A car moving at the 40 km/h is to be stopped by applying brakes in the next 4 m.If the car weighs 2000 kg, what average force must be applied on it?

To determine the average force required to stop a car moving at 40 km/h over a distance of 4 meters, we can use the principles of physics, specifically Newton's second law and the equations of motion. Let's break this down step by step.

Understanding the Problem

First, we need to convert the car's speed from kilometers per hour to meters per second, as this will make our calculations easier. The conversion factor is:

• 1 km/h = 1/3.6 m/s

So, for a speed of 40 km/h:

Speed in m/s = 40 km/h × (1/3.6) ≈ 11.11 m/s

Applying the Equations of Motion

Next, we can use the equation of motion that relates initial velocity, final velocity, acceleration, and distance:

v² = u² + 2as

Where:

• v = final velocity (0 m/s, since the car stops)
• u = initial velocity (11.11 m/s)
• a = acceleration (which we need to find)
• s = distance (4 m)

Rearranging the equation to solve for acceleration (a), we have:

a = (v² - u²) / (2s)

Substituting the known values:

a = (0 - (11.11)²) / (2 × 4)

a = -123.46 / 8 ≈ -15.43 m/s²

The negative sign indicates that this is a deceleration, which makes sense since the car is slowing down.

Calculating the Average Force

Now that we have the acceleration, we can find the average force using Newton's second law:

F = ma

Where:

• F = force
• m = mass of the car (2000 kg)
• a = acceleration (-15.43 m/s²)

Substituting the values into the equation:

F = 2000 kg × (-15.43 m/s²)

F ≈ -30860 N

The negative sign indicates that the force is applied in the opposite direction of the car's motion, which is exactly what we want when stopping the vehicle.

Final Thoughts

In summary, to stop a 2000 kg car moving at 40 km/h over a distance of 4 meters, an average force of approximately 30,860 Newtons must be applied in the opposite direction of the car's motion. This example illustrates the application of basic physics principles to solve real-world problems effectively.