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.
Well, that bullet must have really wanted to get cozy with that tree stump, didn't it? Let's see if we can calculate the force exerted by the stump.
To find the force, we can use Newton's second law, which states that force equals mass times acceleration (F = ma). Since the acceleration is assumed to be constant, we can calculate it using the equation for linear motion.
The bullet initially moves at a speed of 500 m/s, and it comes to rest after penetrating the stump for 0.06 meters (or 6.00 centimeters). We can use the equation v² = u² + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the displacement.
By substituting the values and rearranging the equation, we get a = (v² - u²) / 2s. Plugging in the values, we find a = (0 - 500²) / (2 * 0.06) = -4,166.67 m/s² (the negative sign indicates deceleration).
Now, to find the force, we multiply the mass of the bullet (1.8 x 10^-3 kg) by the acceleration we just calculated: F = (1.8 x 10^-3 kg) * (-4,166.67 m/s²) = -7.5 N.
So, the force exerted by the stump on the bullet is approximately 7.5 Newtons, and it acts in the opposite direction of the bullet's initial velocity. In other words, the stump says, "Nope, you're not coming any further!"
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 acting on an object is equal to the mass of the object multiplied by its acceleration.
Given:
Mass of the bullet (m) = 1.8 x 10^(-3) kg
Initial velocity of the bullet (u) = +500 m/s
Final velocity of the bullet (v) = 0 m/s (since it comes to rest)
Distance covered by the bullet (s) = 6.00 cm = 0.06 m (since 1 cm = 0.01 m)
We need to find the acceleration (a) first. We can use the equation of motion:
v^2 = u^2 + 2as
Rearranging the equation, we get:
a = (v^2 - u^2) / (2s)
Substituting the known values:
a = (0^2 - 500^2) / (2 * 0.06)
a = -1250000 m^2/s^2
The negative sign indicates that the acceleration is in the opposite direction to the initial velocity, i.e., in the direction opposite to the bullet's motion.
Now, we can calculate the force (F) using Newton's second law:
F = m * a
Substituting the known values:
F = (1.8 x 10^(-3)) * (-1250000)
F = -2250 N
The negative sign indicates that the force exerted by the stump on the bullet is in the opposite direction to the initial motion of the bullet.
Therefore, the force exerted by the stump on the bullet is -2250 N.
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 exerted on an object is equal to the mass of the object multiplied by its acceleration:
F = m * a
In this case, we are given the mass of the bullet (m = 1.8 x 10^-3 kg) and we need to find the acceleration. To do this, we can use the following kinematic equation:
v^2 = u^2 + 2 * a * s
where v is the final velocity (zero in this case), u is the initial velocity (+500 m/s), a is the acceleration, and s is the distance penetrated into the stump (6.00 cm = 0.06 m).
Plugging in the known values, we can rearrange the equation to solve for the acceleration:
0 = (500 m/s)^2 + 2 * a * 0.06 m
Simplifying:
0 = 250000 m^2/s^2 + 0.12 a
Rearranging for a:
0.12 a = -250000 m^2/s^2
a = (-250000 m^2/s^2) / 0.12
a ≈ -2.08 x 10^6 m/s^2
The negative sign indicates that the acceleration is in the opposite direction of the bullet's initial velocity. Therefore, the force exerted by the stump on the bullet is given by:
F = (1.8 x 10^-3 kg) * (-2.08 x 10^6 m/s^2)
F ≈ -3.74 N
The force exerted by the stump on the bullet is approximately 3.74 N and the direction of the force is in the opposite direction of the bullet's initial velocity.
Vf^2=Vi^2+2ad
but a= Force/mass
solve for force.