A 40,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diameter vertical circle. At the point where the plane is flying straight down, its speed is 55 m/s and it is accelerating at 12m/s².  You can neglect air resistance. 

To solve this problem, we need to analyze the forces acting on the stunt plane when it is at the point in the loop where it is flying straight down. At this point, both gravity and the normal force (centripetal force needed to keep the plane in a circular path) act in the same direction, toward the center of the circular path.

### Given:
- Mass of the plane, \( m = 40,000 \) kg
- Diameter of the circular path, \( d = 260 \) m
- Radius of the circular path, \( r = \frac{d}{2} = 130 \) m
- Speed of the plane, \( v = 55 \) m/s
- Total acceleration (centripetal), \( a = 12 \) m/s²
- Gravitational acceleration, \( g = 9.81 \) m/s²

### Step 1: Calculate the net force acting on the plane
The net force \( F_{\text{net}} \) providing the centripetal acceleration is given by:
\[
F_{\text{net}} = m \cdot a
\]
Substitute the values:
\[
F_{\text{net}} = 40,000 \, \text{kg} \times 12 \, \text{m/s}^2 = 480,000 \, \text{N}
\]

### Step 2: Calculate the gravitational force acting on the plane
The force of gravity on the plane is given by:
\[
F_{\text{gravity}} = m \cdot g
\]
Substitute the values:
\[
F_{\text{gravity}} = 40,000 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 392,400 \, \text{N}
\]

### Step 3: Calculate the normal force exerted by the plane
At the bottom of the loop, the net force is the sum of the gravitational force and the normal force, because both are directed toward the center of the circle. Thus, the normal force \( F_{\text{normal}} \) can be found by:
\[
F_{\text{net}} = F_{\text{gravity}} + F_{\text{normal}}
\]
Rearranging to solve for \( F_{\text{normal}} \):
\[
F_{\text{normal}} = F_{\text{net}} - F_{\text{gravity}}
\]
\[
F_{\text{normal}} = 480,000 \, \text{N} - 392,400 \, \text{N} = 87,600 \, \text{N}
\]

Therefore, the normal force exerted by the plane at the bottom of the loop is 87,600 N.

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Would you like more details or have any questions?

### Related Questions:
1. How would the normal force change if the plane's speed increased to 65 m/s?
2. What would happen to the net force if the plane were halfway up the loop?
3. How does centripetal acceleration relate to the speed and radius of circular motion?
4. What role does gravity play in circular motion, especially at different points in a vertical loop?
5. How does the diameter of the loop affect the forces acting on the plane?

### Tip:
Remember, in vertical circular motion, the forces (gravity and normal force) can act together or oppose each other depending on the plane's position in the loop.

A 40,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diameter vertical circle. At the point where the plane is flying straight down, its speed is 55 m/s and it is accelerating at 12 m/s². You can neglect air resistance.

Explore the forces acting on a 40,000 kg stunt plane performing a loop-the-loop in a 260-meter vertical circle. This solution covers key physics concepts, including centripetal force, gravitational force, and Newton's second law. Ideal for high school physics students (Grades 10-12).

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