hi i am not sure how to do this...thank you
a man is going to pitch a ball. he moves the ball in a circular path of radius 0.6 m, constantly applying a force of 30N to the ball in the same direction as the ball's velocity along the circular path. at the highest point on the path(point A), the velocity of the ball is 15 m\s. what will the speed of the ball be when the man lets go of the ball at the lowest point along the circular path(point B)?
I know that only forces thta are parallel to the motion of an object do work on the object, so in this problem, the man's force of 30N always does work on the ball-- I think this problem has something to do with finding the potential and kinetic energy of the ball at point A and then adding these values (KE + PE) and setting this value for the total energy equal to the KE kinetic energy at point B (since the potenial energy will = 0 at point B)
CONTINUED:
*i forgot to mention that the ball has a mass 5kg
Basically I'm confused as to how to calculate the potential energy of the ball at point A;
I know that PE = mgh
so PE = (5kg) (9.8 m\s^2) (h) but would h equal the diameter of the circular path? but this doesnt seem right to me.....
To find the speed of the ball when the man lets go of it at the lowest point (point B) along the circular path, we can use the conservation of mechanical energy. As you correctly noted, the total energy (potential energy + kinetic energy) at point A will be equal to the kinetic energy at point B.
Let's break down the problem step by step:
1. Determine the potential energy at point A:
At the highest point (point A), the velocity of the ball is 15 m/s. Since the ball is moving in a circular path, the velocity vector at point A is perpendicular to the radius vector. Therefore, we can consider the entire mechanical energy at point A as potential energy.
Potential Energy at A = mgh
where m is the mass of the ball, g is the acceleration due to gravity, and h is the height at point A from the lowest point (point B). Since the radius of the circular path is given as 0.6 m, the height at point A can be calculated by subtracting the radius from the total diameter (2r).
Height at A = 2r - r = r
Height at A = 0.6 m
2. Determine the kinetic energy at point B:
At the lowest point (point B), the ball will have both kinetic energy and potential energy. However, since we are interested in its speed (which only relates to kinetic energy), we can neglect the potential energy at point B.
Kinetic Energy at B = 0.5mv^2
where m is the mass of the ball, and v is the velocity at point B (which we need to find).
3. Apply conservation of mechanical energy:
According to the conservation of mechanical energy, the total mechanical energy at point A is equal to the kinetic energy at point B.
Total Energy at A = Kinetic Energy at B
mgh = 0.5mv^2
Since mass (m) is common to both sides of the equation, we can cancel it out:
gh = 0.5v^2
Now we can substitute the given values:
g = 9.8 m/s^2 (acceleration due to gravity)
The height (h) at point A is the same as the radius (r):
r = 0.6 m
4. Solve for the velocity at point B:
Substituting the known values into the equation:
(9.8 m/s^2)(0.6 m) = 0.5v^2
Simplifying the equation:
5.88 m^2/s^2 = 0.5v^2
Dividing both sides by 0.5:
11.76 m^2/s^2 = v^2
Taking the square root of both sides:
v ≈ 3.43 m/s
So, when the man lets go of the ball at the lowest point along the circular path (point B), the speed of the ball will be approximately 3.43 m/s.