Work and Energy Legacy Problem #21 Guided Solution

Here we're given a description of Jerome riding a very thrilling amusement park ride. During the ride, riders start from rest and are accelerated in 3.8 seconds to speeds up to 120 mph along a level section of track. Once they've finished being accelerated along this level section, they go into a very steep 420 foot hill. Upon reaching the top of the hill, they curve and they go back down. But part of the problem that we're concerned about is the section in which Jerome accelerates and then begins to go up this vertical hill. We're given a lot of information. First of all, we're told that during the acceleration period along the level section of track that the riders go from 0 to 53.6 mph. We're told that Jerome's mass is 102 kg. And in part A of the problem, we're asked to calculate Jerome's kinetic energy before he's actually accelerated. We wish we would have more problems like this, since Jerome's at rest, his original kinetic energy would be zero joules. In part B, we're asked to determine what Jerome's kinetic energy is after the 3.8 second period of acceleration. To do that, we need to use the final kinetic energy, the final speed, which is 53.6 mph and the mass of 102 kg. We say that ke equals 1 half mv squared. And we substitute 102 in for m, and we substitute 53.6 in for v. When we do that, we get a kinetic energy of 146,521 joules. In part C of the problem, we're asked to calculate the potential energy possessed by Jerome when he reaches the top of this 420 foot hill. To find the potential energy, you always need to know the mass and the height. Here we know the height to be 420 feet. That's not a great unit for calculating potential energy, so we'll need to first convert that to units of meters. Doing so involves taking 420 feet and multiplying by the conversion factor, 1.0 meters divided by 3.28 feet. That gives us 128.0488 meters. Now we can use that figure for height in meters in order to calculate Jerome's potential energy. We would say pe equals mgh, where once more the m is 102, the g is 9.8 newtons per kilogram, and the h is 128.0488 meters. Now that gives us 127,998 joules. Now we have the kinetic energy at the bottom, thus the total mechanical energy at the bottom of this hill, and we have the potential energy at the top of the hill. We're asked to calculate Jerome's kinetic energy at the top of this vertical section. We'll once more presume that the total amount of mechanical energy is conserved, and in doing that, we know that the total mechanical energy, the sum of the two forms, kinetic and plus potential, must equal the original amount we had before going into the hill. That was 146,521 joules. So subtracting the pe from this total amount gives us the ke, and it comes out to be 18,523 joules. And that would be our answer to step d. In step e, we finally get around to calculating the speed at the top of this vertical section. So we equate the kinetic energy, 18,523 joules, to 1 half mv squared. The m is 102, so we could isolate the v and get it by itself by multiplying both sides of the equation by 2, and dividing through by 102. That gives us v squared equals 363.2, and we can take the square root of both sides, and that gives us 19.06 meters per second.