Question

a roller coaster has a mass of 250 kg. it drops from rest at the top of a hill thats 88 m tall. how fast is it going when it reaches the bottom? acceleration due to gravity is ( g = 9.8 , \text{m/s}^2 ).

a. 13.1 m/s
b. 41.6 m/s
c. 52.2 m/s
d. 70.0 m/s

Explanation:

Step1: Apply conservation of mechanical energy

The roller - coaster starts from rest (\(v_0 = 0\)), so its initial kinetic energy \(K_0=\frac{1}{2}mv_0^{2}=0\). Its initial potential energy is \(U_0 = mgh\), and its final potential energy \(U = 0\) (taking the bottom of the hill as the zero - potential level). The final kinetic energy is \(K=\frac{1}{2}mv^{2}\). By the conservation of mechanical energy \(E_0=E\), we have \(K_0 + U_0=K + U\).
Substituting the expressions for kinetic and potential energy: \(0+mgh=\frac{1}{2}mv^{2}+0\).

Step2: Solve for \(v\)

Cancel out the mass \(m\) from both sides of the equation \(mgh=\frac{1}{2}mv^{2}\) (since \(m
eq0\)). We get \(gh=\frac{1}{2}v^{2}\). Then \(v=\sqrt{2gh}\).
Given \(h = 88\ m\) and \(g = 9.8\ m/s^{2}\), substitute these values into the formula: \(v=\sqrt{2\times9.8\times88}\).
First, calculate \(2\times9.8\times88=1724.8\). Then \(v=\sqrt{1724.8}\approx41.6\ m/s\).

Answer:

B. \(41.6\ m/s\)