Question

A car of mass 1000 kg is moving with a velocity of 20 m/s. The driver applies the brakes, and the car comes to rest in 10 seconds. Find the average force exerted by the brakes to stop the car.

• A. \( 2000 \, \text{N} \)
• B. \( 5000 \, \text{N} \)
• C. \( 10000 \, \text{N} \)
• D. \( 4000 \, \text{N} \)

Hint:

When calculating force, always remember to account for the direction of acceleration. A negative acceleration means a decelerating force that acts in the opposite direction of motion.

Answer 1

The Correct Option is A

To find the average force exerted by the brakes, we can use Newton's second law of motion, which states that the force is the product of mass and acceleration: \( F = m \times a \).

Initially, we need to determine the acceleration. Since the car comes to rest, the final velocity (\( v_f \)) is 0 m/s, and the initial velocity (\( v_i \)) is 20 m/s. The time taken (\( t \)) is 10 seconds.

The formula for acceleration (\( a \)) is given by:

\( a = \frac{v_f - v_i}{t} \)

Substituting the values, we get:

\( a = \frac{0 - 20}{10} = \frac{-20}{10} = -2 \, \text{m/s}^2 \)

This negative sign indicates that the car is decelerating.

Now, apply the formula for force:

\( F = m \times a \)

Substitute the known values (mass \( m = 1000 \, \text{kg} \), acceleration \( a = -2 \, \text{m/s}^2 \)):

\( F = 1000 \times (-2) = -2000 \, \text{N} \)

The negative sign denotes that the force is acting in the opposite direction of motion, which corresponds to the braking force. Therefore, the magnitude of the average braking force is:

\( 2000 \, \text{N} \)