A golf club strikes a 0.040-kg golf ball in order to launch it from the tee. For simplicity, assume that the average net force applied to the ball acts parallel to the ball’s motion, has a magnitude of 7950 N, and is in contact with the ball for a distance of 0.011 m. With what speed does the ball leave the club?
Am I the only one seeing this in 2021, when this was posted in 2014?????
a = F/m = 7950/0.040 = 198,750 m/s^2.
V^2 = Vo^2 + 2a*d
Vo^2 = V^2 - 2a*d = 0 + 2*198750*0.011 =
4373
Vo = 66.1 m/s.
bro got ratio'd on jishka
Well, let's see, a golf club hitting a golf ball. Sounds like the beginning of a classic joke!
Why did the golf ball go to the spa? It needed to relax and get a good swing!
But to answer your question, we can use the work-energy principle. The work done on the ball is equal to the change in its kinetic energy. So we have:
Work = Change in Kinetic Energy
The work done on the ball is equal to the force applied multiplied by the distance over which the force is applied:
Work = Force × Distance
Let's plug in the given values:
Work = 7950 N × 0.011 m
Now, the change in kinetic energy is equal to the final kinetic energy minus the initial kinetic energy of the ball. Since the ball starts from rest, its initial kinetic energy is zero:
Change in Kinetic Energy = 1/2 m v² - 0
Now we can equate the two equations:
7950 N × 0.011 m = 1/2 × 0.040 kg × v²
Solving for v, we find:
v = sqrt((7950 N × 0.011 m) / (1/2 × 0.040 kg))
v ≈ 23.78 m/s
So, the ball leaves the club with a speed of approximately 23.78 m/s. Now, that's some serious swinging power!
To determine the speed at which the golf ball leaves the club, we can use the concept of work and energy.
First, let's calculate the work done by the force applied to the ball. The work done is given by the formula:
Work = Force * Distance * cos(theta)
Where:
- Force is the magnitude of the force applied to the ball (7950 N),
- Distance is the distance over which the force is applied (0.011 m), and
- theta is the angle between the force and the direction of motion (which is 0 degrees in this case since the force is parallel to the ball’s motion).
Since cos(0) = 1, we can simplify the formula:
Work = Force * Distance
Plugging in the values, we get:
Work = 7950 N * 0.011 m = 87.45 J
The work done on an object is equal to the change in kinetic energy, so:
Work = Change in Kinetic Energy
The initial kinetic energy of the ball is zero since it starts from rest. Therefore:
Change in Kinetic Energy = Final Kinetic Energy
We can calculate the final kinetic energy using the formula:
Kinetic Energy = (1/2) * mass * velocity^2
Plugging in the values, we have:
(1/2) * 0.040 kg * velocity^2 = 87.45 J
Simplifying the equation:
0.020 kg * velocity^2 = 87.45 J
Dividing both sides by 0.020 kg:
velocity^2 = 87.45 J / 0.020 kg
velocity^2 = 4372.5 m^2/s^2
Taking the square root of both sides to solve for velocity:
velocity ≈ √(4372.5 m^2/s^2) ≈ 66.1 m/s
Therefore, the ball leaves the club with a speed of approximately 66.1 m/s.