel
61.7 cm long to a speed of 362 m/s.
Use the work-energy theorem to find the
average force exerted on the bullet while it is
being accelerated.
Answer in units of N
56
To find the average force exerted on the bullet, we can use the work-energy theorem, which states that the work done on an object is equal to its change in kinetic energy. In this case, the work done on the bullet is equal to the average force exerted on it multiplied by the distance over which the force acts.
First, let's calculate the initial kinetic energy of the bullet. The kinetic energy is given by the equation:
KE = 1/2 * m * v^2
where KE is the kinetic energy, m is the mass, and v is the velocity.
Given:
Mass of the bullet, m = 18 g = 0.018 kg
Initial velocity, v = 0 (since the bullet initially at rest)
So, the initial kinetic energy of the bullet is:
KE_initial = 1/2 * m * v^2 = 1/2 * 0.018 * 0^2 = 0 Joules
Since the bullet starts from rest, its initial kinetic energy is zero.
Next, let's calculate the final kinetic energy of the bullet. The final kinetic energy is given by:
KE_final = 1/2 * m * v^2
Given:
Final velocity, v = 362 m/s
So, the final kinetic energy of the bullet is:
KE_final = 1/2 * m * v^2 = 1/2 * 0.018 * (362)^2 = 1130.2964 Joules
The change in kinetic energy is the difference between the final and initial kinetic energies:
ΔKE = KE_final - KE_initial = 1130.2964 - 0 = 1130.2964 Joules
Now, we can find the work done on the bullet, which is equal to the change in kinetic energy:
Work = ΔKE = 1130.2964 Joules
The work done on the bullet is equal to the average force exerted on it multiplied by the distance over which the force acts. The distance over which the force acts is given as 61.7 cm = 0.617 m.
So, the work done on the bullet is also equal to the average force multiplied by the distance:
Work = Average Force * Distance
From the above equation, we can solve for the average force:
Average Force = Work / Distance = 1130.2964 Joules / 0.617 m = 1833.15 N
Therefore, the average force exerted on the bullet while it is being accelerated is approximately 1833.15 N.