a golf club strikes a 0.045kg 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 6800N and in contact with the ball for a distance of 0.010m. with what speed does the ball leave the club?

To find the speed with which the ball leaves the club, we can use the principles of work and energy. The work done by the force on the ball is equal to the change in its kinetic energy.

The work done is given by the formula: work = force * distance * cos(theta), where theta is the angle between the force and the direction of motion. In this case, since the force is parallel to the ball's motion, the angle theta is 0 degrees and cos(0) is 1.

So, the work done on the ball is: work = 6800N * 0.010m * cos(0) = 68 Joules.

The work done on the ball is equal to the change in its kinetic energy. We know that the initial kinetic energy is zero because the ball is at rest initially.

Therefore, the change in kinetic energy is: change in kinetic energy = final kinetic energy - initial kinetic energy = final kinetic energy - 0.

Since work done = change in kinetic energy, we have: 68 Joules = 0.5 * mass * final velocity^2.

Since the mass of the golf ball is 0.045kg, we can rearrange the equation: final velocity^2 = (2 * 68 Joules) / 0.045kg.

Simplifying, we have: final velocity^2 = 3022.22 m^2/s^2.

Taking the square root of both sides, we get: final velocity = √(3022.22 m^2/s^2).

Therefore, the final velocity at which the ball leaves the club is approximately 55 m/s.