A car with a mass of 1 ton, moving at a speed of 40 km / h, after the start of braking, reduced

A car with a mass of 1 ton, moving at a speed of 40 km / h, after the start of braking, reduced its speed by 2 times. What is the work of friction forces equal to?

Initial data: m (vehicle weight) = 1 t (1 * 10 ^ 3 kg); V1 (the speed with which the car was moving before braking) = 40 km / h (11.11 m / s); V2 (speed after braking) = 0.5V1.
The work of friction forces will be equal to the change in the kinetic energy of the car: Atr. = ΔEk = 0.5 * m * (V1 ^ 2 – V2 ^ 2) = 0.5 * m * (V1 ^ 2 – 0.25V1 ^ 2) = 0.5 * m * 0.75 * V1 ^ 2 …
Calculation: Atr. = 0.5 * 1 * 10 ^ 3 * 0.75 * 11.11 ^ 2 ≈ 46.3 * 10 ^ 3 J (46.3 kJ).
Answer: The work of friction forces is 46.3 kJ.