Question

a roller coaster has a mass of 275 kg. it sits at the top of a hill with height 85 m. if it drops from this hill, how fast is it going when it reaches the bottom? (assume there is no air resistance or friction.)

a. 73.4 m/s
b. 18.0 m/s
c. 40.8 m/s
d. 29.9 m/s

Explanation:

Step1: Apply the conservation of mechanical energy

The roller - coaster's initial potential energy \(U = mgh\) is converted into kinetic energy \(K=\frac{1}{2}mv^{2}\) at the bottom of the hill. Since \(U = K\) (no non - conservative forces like friction or air resistance), we have \(mgh=\frac{1}{2}mv^{2}\).

Step2: Solve for \(v\)

Cancel out the mass \(m\) from both sides of the equation \(mgh=\frac{1}{2}mv^{2}\). We get \(gh=\frac{1}{2}v^{2}\). Then \(v=\sqrt{2gh}\). Given \(g = 9.8\ m/s^{2}\) and \(h = 85\ m\), substitute the values: \(v=\sqrt{2\times9.8\times85}\).

$$v=\sqrt{1666}\approx 40.8\ m/s$$

Answer:

C. \(40.8\ m/s\)