Asked by bob
oes Universal Studio’s claim about the coaster’s height (51 m) appear to be true? According to Universal Studios, the roller coaster’s maximum height at the first hill is 51m. The mass of the roller coaster with riders is approximately 2250 kg. You are to use your knowledge of Physics to determine if this claim is true. Use the data below and answer the following questions.
figure_2_coaster_834bf396-11d3-4764-962d-6deadf97047b
All Answers 1
Answered by
GPT-5 mini
AI
I don’t have the figure or the numeric data from your problem, so I can’t give a definite yes/no. I can, however, tell you exactly what to calculate and how to decide whether the 51 m claim is consistent with the data you have. If you post the figure or the measured numbers (speeds, distances, times, friction info) I’ll do the computation for you.
How to check the 51 m claim (step‑by‑step)
1) Basic (ideal, no friction, no rotation) test — from energy conservation
- If the car’s speed v at the bottom of the first hill is known, then (ignoring losses)
m g h = 1/2 m v^2 => h = v^2 / (2 g).
- Use g = 9.81 m/s^2. If h computed this way is close to 51 m (within experimental uncertainty), the claim is plausible in the frictionless limit. If h << 51 m, then the coaster could not have started from 51 m unless there were large energy inputs; if h >> 51 m it would be impossible.
Example numbers:
- v = 28 m/s → h = 28^2/(2·9.81) ≈ 40.0 m
- v = 30 m/s → h ≈ 45.9 m
- v = 32 m/s → h ≈ 52.2 m
So a measured bottom speed of about 31.8 m/s is required to support a 51 m starting height in the ideal (no‑loss) case.
2) If there are non‑conservative losses (friction, air drag, wheel rotation)
- General energy equation: m g h = 1/2 m v^2 + W_losses + 1/2 I ω^2 (if you include wheel rotation)
- Rearranged for h:
h = v^2/(2 g) + W_losses/(m g) + (rotational KE)/(m g).
- If you can estimate W_losses (for example from measured deceleration, or as μk·m g times contact distance along the drop), include that term. Rotational energy of wheels is usually small but can be included if you know wheel inertia or approximate as a fraction of translational KE (typical rolling bodies add a few percent).
3) Using kinematics instead of direct speed
- If you measure the time t and the vertical distance s (or track length) and can find the average acceleration a, you can get v and/or work done by nonconservative forces and then use step 1 or 2.
4) Practical measurement suggestions
- Measure the coaster’s speed at the bottom from video (frame-by-frame) or timing over a known distance.
- Estimate energy losses by comparing measured speeds at different points or by measuring deceleration on a known flat section.
- Account for measurement uncertainties and roundoff when comparing to 51 m.
If you post the figure or give me the measured speed(s), times, distances or any friction/drag data from your problem, I will do the algebra and numbers and tell you whether 51 m is supported.
How to check the 51 m claim (step‑by‑step)
1) Basic (ideal, no friction, no rotation) test — from energy conservation
- If the car’s speed v at the bottom of the first hill is known, then (ignoring losses)
m g h = 1/2 m v^2 => h = v^2 / (2 g).
- Use g = 9.81 m/s^2. If h computed this way is close to 51 m (within experimental uncertainty), the claim is plausible in the frictionless limit. If h << 51 m, then the coaster could not have started from 51 m unless there were large energy inputs; if h >> 51 m it would be impossible.
Example numbers:
- v = 28 m/s → h = 28^2/(2·9.81) ≈ 40.0 m
- v = 30 m/s → h ≈ 45.9 m
- v = 32 m/s → h ≈ 52.2 m
So a measured bottom speed of about 31.8 m/s is required to support a 51 m starting height in the ideal (no‑loss) case.
2) If there are non‑conservative losses (friction, air drag, wheel rotation)
- General energy equation: m g h = 1/2 m v^2 + W_losses + 1/2 I ω^2 (if you include wheel rotation)
- Rearranged for h:
h = v^2/(2 g) + W_losses/(m g) + (rotational KE)/(m g).
- If you can estimate W_losses (for example from measured deceleration, or as μk·m g times contact distance along the drop), include that term. Rotational energy of wheels is usually small but can be included if you know wheel inertia or approximate as a fraction of translational KE (typical rolling bodies add a few percent).
3) Using kinematics instead of direct speed
- If you measure the time t and the vertical distance s (or track length) and can find the average acceleration a, you can get v and/or work done by nonconservative forces and then use step 1 or 2.
4) Practical measurement suggestions
- Measure the coaster’s speed at the bottom from video (frame-by-frame) or timing over a known distance.
- Estimate energy losses by comparing measured speeds at different points or by measuring deceleration on a known flat section.
- Account for measurement uncertainties and roundoff when comparing to 51 m.
If you post the figure or give me the measured speed(s), times, distances or any friction/drag data from your problem, I will do the algebra and numbers and tell you whether 51 m is supported.
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