A bullet of mass 0.0019 kg moving at 487 m/s embeds itself in a large fixed piece of wood and travels 0.55 m before coming to rest. Assume that the deceleration of the bullet is constant.
What force is exerted by the wood on the bullet?
Consider the work done by the bullet on the block. It gets turned into heat, and equals the initial kinetic energy of the bullet.
Force x distance = Initial Kinetic energy
F = (1/2)M V^2/0.55 = 410 N
To find the force exerted by the wood on the bullet, we can use Newton's second law of motion, which states that the force exerted on an object is equal to the product of its mass and acceleration.
First, we need to find the acceleration experienced by the bullet. The bullet comes to rest, so its final velocity is 0. We are given the initial velocity of the bullet (487 m/s), the distance traveled (0.55 m), and we can assume the deceleration is constant.
We can use the following equation to find the acceleration (a):
v^2 = u^2 + 2as
Where:
- v is the final velocity (0 m/s)
- u is the initial velocity (487 m/s)
- a is the acceleration
- s is the distance traveled (0.55 m)
Rearranging the equation, we have:
a = (v^2 - u^2) / (2s)
Plugging in the values:
a = (0^2 - 487^2) / (2 * 0.55)
a = (-487^2) / 1.1
Calculating this, we find:
a ≈ -117,849 m/s^2
Note: The negative sign indicates that the acceleration is in the opposite direction to the initial velocity, meaning the bullet is decelerating.
Now that we have the acceleration of the bullet, we can use Newton's second law to find the force exerted by the wood on the bullet.
F = ma
F = (0.0019 kg) * (-117,849 m/s^2)
Calculating this, we find:
F ≈ -224.9131 N
The negative sign indicates that the force exerted by the wood on the bullet is in the opposite direction to the initial motion of the bullet. So, the magnitude of the force exerted by the wood on the bullet is approximately 224.9131 Newtons.