To solve this problem, we will use the principles of energy conservation and kinematics:

Part 1: Velocity at the top of the second hill

The roller coaster begins at rest with potential energy at 90.0 m. At the top of the second hill (60.0 m), some potential energy will be converted to kinetic energy.

1. Total energy at the top of the first hill: $E_{\text{total}} = mgh_1$ where $m = 750 \, \text{kg}$, $g = 9.8 \, \text{m/s}^2$, $h_1 = 90.0 \, \text{m}$.

2. Energy at the top of the second hill: At the second hill ($h_2 = 60.0 \, \text{m}$), the energy is split into potential energy and kinetic energy: $E_{\text{total}} = mgh_2 + \frac{1}{2}mv^2$

3. Using energy conservation: $mgh_1 = mgh_2 + \frac{1}{2}mv^2$ Simplify and solve for $v$: $v = \sqrt{2g(h_1 - h_2)}$ Substituting values: $v = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times (90.0 \, \text{m} - 60.0 \, \text{m})}$ $v = \sqrt{2 \times 9.8 \times 30.0}$ $v = \sqrt{588} \approx 24.3 \, \text{m/s}$

So, the velocity at the top of the second hill is approximately 24.3 m/s.

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Part 2: Friction force to stop the car

To bring the car to a stop over a 120-m horizontal track, we calculate the friction force required using the work-energy principle.

1. Initial kinetic energy: The car reaches the ground level from the initial height of 90.0 m. Using energy conservation: $\frac{1}{2}mv^2 = mgh_1$ Solving for $v$: $v = \sqrt{2gh_1}$ Substituting values: $v = \sqrt{2 \times 9.8 \, \text{m/s}^2 \times 90.0 \, \text{m}} = \sqrt{1764} = 42.0 \, \text{m/s}$ So, the car's velocity at ground level is 42.0 m/s.

2. Work done by friction: The work done by friction must equal the car's initial kinetic energy to bring it to rest: $f \cdot d = \frac{1}{2}mv^2$ Solving for $f$: $f = \frac{\frac{1}{2}mv^2}{d}$ Substituting values: $f = \frac{\frac{1}{2} \times 750 \, \text{kg} \times (42.0 \, \text{m/s})^2}{120.0 \, \text{m}}$ $f = \frac{0.5 \times 750 \times 1764}{120}$ $f = \frac{661,500}{120} \approx 5512.5 \, \text{N}$

The required friction force is approximately 5512.5 N.

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Summary of Results:

1. Velocity at the top of the second hill: 24.3 m/s.
2. Friction force to stop the car: 5512.5 N.

Would you like further breakdowns or help with a related scenario?

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Relative Questions:

1. How would the results change if friction were present along the track?
2. What happens to the velocity at the second hill if its height is increased?
3. How would the required friction force vary for different stopping distances?
4. Could air resistance significantly impact the calculated values in real-world conditions?
5. What additional information is needed to model the coaster's velocity as it climbs the second hill with friction?

Tip:

When solving conservation of energy problems, always verify if friction or external forces are acting before assuming energy is conserved.