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A roller coaster train (mass = 4,000 kg) travels 30 m/s at the top of a second hill that is 23 meters tall. What is the total energy of the roller coaster train?

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Final answer:

The total energy of the roller coaster train with a mass of 4,000 kg traveling at 30 m/s at the top of a 23-meter tall hill is calculated by combining its kinetic energy and gravitational potential energy. It amounts to 2,698,520 joules.

Step-by-step explanation:

The question is asking about the total energy of a roller coaster train at the top of a hill. In physics, the total mechanical energy of an object is the sum of its kinetic energy and potential energy. The kinetic energy (KE) can be calculated using the formula KE = 1/2 mv^2, where m is mass and v is velocity. The potential energy (PE), specifically gravitational potential energy, is determined by the formula PE = mgh, where m is mass, g is gravitational acceleration (approximately 9.81 m/s^2 on Earth), and h is height.

For the roller coaster train with a mass of 4,000 kg traveling at 30 m/s at the top of a 23-meter tall hill, the kinetic energy is KE = 1/2 * 4000 kg * (30 m/s)^2 and the potential energy is PE = 4000 kg * 9.81 m/s^2 * 23 m. The total energy is the sum of KE and PE.

Calculating these values, we find:

  • KE = 1/2 * 4000 kg * (30 m/s)^2 = 1,800,000 J (joules)
  • PE = 4000 kg * 9.81 m/s^2 * 23 m = 898,520 J (joules)
  • Total Energy = KE + PE = 1,800,000 J + 898,520 J = 2,698,520 J

Therefore, the total energy of the roller coaster train at the top of the second hill is 2,698,520 joules, assuming no energy is lost due to friction.

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