A car drives 5.0 km north,then 7.3 km east,then 3.4 km northeast,all at a constant velocity.If the car had to perform 2.6 × 10⁶ J of work during this trip,what was the magnitude of the average frictional force on the car?

Question

Quizplus · Accepted Answer

The work done by the frictional force is equal to the negative of the change in kinetic energy of the car. The change in kinetic energy is zero because the car starts and ends at rest. Therefore, the work done by the frictional force is equal to the negative of the total work done by the car. The total work done by the car is equal to the sum of the work done by the car in each leg of the trip. The work done by the car in each leg of the trip is equal to the product of the force applied to the car and the distance traveled by the car in that leg. The force applied to the car is equal to the magnitude of the average frictional force on the car. Therefore, the magnitude of the average frictional force on the car is equal to the negative of the total work done by the car divided by the total distance traveled by the car. The total work done by the car is equal to 2.6 × 10^6 J. The total distance traveled by the car is equal to 5.0 km + 7.3 km + 3.4 km = 15.7 km. Therefore, the magnitude of the average frictional force on the car is equal to 2.6 × 10^6 J / 15.7 km = 1.7 × 10^2 N.

A Car Drives 5