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 kinetic energy of the car at the top of the 10.0 m hill?
6.72×10⁴j
1.06×10⁴j
1.18×10⁵j
4.19×10⁵j

Explanation:

Step1: Apply conservation of mechanical energy

Since the roller - coaster is frictionless, the total mechanical energy \(E = K + U\) is conserved. At the top of the first hill (\(h_1=18.0\ m\)), the initial kinetic energy \(K_1 = 0\) (starts from rest), and the initial potential energy \(U_1=mgh_1\). At the top of the second hill (\(h_2 = 10.0\ m\)), the total mechanical energy is \(E_2=K_2+U_2\), where \(U_2=mgh_2\). By conservation of energy \(E_1 = E_2\), so \(mgh_1=K_2+mgh_2\).

Step2: Solve for \(K_2\)

Rearrange the equation \(K_2=mgh_1 - mgh_2=mg(h_1 - h_2)\). Given \(m = 1500\ kg\), \(g = 9.8\ m/s^2\), \(h_1=18.0\ m\), and \(h_2 = 10.0\ m\). Substitute the values: \(K_2=1500\times9.8\times(18 - 10)\).

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Answer:

\(1.18\times 10^{5}\ J\) (the third option)