Question

problem reference 6.1
a 1,500 kg frictionless roller coaster starts from rest at the top of an 18.0 m hill. the car travels to the bottom of the hill and continues up the next hill that is 10.0 m high.
what is the speed of the car at the top of the 10.0 m hill?
18 m
10.0 m
22.2 m/s
25.9 m/s
18.9 m/s
12.5 m/s

Explanation:

Step1: Apply conservation of mechanical energy

The total mechanical energy \(E = E_{k}+E_{p}\), where \(E_{k}=\frac{1}{2}mv^{2}\) and \(E_{p}=mgh\). At the top of the first hill (\(v_{1} = 0\)), \(E_{1}=mgh_{1}\). At the top of the second hill, \(E_{2}=\frac{1}{2}mv_{2}^{2}+mgh_{2}\). Since there is no friction (\(E_{1} = E_{2}\)), we have \(mgh_{1}=\frac{1}{2}mv_{2}^{2}+mgh_{2}\).

Step2: Solve for \(v_{2}\)

Cancel out the mass \(m\) from the equation \(mgh_{1}=\frac{1}{2}mv_{2}^{2}+mgh_{2}\) (because \(m
eq0\)). We get \(gh_{1}=\frac{1}{2}v_{2}^{2}+gh_{2}\). Rearrange for \(v_{2}\): \(\frac{1}{2}v_{2}^{2}=g(h_{1}-h_{2})\), so \(v_{2}=\sqrt{2g(h_{1}-h_{2})}\). Given \(g = 9.8\ m/s^{2}\), \(h_{1}=18.0\ m\), \(h_{2}=10.0\ m\). Substitute the values: \(v_{2}=\sqrt{2\times9.8\times(18 - 10)}\).

$$v_{2}=\sqrt{16\times9.8}=\sqrt{156.8}\approx 12.5\ m/s$$

Answer:

\(12.5\ m/s\)