To find the force exerted when the school bus stops, we can use the work-energy principle and Newton's second law:

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Given:

• Mass of the bus: $m = 13,500 \, \text{kg}$
• Initial velocity: $v = 60 \, \text{km/h} = 16.67 \, \text{m/s}$ (converted to meters per second)
• Final velocity: $v_f = 0 \, \text{m/s}$ (the bus stops)
• Stopping distance: $d = 5 \, \text{m}$

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Step 1: Deceleration of the bus

Using the kinematic equation: $v_f^2 = v^2 + 2ad$ Rearranging for $a$: $a = \frac{v_f^2 - v^2}{2d}$ Substitute the known values: $a = \frac{0^2 - (16.67)^2}{2(5)} = \frac{-277.89}{10} = -27.79 \, \text{m/s}^2$

The deceleration is $a = -27.79 \, \text{m/s}^2$.

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Step 2: Force acting on the bus

Using Newton's second law: $F = ma$ Substitute $m = 13,500 \, \text{kg}$ and $a = -27.79 \, \text{m/s}^2$: $F = 13,500 \times (-27.79) = -375,165 \, \text{N}$

The force exerted is approximately $-375,165 \, \text{N}$ (negative sign indicates the direction of the force opposes motion).

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Final Answer:

The force required to stop the bus is approximately 375,165 N.

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Would you like further details or have any questions?
Here are five related questions to deepen your understanding:

1. How would the force change if the stopping distance were doubled?
2. What is the work done by the brakes to stop the bus?
3. How does the stopping force compare to the force of gravity on the bus?
4. What happens to the energy of the bus as it stops?
5. How does this problem illustrate Newton's laws of motion?

Tip: Always convert speeds to SI units (meters per second) in physics problems for consistency.