Question

question 25 of 27
a roller coaster has a mass of 450 kg. it sits at the top of a hill with height 49 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. 25.4 m/s
b. 93.9 m/s
c. 45.8 m/s
d. 31.0 m/s

Explanation:

Step1: Apply conservation of energy

At the top, the roller - coaster has only potential energy $U = mgh$. At the bottom, it has only kinetic energy $K=\frac{1}{2}mv^{2}$. Since energy is conserved ($U = K$), we have $mgh=\frac{1}{2}mv^{2}$.

Step2: Solve for velocity $v$

First, 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, solve for $v$: $v = \sqrt{2gh}$. Given $g = 9.8\ m/s^{2}$ and $h = 49\ m$, we substitute these values into the formula. $v=\sqrt{2\times9.8\times49}=\sqrt{960.4}\approx 31.0\ m/s$.

Answer:

D. 31.0 m/s