A roller coaster car has mass of 1300 kg when fully loaded with passengers. As the car passes over the top of a circular hil of radius 20m, its speed is not changing. At the top of the hill, what are the (a) magnitude `F_(N)` and (b) direction (up or down) of the normal force on the car from the track if the car's speed is v=11m/s? What are (C) `F_(N)` and (d) the direction if v=14m/s?

Answer

To solve the problem, we need to analyze the forces acting on the roller coaster car as it passes over the top of a circular hill. We will use the concepts of centripetal force and Newton's second law.

### Given Data:
- Mass of the roller coaster car, \( m = 1300 \, \text{kg} \)
- Radius of the circular hill, \( r = 20 \, \text{m} \)
- Gravitational acceleration, \( g = 9.8 \, \text{m/s}^2 \)

### Part (a): Finding the Normal Force \( F_N \) when \( v = 11 \, \text{m/s} \)

1. **Calculate the gravitational force \( F_g \)**:
\[ F_g = m \cdot g = 1300 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 12740 \, \text{N} \]

2. **Calculate the centripetal force \( F_c \)** required to keep the car moving in a circular path:
\[ F_c = \frac{m \cdot v^2}{r} = \frac{1300 \, \text{kg} \cdot (11 \, \text{m/s})^2}{20 \, \text{m}} = \frac{1300 \cdot 121}{20} = 7875 \, \text{N} \]

3. **Set up the equation for forces at the top of the hill**:
At the top of the hill, the gravitational force acts downward, and the normal force \( F_N \) acts upward. The net force towards the center of the circular path is given by:
\[ F_g - F_N = F_c \]
Rearranging gives:
\[ F_N = F_g - F_c \]

4. **Substitute the values**:
\[ F_N = 12740 \, \text{N} - 7875 \, \text{N} = 4865 \, \text{N} \]

### Part (b): Direction of the Normal Force
The normal force \( F_N \) is positive, indicating that it acts upward.

### Part (c): Finding the Normal Force \( F_N \) when \( v = 14 \, \text{m/s} \)

1. **Calculate the centripetal force \( F_c \)** for \( v = 14 \, \text{m/s} \):
\[ F_c = \frac{m \cdot v^2}{r} = \frac{1300 \, \text{kg} \cdot (14 \, \text{m/s})^2}{20 \, \text{m}} = \frac{1300 \cdot 196}{20} = 12740 \, \text{N} \]

2. **Set up the equation for forces at the top of the hill**:
\[ F_g - F_N = F_c \]
Rearranging gives:
\[ F_N = F_g - F_c \]

3. **Substitute the values**:
\[ F_N = 12740 \, \text{N} - 12740 \, \text{N} = 0 \, \text{N} \]

### Part (d): Direction of the Normal Force
Since \( F_N = 0 \, \text{N} \), there is no normal force acting on the car at this speed. Thus, the direction is undefined.

### Summary of Results:
- (a) Magnitude of \( F_N \) when \( v = 11 \, \text{m/s} \): \( 4865 \, \text{N} \)
- (b) Direction of \( F_N \): Upward
- (c) Magnitude of \( F_N \) when \( v = 14 \, \text{m/s} \): \( 0 \, \text{N} \)
- (d) Direction of \( F_N \): Undefined (no normal force)