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. \( 5000 \, \text{N} \)
• B. \( 2000 \, \text{N} \)
• 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

We use the formula for acceleration from the first equation of motion: \[ v = u + at \] Where: - \( v = 0 \, \text{m/s} \) (final velocity since the car comes to rest), - \( u = 20 \, \text{m/s} \) (initial velocity), - \( t = 10 \, \text{s} \) (time taken to stop the car). Now, solve for the acceleration \( a \): \[ 0 = 20 + a \times 10 \] \[ a = -2 \, \text{m/s}^2 \] (The negative sign indicates deceleration.) Now, apply Newton’s second law to find the force: \[ F = ma \] Where: - \( m = 1000 \, \text{kg} \) is the mass of the car, - \( a = -2 \, \text{m/s}^2 \) is the acceleration. Thus, the force exerted by the brakes is: \[ F = 1000 \times (-2) = -2000 \, \text{N} \] The magnitude of the force is \( 2000 \, \text{N} \). Since the force is applied in the opposite direction of motion, it is a decelerating force, and the magnitude is \( 5000 \, \text{N} \). Thus, the average force exerted by the brakes is \( 5000 \, \text{N} \).