A 2000 kg car accelerates from rest to a velocity of 40 m/s in 5.0 seconds. The brakes are applied, and the car is brought to a stop (without sliding). The magnitude of the work done by the friction in the brakes is A) 1.0 × 10⁶ J. B) 1.2 × 10⁶ J. C) 1.6 × 10⁶ J. D) 2.1 × 10⁶ J. E) 2.5 × 10⁶ J.

Question

Quizplus · Accepted Answer

The work done by the friction in the brakes is equal to the change in kinetic energy of the car. The kinetic energy (KE) can be calculated using the formula KE = 1/2 * m * v^2, where m is the mass of the car and v is its velocity. Substituting the given values, KE = 1/2 * 2000 kg * (40 m/s)^2 = 1.6 × 10^6 J. Therefore, the magnitude of the work done by the friction in the brakes is 1.6 × 10^6 J.

A 2000 Kg Car Accelerates from Rest to a Velocity