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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?

  • Physics - ,

    W= change in Kinetic Energy
    F*s = 1/2mv^2
    Solve by v:
    v= sqrt((2Fs)/m))
    v= sqrt((2*7760N*0.014m)/0.061kg)
    v=59.68 m/s

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