Question

a roller coaster has a mass of 425 kg. it sits at the top of a hill with height 66 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. 36.0 m/s
b. 56.8 m/s
c. 12.5 m/s
d. 91.3 m/s

Explanation:

Step1: Apply conservation of energy

Initial potential energy $U = mgh$ and final kinetic energy $K=\frac{1}{2}mv^{2}$. Since there is no air - resistance or friction, $U = K$. So $mgh=\frac{1}{2}mv^{2}$.

Step2: Solve for velocity $v$

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 $h = 66\ m$, substitute these values: $v=\sqrt{2\times9.8\times66}$.

Step3: Calculate the value of $v$

$v=\sqrt{1293.6}\approx36.0\ m/s$.

Answer:

A. 36.0 m/s