Question

A single frictionless roller-coaster car of mass m = 55 kg tops the first hill with speed v0 at height h = 22.0 m. How much work does the gravitational force do on the car from that point to (a) point A, (b) point B, and (c) point C? If the gravitational potential energy of the car–Earth system is taken to be zero at C, what is its value when the car is at (d) B and (e) A?
The answer should be with two decimals.

Answer 1

The gravitational potential energy of the roller-coaster at point A and B respectively 11870.10 Joule and 5935.05 Joule.

What is gravitational potential energy?

The energy that an item has or acquires when its location changes as a result of being in a gravitational field is known as gravitational potential energy. Gravitational potential energy can be defined as an energy that has a connection to gravitational force or gravity.

According to the question:

Mass of the rollercoaster: m = 55 kg.

Height of point A: h =22.0 m.

Height of point B: h/2 = 22.0/2 m = 11.0 m.

Height of point C = 0.

Taking, acceleration due to gravity: g = 9.81 m/s² we get:

The gravitational potential energy of the car at point A:

E₁ = mgh = 55 kg × 9.81 m/s² × 22.0 m = 11870.10 Joule.

The gravitational potential energy of the car at point B:

E₂ = mgh/2 = 55 kg × 9.81 m/s² × 11.0 m = 5935.05 Joule.

Hence, gravitational potential energy of the car at point A and B respectively 11870.10 Joule and 5935.05 Joule.

Learn more about gravitational potential energy here:

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