Asked by bv
(a)
Does Universal Studio’s claim about the coaster’s height (51 m) appear to be true? Answer the question with a claim using your knowledge of Conservation of Energy and support it with evidence and reasoning using data and calculations.
with reasoning
All Answers 1
Answered by
GPT-5 mini
AI
I can’t decide whether the 51 m claim is true until I know the experimental data (in particular the coaster’s speed at the bottom of the drop or any measured times). Below I give the correct energy-based test you should use, the formulas, how to apply them to your data, and two short worked examples so you can see how to interpret the result.
Method (Conservation of Energy)
- Neglecting non-conservative losses (friction, air drag), mechanical energy conservation gives
m g h = 1/2 m v^2 => h = v^2 / (2 g).
Here h is the initial height above the point where speed v is measured, g ≈ 9.80 m/s^2.
- If friction/drag do work W_nc (positive number representing energy lost), then
m g h = 1/2 m v^2 + W_nc => h = v^2/(2 g) + W_nc/(m g).
So the frictionless formula h = v^2/(2g) gives a lower bound on the true height: measured v yields h_min = v^2/(2g). The actual height is h_min plus any height-equivalent of energy lost to non-conservative forces.
How to decide about the 51 m claim
- Compute h_min = v^2/(2 g) from your measured speed v at the bottom.
- If h_min ≈ 51 m (within your measurement uncertainty), the claim is consistent with your data.
- If h_min > 51 m by a significant margin, the claim is inconsistent (the coaster would have to be taller than claimed).
- If h_min < 51 m, the claim could still be true if there were non-conservative losses large enough to make up the difference; you cannot disprove the claim from v alone without estimating losses.
Worked examples
1) Suppose measured v = 30.0 m/s at the bottom:
h_min = v^2/(2 g) = (30.0^2)/(2·9.80) = 900/19.6 = 45.9 m.
Interpretation: 45.9 m < 51 m. Neglecting losses, the height would be about 45.9 m, so the 51 m claim is not supported by a frictionless calculation. But because real rides lose energy, the true height could still be 51 m if the ride lost an extra m g (51 − 45.9) = m·9.8·5.1 ≈ 49.98 J/kg (≈50 J per kg of coaster mass) to friction/drag. If your measured v is accurate and you estimate energy losses are small, then the claim is unlikely.
2) Suppose measured v = 31.5 m/s:
h_min = (31.5^2)/(19.6) = 992.25/19.6 ≈ 50.6 m.
Interpretation: h_min ≈ 50.6 m, which is very close to 51 m. Within typical experimental uncertainty this would support the 51 m claim (or at least make it plausible even with small losses).
What I need from you to give a definite answer
- The measured speed (or the times/distances you used to infer speed) at the bottom of the drop, and the uncertainty in that measurement.
- Any estimate of non-conservative energy losses (or a statement that they are negligible).
If you give me your measured v (and its uncertainty), I will compute h_min, the required energy loss per kilogram to reach 51 m, and a clear yes/no assessment about whether the 51 m claim is supported by your data.
Method (Conservation of Energy)
- Neglecting non-conservative losses (friction, air drag), mechanical energy conservation gives
m g h = 1/2 m v^2 => h = v^2 / (2 g).
Here h is the initial height above the point where speed v is measured, g ≈ 9.80 m/s^2.
- If friction/drag do work W_nc (positive number representing energy lost), then
m g h = 1/2 m v^2 + W_nc => h = v^2/(2 g) + W_nc/(m g).
So the frictionless formula h = v^2/(2g) gives a lower bound on the true height: measured v yields h_min = v^2/(2g). The actual height is h_min plus any height-equivalent of energy lost to non-conservative forces.
How to decide about the 51 m claim
- Compute h_min = v^2/(2 g) from your measured speed v at the bottom.
- If h_min ≈ 51 m (within your measurement uncertainty), the claim is consistent with your data.
- If h_min > 51 m by a significant margin, the claim is inconsistent (the coaster would have to be taller than claimed).
- If h_min < 51 m, the claim could still be true if there were non-conservative losses large enough to make up the difference; you cannot disprove the claim from v alone without estimating losses.
Worked examples
1) Suppose measured v = 30.0 m/s at the bottom:
h_min = v^2/(2 g) = (30.0^2)/(2·9.80) = 900/19.6 = 45.9 m.
Interpretation: 45.9 m < 51 m. Neglecting losses, the height would be about 45.9 m, so the 51 m claim is not supported by a frictionless calculation. But because real rides lose energy, the true height could still be 51 m if the ride lost an extra m g (51 − 45.9) = m·9.8·5.1 ≈ 49.98 J/kg (≈50 J per kg of coaster mass) to friction/drag. If your measured v is accurate and you estimate energy losses are small, then the claim is unlikely.
2) Suppose measured v = 31.5 m/s:
h_min = (31.5^2)/(19.6) = 992.25/19.6 ≈ 50.6 m.
Interpretation: h_min ≈ 50.6 m, which is very close to 51 m. Within typical experimental uncertainty this would support the 51 m claim (or at least make it plausible even with small losses).
What I need from you to give a definite answer
- The measured speed (or the times/distances you used to infer speed) at the bottom of the drop, and the uncertainty in that measurement.
- Any estimate of non-conservative energy losses (or a statement that they are negligible).
If you give me your measured v (and its uncertainty), I will compute h_min, the required energy loss per kilogram to reach 51 m, and a clear yes/no assessment about whether the 51 m claim is supported by your data.
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