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?
force*time=mass*velocity solve for velocity
But how do you go about finding time since it is not given and you are only given a magnitude of the force and not the velocity
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 golf ball can be calculated using the formula:
Work = Force * Distance * cos(theta)
In this case, the force is 6400 N and the distance is 0.012 m. Since the force is parallel to the motion of the ball, the angle theta between the force and displacement is 0 degrees, so cos(theta) = 1.
Therefore, the work done on the ball is:
Work = 6400 N * 0.012 m * 1 = 76.8 J
According to the work-energy principle, the work done is equal to the change in kinetic energy of the ball:
Work = Change in Kinetic Energy
The initial kinetic energy of the ball is zero since it is at rest, so the change in kinetic energy is equal to the final kinetic energy.
Using the formula for kinetic energy:
Kinetic Energy = (1/2) * mass * velocity^2
Since the ball has a mass of 0.041 kg and the initial velocity is zero, the final kinetic energy is:
76.8 J = (1/2) * 0.041 kg * velocity^2
Now, we can solve for the final velocity of the ball.
76.8 J = 0.02005 kg * velocity^2
Dividing both sides by 0.02005 kg gives:
velocity^2 = 76.8 J / 0.02005 kg
velocity^2 = 3840 m^2/s^2
Taking the square root of both sides gives the final velocity:
velocity = sqrt(3840 m^2/s^2) ≈ 61.98 m/s
Therefore, the ball leaves the club with a speed of approximately 61.98 m/s.