A bullet of mass 0.0022 kg moving at 568 m/s
impacts a large fixed block of wood and travels
5.2 cm before coming to rest.
Assuming that the deceleration of the bullet
is constant, find the force exerted by the wood
on the bullet.
Answer in units of kN
Please show the steps to get the answer, I want to learn!
To find the force exerted by the wood on the bullet, you can use Newton's second law of motion, which states that force (F) is equal to mass (m) times acceleration (a), or F = ma.
Given:
Mass of the bullet (m) = 0.0022 kg
Initial velocity of the bullet (u) = 568 m/s
Distance traveled by the bullet (s) = 5.2 cm = 0.052 m
Deceleration of the bullet (a) = ?
First, we need to calculate the deceleration of the bullet using the formula:
v^2 = u^2 + 2as
Rearranging the formula, we have:
a = (v^2 - u^2) / (2s)
Substituting the known values:
a = (0 - (568 m/s)^2) / (2 * 0.052 m)
Simplifying:
a = (-322624 - 322624) / 0.104 = -6195323.08 m/s^2
The negative sign indicates that the bullet is experiencing deceleration.
Now, we can find the force exerted by the wood on the bullet using Newton's second law:
F = ma
Substituting the values:
F = 0.0022 kg * -6195323.08 m/s^2
Simplifying:
F = -13612.11 N
The given value is in newtons (N), but we need to convert it to kilonewtons (kN).
1 kN = 1000 N
F = -13612.11 N / 1000 = -13.61 kN
So, the force exerted by the wood on the bullet is approximately -13.61 kN (negative sign indicates that the force is opposing the motion).