A golf club strikes a 0.041-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 6400 N, and is in contact with the ball for a distance of 0.012 m. With what speed does the ball leave the club?
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To find the speed at which the ball leaves the club, we can use the principle of work and energy. The work done on an object is equal to the change in its kinetic energy.
The work done on the golf ball can be calculated using the formula:
Work = Force * Distance * cosθ
Where:
- Force is the magnitude of the net force applied to the ball (6400 N).
- Distance is the distance over which the force is applied (0.012 m).
- θ is the angle between the direction of the force and the direction of motion (assuming they are parallel, cosθ = 1).
Now we can calculate the work:
Work = 6400 N * 0.012 m * 1 = 76.8 joules
The work done on the golf ball is equal to the change in its kinetic energy:
Work = ΔK.E.
The initial kinetic energy of the golf ball is zero because it starts from rest:
Initial K.E. = 0
The final kinetic energy of the golf ball after leaving the club is:
Final K.E. = 1/2 * m * v^2
Where:
- m is the mass of the golf ball (0.041 kg).
- v is the final speed of the golf ball.
Since the initial kinetic energy is zero, we can now set up the equation:
76.8 J = 1/2 * 0.041 kg * v^2
Simplifying the equation:
76.8 J = 0.0205 kg * v^2
Divide both sides by 0.0205 kg:
v^2 = 76.8 J / 0.0205 kg
v^2 ≈ 3751.22 m^2/s^2
Take the square root of both sides to find v:
v ≈ √3751.22 m^2/s^2
v ≈ 61.25 m/s
Therefore, the ball leaves the club with a speed of approximately 61.25 m/s.