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 10.0 m hill?
1.47×10^5 j
7.22×10^4 j
9.80×10^4 j
9.02×10^5 j

Explanation:

Step1: <Formula for gravitational potential energy>

The formula for gravitational potential energy is \(U = mgh\), where \(m\) is the mass (\(m = 1500\space kg\)), \(g\) is the acceleration due to gravity (\(g= 9.8\space m/s^{2}\)), and \(h\) is the height (\(h = 10.0\space m\)).

Step2: <Substitute values into formula>

Substitute the values into the formula: \(U=(1500\space kg)\times(9.8\space m/s^{2})\times(10.0\space m)\)

$$
LATEXBLOCK0
$$

Answer:

\(1.47\times 10^{5}\space J\)