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.
