Answer:
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
