A bullet of mass 1.8 X 10-3 kg is moving at +500. m/s when it impacts a tree stump. It penetrates into the stump 6.00 centimeters before coming to rest.
1. Assuming the acceleration to be constant, calculate the force (including direction) exerted by the stump on the bullet.
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gfs 345
To calculate the force exerted by the stump on the bullet, we can use Newton's second law of motion, which states that the force on an object is equal to its mass multiplied by its acceleration.
In this case, the bullet's initial velocity is +500 m/s, and it comes to rest after penetrating into the stump for a distance of 6.00 centimeters or 0.06 meters. We can calculate the acceleration of the bullet using the equation for constant acceleration:
v^2 = u^2 + 2as
where:
v = final velocity (0 m/s, as the bullet comes to rest)
u = initial velocity (+500 m/s)
s = displacement (0.06 meters)
Rearranging the equation, we have:
0^2 = (+500)^2 + 2a(0.06)
250000 = 0.12a
a = 250000 / 0.12
a = 2,083,333.33 m/s^2
Now we have the acceleration of the bullet. To calculate the force exerted by the stump, we use Newton's second law:
F = ma
where:
F = force exerted by the stump on the bullet
m = mass of the bullet (1.8 x 10^-3 kg)
a = acceleration of the bullet (2,083,333.33 m/s^2)
Substituting the values:
F = (1.8 x 10^-3) kg * (2,083,333.33 m/s^2)
F ≈ 3,750 N
The force exerted by the stump on the bullet is approximately 3,750 Newtons. The direction of the force is opposite to the direction of the bullet's initial velocity, which means the force is negative or in the opposite direction of the bullet's motion.