Question

read the following passage and answer the questions.
roller coasters
the first
oller coaster\ was originally built to transport coal through the mountains of mauch chunk, pennsylvania. the railroad changed its cargo to passengers in 1873 when a more efficient railway for coal was built. people were able to pay a small fee to coast down a treacherous mountain terrain. this new attraction came to be known as the \switzerland of america.\ the coaster drew more than 35,000 customers a year. in 1976, the coasters skeleton was declared a historic monument. although its original purpose was not for entertainment, the railway has gone into the record books as the longest and highest coaster ever built. the total drop was 1,126 feet and over 18 miles out. the roller coaster continued to evolve with added loops, scenic experiences, and other added components to appeal to thrill seekers.
it can be said that learning the physics of roller coasters can enhance your experience on one. there are many variations on roller coaster design. some involve massive loops, sharp turns, and others steep drops and backwards motions.
a rollercoaster works with gravity and shockingly there are no motors used to power it during the ride. roller coasters simply descend down steep hills, and convert stored gravitational potential energy into kinetic energy. this happens as the coaster gains speed.
5 lets say i wanted to determine the velocity of the coaster but did not have the mass of the coaster. what formula could i derive to find the velocity knowing what i know about kinetic energy and gravitational potential energy?
a ( v = mgh )
b ( v=\frac{1}{2}mv^{2} )
c ( v=sqrt{2gh} )
d ( v = 2gh )

Explanation:

Step1: Recall energy conservation

By the law of conservation of energy, gravitational potential energy \(U = mgh\) (where \(m\) is mass, \(g\) is acceleration due to gravity, \(h\) is height) is converted into kinetic energy \(K=\frac{1}{2}mv^{2}\) (where \(v\) is velocity). So \(mgh=\frac{1}{2}mv^{2}\).

Step2: Solve for \(v\)

Since \(m\) appears on both sides of the equation \(mgh=\frac{1}{2}mv^{2}\), we can cancel out \(m\) (because \(m
eq0\)). We get \(gh = \frac{1}{2}v^{2}\). Then multiply both sides by \(2\) to get \(2gh=v^{2}\). Taking the square - root of both sides gives \(v=\sqrt{2gh}\).

Answer:

C. \(V = \sqrt{2gh}\)