A roller coaster is moving at 6 m/s at the top of the first hill (h= 56 m). Ignoring friction and air resistance, how fast will the roller coaster be moving at the top of a subsequent hill, which is 24 m?

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Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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**Problem: Roller Coaster Dynamics**

A roller coaster is moving at 6 m/s at the top of the first hill (h = 56 m). Ignoring friction and air resistance, how fast will the roller coaster be moving at the top of a subsequent hill, which is 24 m?

**Explanation:**

To solve this problem, we can use the principle of conservation of mechanical energy. The total mechanical energy at the top of the first hill (sum of potential and kinetic energy) will equal the total mechanical energy at the top of the subsequent hill, since no energy is lost to friction or air resistance. 

***Steps:***

1. **Calculate Initial Total Energy:**
   - Potential Energy (PE) at the first hill:
     \[ PE_1 = m \cdot g \cdot h_1 \]
   - Kinetic Energy (KE) at the first hill:
     \[ KE_1 = \frac{1}{2} m \cdot v_1^2 \]

2. **Calculate Total Energy at the Subsequent Hill:**
   - Potential Energy (PE) at the subsequent hill:
     \[ PE_2 = m \cdot g \cdot h_2 \]
   - Kinetic Energy (KE) at the subsequent hill:
     \[ KE_2 = \frac{1}{2} m \cdot v_2^2 \]

3. **Equating the Total Energies:**
   \[ PE_1 + KE_1 = PE_2 + KE_2 \]

4. **Substitute the Values:**
   - Use the known values to calculate \( v_2 \), the speed at the top of the subsequent hill.

These calculations will give you the new speed at the top of the second hill.
Transcribed Image Text:**Problem: Roller Coaster Dynamics** A roller coaster is moving at 6 m/s at the top of the first hill (h = 56 m). Ignoring friction and air resistance, how fast will the roller coaster be moving at the top of a subsequent hill, which is 24 m? **Explanation:** To solve this problem, we can use the principle of conservation of mechanical energy. The total mechanical energy at the top of the first hill (sum of potential and kinetic energy) will equal the total mechanical energy at the top of the subsequent hill, since no energy is lost to friction or air resistance. ***Steps:*** 1. **Calculate Initial Total Energy:** - Potential Energy (PE) at the first hill: \[ PE_1 = m \cdot g \cdot h_1 \] - Kinetic Energy (KE) at the first hill: \[ KE_1 = \frac{1}{2} m \cdot v_1^2 \] 2. **Calculate Total Energy at the Subsequent Hill:** - Potential Energy (PE) at the subsequent hill: \[ PE_2 = m \cdot g \cdot h_2 \] - Kinetic Energy (KE) at the subsequent hill: \[ KE_2 = \frac{1}{2} m \cdot v_2^2 \] 3. **Equating the Total Energies:** \[ PE_1 + KE_1 = PE_2 + KE_2 \] 4. **Substitute the Values:** - Use the known values to calculate \( v_2 \), the speed at the top of the subsequent hill. These calculations will give you the new speed at the top of the second hill.
Expert Solution
Step 1

Write the given values with suitable variables.

u=6 m/sh=56 mh'=24 m

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