Work and Energy Legacy Problem #18 Guided Solution

This problem involves the motion of IMA scared through a roller coaster loop. We're given information about our speed and our height at the very top of the loop, and we're asked to ultimately determine our speed at the bottom of the loop. In the process of solving the problem, we're told that we can assume that there are negligible losses of energy due to friction and air resistance, and as such, the total mechanical energy of IMA would be conserved. The problem steps us through the solution in getting from the givens to ultimately the speed at the bottom of the loop. The first part of the problem asks us to calculate IMA's kinetic energy at the top of the loop. We understand kinetic energy to be dependent upon the mass of an object and the speed at which it moves. According to the equation, KE equals one-half MV squared. The M in this problem is 56.2 kilograms, and the V is 12.8 meters per second. So, at the top of the loop, I can substitute those numbers into the equation, and I can calculate IMA's kinetic energy. When I do, I get 4,603.9 joules. In part B of the problem, we're asked to calculate the potential energy at the top of the loop. I understand potential energy to be dependent upon the mass of the object and the height of the object. According to the equation, PE equals MGH. Once more, I can substitute 56.2 kilograms into the equation for M, 9.8 for newtons per kilogram for G, and 19.5 meters for the H. When I do, I get a result of 10,739.8 joules as my amount of potential energy at the top of the loop. In part C, we're asked to calculate IMA's total mechanical energy at the bottom of the loop, again, assuming there's no loss of energy from the top to the bottom. And, as such, her total mechanical energy at the bottom of the loop will be the same as it is at the top of the loop. So, we need to find the total mechanical energy at the top of the loop, which would be simply the sum of the two forms of energy. The PE of 10,739.8 joules plus the KE of 4,603.9 joules gives us a total amount of energy there of 15,343.7 joules. So, the total mechanical energy at the bottom of the loop is 15,343.7 joules, and all of this energy is now in the form of kinetic. So, to find the speed at the bottom of the loop, we have to associate this figure for total energy to one-half MV squared, since that's kinetic energy. So, we say 15,343.7 joules is equal to one-half times the mass of 56.2 times V squared. We have to solve this equation for V. Doing so means we have to multiply both sides of the equation by two, and then divide through by 56.2. This would give us V squared equal 546.04. Now that we have V squared by itself, we need to take the square root of both sides, and we get 23.3675, or rounded to three digits, is 23.4 meters per second.