a 1500 kg car is traveling with a speed of 11.8 m/s. What is the magnitude of the horizontal net force that is required to bring the car to a halt in a distance of 65.3 m?

To find the magnitude of the horizontal net force required to bring the car to a halt, we can use the **work-energy principle**, which states that the work done on the car is equal to its change in kinetic energy. The net force will do work to bring the car's kinetic energy to zero as it comes to a stop.

The steps are as follows:

### 1. **Kinetic Energy**:
The initial kinetic energy \( KE_i \) of the car is given by:

\[
KE_i = \frac{1}{2} m v^2
\]

where:
- \( m = 1500 \, \text{kg} \) (mass of the car),
- \( v = 11.8 \, \text{m/s} \) (initial velocity).

\[
KE_i = \frac{1}{2} \times 1500 \, \text{kg} \times (11.8 \, \text{m/s})^2 = 104580 \, \text{J} \, (\text{joules})
\]

### 2. **Work-Energy Theorem**:
The work done \( W \) by the net force to stop the car is equal to the change in kinetic energy:

\[
W = \Delta KE = KE_f - KE_i
\]

Since the final kinetic energy \( KE_f = 0 \) (because the car comes to a stop):

\[
W = 0 - 104580 = -104580 \, \text{J}
\]

### 3. **Work Done by the Net Force**:
Work is also given by the force multiplied by the distance:

\[
W = F \cdot d
\]

where:
- \( F \) is the net force,
- \( d = 65.3 \, \text{m} \) is the distance over which the car comes to a stop.

Solving for \( F \):

\[
F = \frac{W}{d} = \frac{-104580 \, \text{J}}{65.3 \, \text{m}} = -1601.2 \, \text{N}
\]

The negative sign indicates that the force is acting in the opposite direction of the car's motion (i.e., it's a decelerating force). The **magnitude** of the force is:

\[
F = 1601.2 \, \text{N}
\]

### Final Answer:
The magnitude of the horizontal net force required to bring the car to a halt is **1601.2 N**.

Would you like further details on this, or have any questions?

Here are five related questions to expand on the topic:
1. How does the stopping distance change if the mass of the car is doubled?
2. What happens to the stopping force if the speed of the car increases by 50%?
3. How is kinetic energy related to the stopping distance for different vehicles?
4. What role does friction play in real-world braking scenarios?
5. How would you calculate the force if the car was moving on an incline?

**Tip**: Always ensure that units are consistent throughout calculations to avoid errors.

A 1500 kg car is traveling with a speed of 11.8 m/s. What is the magnitude of the horizontal net force that is required to bring the car to a halt in a distance of 65.3 m?

Learn how to calculate the net force required to bring a 1500 kg car traveling at 11.8 m/s to a halt over a distance of 65.3 m using the work-energy principle. This detailed solution covers key concepts like kinetic energy, the work-energy theorem, and force calculations. Suitable for Grades 10-12.

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