High School

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An 850-kg roller coaster car drops from rest at a height of 93 m along a frictionless track.

a) What is the velocity of the roller coaster at the top of a second hill that is 65 m high?

Answer :

Final answer:

The velocity of an 850-kg roller coaster car at the top of a second hill that is 65 m high, when it drops from rest at a height of 93 m on a frictionless track, is approximately 24.0 m/s, determined by using the conservation of energy principle.

Explanation:

The student is asking about the velocity of a roller coaster car at a certain point on the track, which is a classic physics problem involving conservation of energy. As the car is on a frictionless track, we know that mechanical energy is conserved. The initial potential energy at the height of 93 m is converted into kinetic and potential energy at the height of 65 m.

To find the velocity at the 65 m height, we can set up the conservation of energy equation:

  • Initial potential energy at 93 m = kinetic energy + potential energy at 65 m
  • (m * g * h1) = (0.5 * m * v^2) + (m * g * h2)
  • (850 kg * 9.8 m/s^2 * 93 m) = (0.5 * 850 kg * v^2) + (850 kg * 9.8 m/s^2 * 65 m)

By solving this equation for v, the velocity at the top of the second hill (65 m), we get:

  • v = sqrt(2 * g * (h1 - h2))
  • v = sqrt(2 * 9.8 m/s^2 * (93 m - 65 m))
  • v ≈ 24.0 m/s

Therefore, the velocity of the roller coaster at the top of the second hill that is 65 m high is approximately 24.0 m/s.

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