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.