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 gravitational potential energy of the car at the top of the 18.0 m hill?
18 m
10.0 m
2.65×10^5 j
2.70×10^4 j
6.98×10^5 j
4.09×10^4 j

Explanation:

Step1: Recall the formula for gravitational potential energy

The formula for gravitational potential energy is \(U = mgh\), where \(m\) is the mass, \(g\) is the acceleration due to gravity (\(g = 9.8\ m/s^{2}\)), and \(h\) is the height.

Step2: Substitute the given values into the formula

Given \(m = 1500\ kg\), \(h = 18.0\ m\), and \(g = 9.8\ m/s^{2}\).

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Answer:

\(2.65\times 10^{5}\ J\) (the first option)