A 8kg ball is thrown up with an initial speed of 20m/s. Suppose there is a constant air resistence of 30N opposing the motion during its entire up and down path. How long will it take to get to the highest point in its path? how high is the highest point? How much work was done by the air resistence during the entire path up and down? What is the kinectic energy when it gets back to its original point?
I will be glad to critique your work.
I don't how to start it off
When the ball reaches its highest point H, potential energy will be
M g H = (1/2)MVo^2 - f*H
where f is the friction force, 30N.
H = (1/2)MVo^2/(Mg + f)
Solve for H.
The time to get to that high point will be
T = H/Vo, since it will be decelerating at a constant rate.
The KE when it hits the ground will be M g H - f*H. This will equal
(1/2)MVo^2 - 2fH
Thank you. one more question what would be the acceleration on the way up? on the way down?
acceleration due to gravity
The acceleration is
g + (f/M) on the way up, and g - (f/M) on the way down. It is pointed down in both cases.
They are different because both g and friction point down on the way up, but are in opposite directions on the way down. For that reason, it arrives at the ground at a lower velocity than it took off.
To answer the first question - how long will it take to get to the highest point in its path - we can use the concept of projectile motion.
The first step is to calculate the acceleration of the ball while it is moving upwards. We know that the force of gravity acting on it is equal to its weight, which is given by the formula weight = mass x acceleration due to gravity. Therefore, weight = 8kg x 9.8m/s² = 78.4N.
Since there is a constant air resistance opposing the motion, we need to subtract the force of air resistance from the weight to get the net force acting on the ball. Net force = weight - air resistance = 78.4N - 30N = 48.4N.
Next, we can use the second law of motion, which states that force equals mass times acceleration (F = m x a), to find the acceleration of the ball. Rearranging the formula, acceleration = force / mass = 48.4N / 8kg = 6.05m/s².
Now we can use the kinematic equation v = u + at, where v is final velocity, u is initial velocity, a is acceleration, and t is time. In this case, the final velocity at the highest point is 0 (since the ball reaches its maximum height and then starts to fall). The initial velocity is 20m/s, and the acceleration is -6.05m/s² (negative because it is opposing the motion). Plugging these values into the equation:
0 = 20m/s + (-6.05m/s²) x t.
Simplifying the equation, we get:
6.05t = 20.
t = 20 / 6.05.
t ≈ 3.31 seconds.
Therefore, it will take approximately 3.31 seconds to get to the highest point in its path.
To find the highest point, we can use the formula for displacement in vertical motion, given by the equation:
s = ut + (1/2)at².
Here, u is the initial velocity, t is the time taken to reach the highest point (which we found to be 3.31 seconds), a is the acceleration. Since the ball is moving upwards, the acceleration is -6.05m/s² (opposite to the direction of motion). Plugging these values into the equation:
s = (20m/s)(3.31s) + (1/2)(-6.05m/s²)(3.31s)².
s = 66.2m - 32.02m.
s ≈ 34.18 meters.
Therefore, the highest point in its path is approximately 34.18 meters above the initial point.
To calculate the work done by air resistance during the entire path up and down, we need to determine how far the ball travels in total (up and then back down). The total distance covered is twice the distance from the initial point to the maximum height, which is 2 x 34.18 meters = 68.36 meters.
Work = force x distance.
Since the force of air resistance is constant and equal to 30N, and the distance is 68.36 meters, we can calculate the work done:
Work = 30N x 68.36m.
Work ≈ 2050.8 Joules.
Therefore, the work done by air resistance during the entire path up and down is approximately 2050.8 Joules.
Finally, to calculate the kinetic energy when the ball gets back to its original point, we can use the formula for kinetic energy:
Kinetic energy = (1/2)mv².
Here, m is the mass of the ball (8kg), and v is the velocity. The velocity at the highest point is 0, so we need to find the velocity at the original point. To do this, we can use the equation v = u + at, where u is the initial velocity (20m/s) and a is the acceleration (-6.05m/s²). Plugging these values into the equation:
v = 20m/s - (6.05m/s²)(3.31s).
v = 20m/s - 20.02m/s.
v ≈ -0.02m/s.
Since the ball is returning to the original point, we take the velocity as negative. Now we can calculate the kinetic energy:
Kinetic energy = (1/2)(8kg)(-0.02m/s)².
Kinetic energy = 0.002 Joules.
Therefore, the kinetic energy when the ball gets back to its original point is approximately 0.002 Joules.