Work and Energy Legacy Problem #29 Guided Solution

A good problem solver reads the problem carefully and develops a mental picture of what's going on, maybe even constructs a diagram representing the motion. A good problem solver identifies known and unknown information, and begins to use physics understanding in order to plot out a strategy to get from the known to the unknown. Here in this problem, we're given the motion of Connor, a diver, who jumps off the springboard with an initial upward velocity, having a height of three meters above the water, travels upward and then begins to travel back downwards, hits the water surface, and upon hitting the water surface, 3.0 meters from the board, begins to slow down, finally coming to a stop, 2.15 meters below the water. I constructed a picture showing the initial height, 3.0 meters above the water surface, and the final position, 2.15 meters below the water surface. I labeled the VI for the initial location, 5.94 meters per second, and the VF for that final location underneath the water, 0 meters per second. What I know is that the only time that work is being done upon our diver is when traveling through the water. Work is being done by the water to slow the diver down. All other times, the work that's being done is being done by conservative forces, and as such, is ignored when we do our analysis. So we're going to use this five-term equation to solve for two different unknowns. We're going to solve for the speed at which the diver strikes the water, and then we're going to repeat another calculation to find out the force of water resistance acting upon our diver. So I'm going to begin with part A, finding the speed when the diver hits the water, and my initial state is simply when leaving the diving board with a HI of 3.00 meters and a VI of 5.94 meters per second. Here I'm calling zero height the water surface, such that when the diver hits the water, the HF is zero, and we wish to find the VF. In that five-term equation that you see for this part A, we're going to cancel two terms. PEF, since we're calling the water surface a zero height, and WNC, since we're assuming that there's negligible air resistance doing work, and that the only force doing work during this part of the motion is the force of gravity, a conservative force. So I'll calculate KEI as 1.5 MVI squared, that is 1.5 times 76 times 5.94 meters per second, that quantity squared, and I'll get 1,340.78 joules of energy. Then I'll calculate PE initial using 3.0 for my initial height, and 76 kilograms is my M, and 9.8 newtons per kilogram is my G value, and I get 2,234.4 joules of energy. Together this gives me a total initial energy of 3,575.18 joules. Now this total initial energy will be the kinetic energy upon hitting the water. So I can say 3,575 joules of kinetic energy is equal to 1.5 MVI squared. I can put in 76 for M, and then solve for VF. I get 9.6997 meters per second, rounded to three digits, that's 9.70 meters per second. In Part B, I have to analyze the motion of the diver from the water surface to a location 2.15 meters below the water surface where the diver has finally stopped. To do this, I have to define zero height, and I'm going to define the zero height once more as the water surface. Defining the zero height as the water surface means that the final height is actually below it with a height of negative 2.15 meters. So as I use my five term work and energy equation that you see on this page, what I'm going to do is say KEI plus the PEI is zero, zero initially since we're on the water surface, plus the W and C equal KEF, which would be zero since the diver has finally stopped, and PEF, which is going to be a negative number. Now I'm going to calculate the KEI. Actually, I've already calculated the KEI because it's the same as the initial kinetic energy when the diver left the springboard. We did that in Part A, and it came out to be 3,575.18 joules. That KEI plus the W and C will be equal to the final potential energy. The final potential energy can be calculated as MGHF, putting in M for 76 and G is 90.8, and HF as, get this, negative 2.15 meters since the final height is below the water surface. That gives me a value of 1,601.32 joules for the final potential energy, and that was a negative 1,601.32 joules. Now I have an equation that goes something like this. 3,575.18 plus W and C equal negative 1,601.32 joules. Subtracting the 3,575.18 joules from both sides of the equation would give me W and C equal negative 5,176.50 joules. Now this negative value for work is going to be equal to the F times the D times the cosine of the angle between F and D. We'll presume that the diver goes straight down vertically into the water such that the force is up while the motion is down, and the angle theta in this equation is 180 degrees. So we'll go F times the distance over which the force acts, which is the 2.15 meters, times the cosine of 180 equal negative 5,176.50 joules. Now we can solve for F by dividing through by the 2.15, and we get 2,407.67 newtons. We can round that to three significant digits, and we'd have 2.41 times 10 to the third newtons.