A 1000 -kg car moving at \( 10 \mathrm{~m} / \mathrm{s} \) brakes to a stop in 5 s . The average braking force is (A) a) 5000 N . (B) b) 2000 N . (C) c) 1000 N . (D) d) 4000 N .

To determine the average braking force exerted on the car, we'll use the principles of **Newton's Second Law of Motion**, which states: \[ F = m \cdot a \] Where: - \( F \) is the force, - \( m \) is the mass of the car, - \( a \) is the acceleration. **Given:** - Mass of the car, \( m = 1000 \, \text{kg} \) - Initial velocity, \( u = 10 \, \text{m/s} \) - Final velocity, \( v = 0 \, \text{m/s} \) (since the car comes to a stop) - Time taken to stop, \( t = 5 \, \text{s} \) **Step 1: Calculate Acceleration** Acceleration (\( a \)) is the change in velocity over time: \[ a = \frac{v - u}{t} = \frac{0 - 10}{5} = -2 \, \text{m/s}^2 \] The negative sign indicates that the acceleration is in the opposite direction of the motion (i.e., it's a deceleration). **Step 2: Calculate the Braking Force** Using Newton's Second Law: \[ F = m \cdot a = 1000 \, \text{kg} \times (-2 \, \text{m/s}^2) = -2000 \, \text{N} \] The negative sign indicates that the force is acting in the opposite direction to the car's motion. However, since we're interested in the **magnitude** of the force: \[ |F| = 2000 \, \text{N} \] **Answer:** **(B) 2000 N**