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?
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To find the speed at which the ball leaves the club, we can use the work-energy principle. The work done on an object is equal to the change in its kinetic energy.
The work done on the ball can be calculated using the formula:
Work = Force × Distance × cos(θ)
In this case, the force applied to the ball is given as 7950 N, the distance is given as 0.011 m, and the angle (θ) between the force and the direction of motion is assumed to be 0°, so cos(θ) = 1.
Thus, the work done on the ball is:
Work = 7950 N × 0.011 m × 1 = 87.45 J
According to the work-energy principle, the work done on the ball is equal to the change in its kinetic energy. So the work done is also equal to (1/2)mv^2, where m is the mass of the ball and v is its final velocity.
87.45 J = (1/2) × 0.040 kg × v^2
Now, we can solve for v^2:
v^2 = (87.45 J × 2) / 0.040 kg
v^2 = 4372.5 m^2/s^2
Taking the square root of both sides to find the velocity:
v = sqrt(4372.5 m^2/s^2)
v = 66.12 m/s
Therefore, the ball leaves the club with a speed of approximately 66.12 m/s.