a 56-g tennis ball is accelerated horizontally from rest to a speed of 39 m/s. assuming that the acceleration is uniform when the racque is applied over a distance of 0.55m, what is the magnitude of the force exerted on the ball by the racquet?
You can use the work kinetic energy theorem which says the change in the kinetic energy of a body, or .5mv^2, is aways equal to the work done to the body, and work is equal to force times the distance the force is applied over. In this case the ball started from rest so it had no kinetic energy so the total change in kinetic energy is equal to the final kinetic energy so
.5mv^2 = Fd
plugging in the numbers we have
.5*.056*(39)^2 = F*.55
then solve for F so
F = 77.4 Newtons. The grams were converted to kilograms to be in standard units.
Oct 9th, 2014
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