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

55$m / s$

Physics 101 Mechanics

Chapter 6

Work and Energy

Work

Kinetic Energy

University of Michigan - Ann Arbor

Simon Fraser University

University of Sheffield

University of Winnipeg

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so we have work. Net is equal to Delta K and there's only one force here which is the force applied by the club catcher equal to 1/2 and re squared and the initial kinetic energy off the ball is zero. So we can now start putting some numbers and so we have 6800 is the amount of force which is applied over a distance of your 68000.10 meters. And that should equal 1/2 the mass off the ball, a 0.45 kilograms and times V squared. So when you saw this for V, you get V is equal to 55.0 meter per second.

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