Power Equations

Question

When an automobile moves with constant velocity, the power developed is used to overcome the frictional forces exerted by the air and the road. If the engine develops 30 hp, what total frictional force acts on the car at 140 mph? One horsepower equals 746 W, and one mile is 1609 m. Answer in units of N.

Answer 1

Hello Samuel,

To answer this question, let's approach it from a work and energy perspective.

The work done by a force on an object is equal to Fscosθ where F is the magnitude of the force, s is the displacement of the object during the time that the force is applied, and θ is the angle between the force and displacement vectors.

If we treat the frictional forces exerted by the air and the road as one, then two forces do work on the car as it travels: the total frictional force and the force the engine exerts on the car. According to the Work-Energy theorem, since the car is moving at a constant velocity, the total work done on the car at all times must be equal to zero. Thus, the magnitudes of the work done by friction and the engine are equal. This can be represented by:

Work_friction + Work_engine = 0

Work_engine = -Work_friction

Work_engine = -F_frictionscos(180°)

Now, we have an equation we can use to solve for F_friction.

The question states that for the car to maintain a constant velocity of 140mph (62.57m/s), the engine must develop 30 horse power. Since power is the rate of change of work with respect to time, we can find the amount of work the engine exerts on the car per second by doing the following calculation:

30 hp x 746 Watt/1hp = 22,380 Watt = 22,380 Joules/second

Knowing that both forces must do the same magnitude of work in a second, we can now solve for F_friction:

Work_engine = -F_frictionscos(180°)

22,380J = -F_friction(62.57m)cos(180°)

22,380J = F_friction(62.57m)

F_friction = 22,380J/62.57m = 358N