Roller coasters are an easy way to demonstrate the Law of Conservation of Energy, but everything obeys this law. The law states that energy cannot be created or destroyed, only transformed from one form to another. This can be shown using the equation \(\small\mathsf{ \Sigma \text{TE}_{before} = \Sigma \text{TE}_{after} }\), where TE is the total energy of the defined system. For the example of a roller coaster, among others, the total energy before is the sum of the gravitational potential energy and the kinetic energy. Look at each of the problems below. Solve them on your own and then come back and check your work.
The Fahrenheit
The Storm Runner
The Blob Jump
Fahrenheit, which opened in 2008, is one of Hersheypark's most exciting roller coasters. Its main feature is a 97-degree drop at the beginning of the ride. At the start of Fahrenheit, thrill-seekers are pulled straight up, 121 feet (36.9 m) in the air. They stop briefly at the top of the first hill before plummeting down the 97-degree drop. If the mass of the fully-loaded coaster is 8350 kg, what is the car's velocity at the bottom of the hill? (Ignore air resistance and friction) How much work is done to the coaster car?
Using the concepts of the Law of Conservation of Energy, you know that the sum of the energy at any point along the path of the roller coaster will be the same.
\(\mathsf{ \Sigma \text{TE}_{before} = \Sigma \text{TE}_{after} }\)
\(\mathsf{ \text{GPE}_{before} + \text{KE}_{before} = \text{GPE}_{after} + \text{KE}_{after} }\)
\(\mathsf{ (mgh)_{before} + 0 = 0 + (\frac{1}{2}mv^2)_{after} }\)
\(\mathsf{ (8350 \text{ kg})(9.81 \text{ m/s}^2)(36.9 \text{ m}) = \frac{1}{2}(8350 \text{ kg})v^2 }\)
\(\mathsf{v = \sqrt{\frac{3022608.15 \text{ J}}{4175 \text{ kg}}} }\)
\(\mathsf{v = 26.9 \text{ m/s} }\)
There was 3022608.15 J of work done on the car to change its speed.
The Storm Runner, another very exciting roller coaster at Hersheypark, is the first hydraulic launch roller coaster to feature inversions. Storm Runner launches riders from 0 to 72 miles-per-hour (32 m/s) in two seconds. If the mass of the loaded Storm Runner cart is 4200 kg, how high above the ground could the coaster travel? (Ignore air resistance and friction) How much work is done to the coaster car?
Using the concepts of the Law of Conservation of Energy, you know that the sum of the energy at any point along the path of the roller coaster will be the same.
\(\mathsf{ \Sigma \text{TE}_{before} = \Sigma \text{TE}_{after} }\)
\(\mathsf{ \text{GPE}_{before} + \text{KE}_{before} = \text{GPE}_{after} + \text{KE}_{after} }\)
\(\mathsf{ 0 + (\frac{1}{2}mv^2)_{before} = (mgh)_{after} + 0 }\)
\(\mathsf{ \frac{1}{2}(4200 \text{ kg})(32 \text{ m/s})^2 = (4200 \text{ kg})(9.81 \text{ m/s}^2)(h) }\)
\(\mathsf{ 2150400 \text{ J} = (41202 \text{ kg m/s}^2)h }\)
\(\mathsf{ h = \frac{2150400 \text{ J}}{41202 \text{ kg m/s}^2} }\)
\(\mathsf{ h = 52 \text{ m}}\)
There was -2150400 J of work done on the car to change its height. The work done is negative because the velocity of the car decreased.
A 70.0 kg man jumps from 6.50 meters in the air to a blob jump (as shown in the picture). When the first guy hits the blob the second guy who was laying on the blob is launched into the air. If all of the first guy's energy is tranferred to the second (neglecting air resistance), what is the maximum height the second guy could reach if his mass is 62.5 kg? How much work was done on the second man?
Using the concepts of the Law of Conservation of Energy, you know that the sum of the energy at any point along the path of the roller coaster will be the same.
\(\mathsf{ \Sigma \text{TE}_{before} = \Sigma \text{TE}_{after} }\)
\(\mathsf{ \text{GPE}_{before} + \text{KE}_{before} = \text{GPE}_{after} + \text{KE}_{after} }\)
\(\mathsf{ (mgh)_{man 1 before} + 0 = (mgh)_{man 2 after} + 0 }\)
\(\mathsf{ (70.0 \text{ kg})(9.81 \text{ m/s}^2)(6.50 \text{ m}) = (62.5 \text{ kg})(9.81 \text{ m/s}^2)(h) }\)
\(\mathsf{ 4463.55 \text{ J} = (613.125 \text{ kg m/s}^2)(h) }\)
\(\mathsf{ h = \frac{4463.55 \text{ J}}{613.125 \text{ kg m/s}^2} }\)
\(\mathsf{ h =7.3 \text{ m}}\)
There was -4463.55 J of work done on the second man (i.e. to change his velocity and then his height. It is negative during the "up" travel of the man since his velocity is decreasing.
