A 25 g bullet is fired from a rifle with a barrel of length 65 cm. The bullet starts from rest at
one end of the barrel and experiences an average force of 1200 N due to the expanding gases as
it moves down the length of the barrel.
a) Calculate the work done on the bullet by the expanding gases over the length of the
barrel.
b) Assuming that this is the only force doing work on the bullet, use the Work-Energy
theorem to determine the speed of the bullet as it leaves the gun.
work=force*distance
1/2 mv^2=force*distance
a) To calculate the work done on the bullet by the expanding gases over the length of the barrel, we can use the formula for work:
Work = Force * Distance * cos(theta)
Where:
- Force is the average force exerted by the gases (1200 N)
- Distance is the length of the barrel (65 cm or 0.65 m)
- theta is the angle between the direction of the force and the direction of the displacement (which is 0 degrees since the force and displacement are along the same line)
Therefore, the work done on the bullet by the expanding gases is:
Work = 1200 N * 0.65 m * cos(0 degrees)
Work = 1200 N * 0.65 m * 1 (since cos(0 degrees) = 1)
Work = 780 J
b) According to the Work-Energy theorem, the work done on an object is equal to the change in its kinetic energy:
Work = Change in Kinetic Energy
Change in Kinetic Energy = (1/2) * m * v^2
Where:
- m is the mass of the bullet (25 g or 0.025 kg)
- v is the velocity of the bullet
Rearranging the equation, we can solve for the velocity of the bullet as it leaves the gun:
v = sqrt((2 * Work) / m)
v = sqrt((2 * 780 J) / 0.025 kg)
v = sqrt(62400 J / kg)
v = 249.6 m/s
Therefore, the speed of the bullet as it leaves the gun is approximately 249.6 m/s.
To answer these questions, we need to use the concepts of work done and the work-energy theorem.
a) Work done on an object is defined as the product of the force acting on the object and the distance over which the force is applied. In this case, the force acting on the bullet is 1200 N and the distance is the length of the barrel, which is 65 cm (or 0.65 m). So, we can calculate the work done as follows:
Work = Force * Distance
Work = 1200 N * 0.65 m
Work = 780 J
Therefore, the work done on the bullet by the expanding gases over the length of the barrel is 780 Joules.
b) The work-energy theorem states that the work done on an object is equal to the change in its kinetic energy. In this case, the work done by the expanding gases is the only force doing work on the bullet. Therefore, we can equate the work done to the change in kinetic energy.
Work = Change in Kinetic Energy
The initial kinetic energy of the bullet is zero since it starts from rest. The final kinetic energy of the bullet can be calculated using the formula:
Kinetic Energy = (1/2) * mass * velocity^2
Here, the mass of the bullet is 25 g (or 0.025 kg), and we need to find the final velocity.
Using the equation,
Work = Change in Kinetic Energy
780 J = (1/2) * 0.025 kg * (final velocity)^2
Re-arranging the equation, we get:
(final velocity)^2 = (2 * 780 J) / 0.025 kg
(final velocity)^2 = 62400 m^2/s^2
Taking the square root on both sides, we find:
final velocity = √(62400 m^2/s^2)
Therefore, the speed of the bullet as it leaves the gun is approximately 249.8 m/s.