4. a car of mass 2000 kg is traveling at 45 m/s when the driver spots a policeman anead. i ne univer apprivo
the brakes lightly for 3.0 seconds until he slows down below the speed limit. if the average force applied
by the brakes was 1.4 x 104 n, by how much did the kinetic energy of the car change?

The change in the Kinetic Energy of the car, E = 1449000 joules

Explanation:

Given,

The mass of the car, m = 2000 Kg

The speed of the car, v = 45 m/s

The brake applied on the car for a duration, t = 3 s

The average force applied by the brake, F = 1.4 x 10⁴ N

The kinetic energy of the body is given by the relation,

                                     K.E = 1/2 mv²

The initial kinetic energy of the car,

                                   K.E = 0.5 x 2000 x 45

                                          = 2025000 J

The force applied by the brakes

                                   F = m x a

Therefore, the deceleration of the car

                                     a = F / m

                                        = 1.4 x 10⁴ / 2000

                                       = 7 m/s²

Using the first equations of motion,

                                 v = u + at

                                 v = 45 + (-7) (3)                      ∵  (-7 ) car is decelerating

                                  v =24 m/s

The final kinetic energy of the car

                                 k.e = 0.5 x 2000 x 24

                                        = 576000 J

The difference in the kinetic energy,

                           E = K.E - k.e

                               = 2025000 J - 576000 J

                               = 1449000 joules

Hence, the change in the Kinetic Energy of the car, E = 1449000 joules

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