A bullet of mass 10 g has a speed of 500 m/s. a) What is its kinetic energy? b) if it penetrates a block of wood a distance of 10 cm before coming to rest, what is the average frictional force stopping the bullet? A second bullet identical to the first is shot through a sheet of wood 5 cm thick. c) With what speed does the second bullet exit the wood?
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To answer these questions, we'll use the principles of conservation of energy and Newton's laws of motion.
a) To find the kinetic energy of the bullet, we can use the formula:
Kinetic energy (K.E.) = (1/2) * mass * velocity^2
1. Convert the mass of the bullet from grams to kilograms:
Mass = 10 g = 10 * (1/1000) kg = 0.01 kg
2. Plug in the values into the formula:
K.E. = (1/2) * 0.01 kg * (500 m/s)^2
= 0.01 kg * 250000 m^2/s^2
= 2500 joules
So, the kinetic energy of the bullet is 2500 joules.
b) To find the average frictional force stopping the bullet, we need to calculate the work done by the friction force on the bullet.
The work done by the friction force is equal to the change in kinetic energy.
Work done (W) = Change in kinetic energy (ΔK.E.)
Since the bullet comes to rest, the change in kinetic energy is equal to the initial kinetic energy (2500 J).
W = ΔK.E. = 2500 J
The work done is also equal to the force applied (frictional force) multiplied by the distance over which it acts (displacement).
W = Frictional force * Distance
We are given the distance the bullet penetrates the wood, which is 10 cm or 0.1 m.
Frictional force * 0.1 m = 2500 J
Therefore, the average frictional force stopping the bullet is:
Frictional force = 2500 J / 0.1 m = 25000 N
So, the average frictional force stopping the bullet is 25000 Newtons.
c) To find the speed of the second bullet as it exits the wood, we can again use the principle of conservation of energy.
We know the thickness of the wood is 5 cm or 0.05 m.
The second bullet loses its initial kinetic energy as it penetrates the wood. This loss of energy is equal to the work done by the friction force:
Work done (W) = Change in kinetic energy (ΔK.E.)
W = Frictional force * Distance
Here, the distance over which the frictional force acts is the thickness of the wood (0.05 m), and the frictional force is 25000 N (as calculated earlier).
W = 25000 N * 0.05 m
= 1250 J
The loss in kinetic energy is 1250 J.
Since the bullet comes to rest after exiting the wood, the final kinetic energy is zero.
So, we have:
Initial kinetic energy - Loss in kinetic energy = Final kinetic energy
(1/2) * mass * velocity^2 - 1250 J = 0
Since the second bullet is identical to the first one, the mass and initial kinetic energy are the same (0.01 kg and 2500 J, respectively).
(1/2) * 0.01 kg * velocity^2 - 1250 J = 0
Simplifying the equation:
0.005 kg * velocity^2 = 1250 J
Velocity^2 = 1250 J / 0.005 kg
= 250000 m^2/s^2
Taking the square root of both sides to find the velocity:
Velocity = √(250000 m^2/s^2)
= 500 m/s
Therefore, the second bullet exits the wood with a speed of 500 m/s.