Work and Energy Legacy Problem #22 Guided Solution

Here we are given the description of a volleyball moving through the air at which time it gets spiked and ultimately lands on the ground. The problem steps us through the process of determining the speed of the ball just before it hits the floor on the opponent's side of the net. In the first part of the problem, we're asked to determine the potential energy of the ball before the player spikes it. We're given the mass of the ball, 0.226 kilograms, the height of the ball above the ground, 2.29 meters, and we're given the speed of the ball, 1.06 meters per second. To calculate the potential energy at this location, we need to go mass times 9.8 newtons per kilograms times h. We need to plug into the equation 0.226 for m and 2.29 for h, and when we do, we calculate that the potential energy of the ball at this location is 5.0719 joules. In part B, we're asked to determine the kinetic energy of the ball at this location at which it's moving, 1.06 meters per second. So we need to use the equation ke equals one-half mv squared, where the m is 0.226 kilograms and the v is 1.06. Remembering to square the v would give us as a value for kinetic energy, 0.1270 joules. In part C, we're asked to sum the two forms of mechanical energy, and thus to determine the total mechanical energy at this point before Page actually spikes it. Also summing the answers to part A and part B gives us a total mechanical energy value of 5.1989 joules. In part D, we're asked to determine the total mechanical energy of the ball just before it hits the floor on the opponent's side of the net. The fact that Page spikes the ball means that the energy that it will have before hitting the floor will be bigger than the energy that it had before the spike. It had 5.1989 joules of energy, but then Page spikes it doing 9.89 joules of work upon it, thus contributing to the original energy and giving it a final energy that is equal to the sum of the original energy plus the work done. That gives us 15.0889 joules of energy just before striking the ground. In the last step of the problem, we're asked to determine the speed of the ball at this location just before hitting the ground. So we need to equate the 15.0889 joules of energy to one-half mv squared, since all the energy now is in the form of kinetic, since there's no height just above the ground. So we set 15.0889 joules equal to one-half times 0.226 kilograms times v squared. Solving for v gives us 11.555 meters per second as our value, and we can round that to three digits, 11.6 meters per second.