Math Problem Statement

You have been hired to design a spring-launched roller coaster that will carry two passengers per car. The car goes up a 11-m-high hill, then descends 20 m to the track's lowest point. You've determined that the spring can be compressed a maximum of 2.5 m and that a loaded car will have a maximum mass of 450 kg. For safety reasons, the spring constant should be 13 % larger than the minimum needed for the car to just make it over the top. k =1.8×10 4 Nm

Solution

To design a safe spring-launched roller coaster, let’s calculate the minimum spring constant required for the car to reach the top of the hill, and then determine the adjusted spring constant with the 13% safety margin.

Step 1: Determine the Potential Energy at the Top of the Hill

The potential energy (PEPE) needed to reach the top of an 11-meter hill is calculated using: PE=mghPE = m \cdot g \cdot h where:

  • m=450kgm = 450 \, \text{kg} (mass of the loaded car),
  • g=9.81m/s2g = 9.81 \, \text{m/s}^2 (acceleration due to gravity),
  • h=11mh = 11 \, \text{m} (height of the hill).

PE=450×9.81×11=48510JPE = 450 \times 9.81 \times 11 = 48510 \, \text{J}

Step 2: Determine the Minimum Spring Constant kmink_{\text{min}}

The spring energy required to get the car up the hill is given by: Spring Energy=12kx2\text{Spring Energy} = \frac{1}{2} k \cdot x^2 where x=2.5mx = 2.5 \, \text{m} (maximum compression distance of the spring). To just make it over the hill, this energy must be equal to the potential energy at the top:

12kmin(2.5)2=48510\frac{1}{2} k_{\text{min}} \cdot (2.5)^2 = 48510

Solving for kmink_{\text{min}}: kmin=2×48510(2.5)2k_{\text{min}} = \frac{2 \times 48510}{(2.5)^2} kmin=970206.25=15523.2N/mk_{\text{min}} = \frac{97020}{6.25} = 15523.2 \, \text{N/m}

Step 3: Apply the 13% Safety Margin

For safety, the spring constant should be 13% larger than kmink_{\text{min}}:

ksafe=kmin×1.13k_{\text{safe}} = k_{\text{min}} \times 1.13 ksafe=15523.2×1.13=17541.2N/mk_{\text{safe}} = 15523.2 \times 1.13 = 17541.2 \, \text{N/m}

Conclusion:

The spring constant for the roller coaster should be designed to k=17541.2N/m\mathbf{k = 17541.2 \, \text{N/m}} to ensure the car safely reaches the top of the hill.

Would you like a more detailed breakdown of any part of this calculation?


Here are five related questions:

  1. How would the calculation change if the hill were 15 meters instead of 11 meters?
  2. What would happen if the spring compression limit were increased to 3 meters?
  3. How does the safety margin affect the spring constant in other amusement park rides?
  4. What impact would doubling the car's mass have on the required spring constant?
  5. How could potential energy losses (like friction) be factored into the design?

Tip: Always factor in real-world inefficiencies like friction or air resistance when designing systems with minimum energy requirements.

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Math Problem Analysis

Mathematical Concepts

Potential Energy
Spring Constant
Energy Conservation
Physics of Motion

Formulas

Potential Energy formula: PE = m * g * h
Spring Energy formula: (1/2) * k * x^2
Adjusted spring constant with safety margin: k_safe = k_min * (1 + safety_margin)

Theorems

Conservation of Energy
Hooke's Law for Springs

Suitable Grade Level

Grades 11-12