A car (m=1606 kg) is travelling down the road at a speed of 10.3 m/s.  The car applies its brakes, which exert a constant force and slows to 6 m/s in a distance of 37.9 metres.  What is the magnitude of the braking force?  Express your answer in Newtons.


To find the magnitude of the braking force, we can use the work-energy principle. According to this principle, the work done by the braking force is equal to the change in kinetic energy of the car. The work done by the force can also be expressed as \( W = F \cdot d \), where \( F \) is the force and \( d \) is the distance over which the force acts.

### Step 1: Calculate the initial and final kinetic energies

The kinetic energy \( KE \) is given by the formula:

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

where:
- \( m = 1606 \, \text{kg} \) is the mass of the car,
- \( v \) is the velocity.

#### Initial kinetic energy (at 10.3 m/s):
\[
KE_{\text{initial}} = \frac{1}{2} \times 1606 \times (10.3)^2
\]

#### Final kinetic energy (at 6 m/s):
\[
KE_{\text{final}} = \frac{1}{2} \times 1606 \times (6)^2
\]

### Step 2: Calculate the change in kinetic energy
\[
\Delta KE = KE_{\text{final}} - KE_{\text{initial}}
\]

### Step 3: Work done by the braking force
The work-energy principle states that the work done by the braking force is equal to the change in kinetic energy:
\[
W = F \cdot d = \Delta KE
\]

Where \( d = 37.9 \, \text{m} \) is the distance over which the braking force is applied. We can rearrange the equation to solve for the force \( F \):
\[
F = \frac{\Delta KE}{d}
\]

Let's calculate it step by step.### Step-by-step calculations:

1. **Initial kinetic energy** at \( v = 10.3 \, \text{m/s} \):
   \[
   KE_{\text{initial}} = 85,190.27 \, \text{J}
   \]

2. **Final kinetic energy** at \( v = 6 \, \text{m/s} \):
   \[
   KE_{\text{final}} = 28,908.00 \, \text{J}
   \]

3. **Change in kinetic energy**:
   \[
   \Delta KE = KE_{\text{final}} - KE_{\text{initial}} = 28,908.00 - 85,190.27 = -56,282.27 \, \text{J}
   \]

4. **Braking force**:
   \[
   F = \frac{\Delta KE}{d} = \frac{-56,282.27}{37.9} \approx -1,485.02 \, \text{N}
   \]

The negative sign indicates that the force is acting in the opposite direction to the car's motion. Therefore, the **magnitude of the braking force** is \( \boxed{1,485.02 \, \text{N}} \).

Do you want further details or have any questions?

### Related Questions:
1. What would be the stopping force if the distance was halved?
2. How would the braking force change if the car's initial speed increased to 15 m/s?
3. What is the time taken to decelerate from 10.3 m/s to 6 m/s using this force?
4. How does this braking force compare to the gravitational force acting on the car?
5. What power is dissipated by the braking force over the duration of the deceleration?

### Tip:
When applying the work-energy principle, always consider the direction of forces to avoid confusion between magnitudes and negative signs.

A car (m=1606 kg) is travelling down the road at a speed of 10.3 m/s. The car applies its brakes, which exert a constant force and slows to 6 m/s in a distance of 37.9 metres. What is the magnitude of the braking force? Express your answer in Newtons.

Learn how to calculate the braking force of a car (1606 kg) decelerating from 10.3 m/s to 6 m/s over a distance of 37.9 meters. This problem involves key physics concepts such as kinetic energy and the work-energy principle, suitable for high school students in Grades 10-12.

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