A car drives 5.0 km north, then 7.3 km east, then 5.3 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?

147.73N

Explanation:

Work is said to be done if an applied force causes a body to move through a distance.

Work done by the car = Force * distance.

Given:

Work done = 2.6 * 10^6 Joules

Total distance covered by the car = 5.0km + 7.3km + 5.3km

= 17.6km

= 17600m

From the formula:

Force = Work done/Total distance

Force = 2.6*10^6/17600

Force experienced by the car = 147.73N

Magnitude of the average frictional force on the car is 147.73N

147.73 N

Explanation:

Since Work, W = Force, F * distance, d

W = Fd

F = W/d W = 2.6 * 10⁶ J and d = 5.0 km + 7.3 km + 5.3 km = 17.6 km = 17600 m

F = W/d = 2.6 * 10⁶ J/17600 m = 147.73 N. Which is the force causing the car to move.

Now if f is the average frictional force on the car,

F - f = ma

Since the car moves at constant velocity, a = 0

F - f = 0

F = f = 147.73 N