A golf club strikes a 0.061-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 7760 N, and is in contact with the ball for a distance of 0.014 m. With what speed does the ball leave the club?
W= change in Kinetic Energy
F*s = 1/2mv^2
Solve by v:
v= sqrt((2Fs)/m))
substitution:
v= sqrt((2*7760N*0.014m)/0.061kg)
v=59.68 m/s
To find the speed with which the golf ball leaves the club, we can use the principle of work done. Work done is given by the formula:
Work = Force × Distance × cos(θ)
In this case, the force acting on the ball is 7760 N and the distance over which this force acts is 0.014 m. The angle θ is not given, but since the force acts parallel to the ball's motion, the angle is 0 degrees and hence the cosine of 0 degrees is 1. So we have:
Work = 7760 N × 0.014 m × 1
Work = 108.64 Joules
Now, we can use the work-energy principle, which states that the work done on an object is equal to the change in its kinetic energy. The initial kinetic energy of the ball is zero since it is at rest. Therefore:
Work = ΔKE
where ΔKE is the change in kinetic energy. The final kinetic energy of the ball is given by the formula:
KE = (1/2)mv^2
where m is the mass of the ball and v is its final velocity. Rearranging this formula, we get:
v^2 = 2KE / m
Substituting the value of work done (108.64 Joules) for ΔKE and the mass of the ball (0.061 kg) for m, we can solve for v:
v^2 = 2(108.64 J) / 0.061 kg
v^2 = 3559.34 m^2/s^2
Finally, taking the square root of both sides, we find:
v ≈ 59.74 m/s
Therefore, the ball leaves the club with a speed of approximately 59.74 m/s.