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

Answer:

Explanation:

Given:

Power, p = 30 hp

1 hp = 746 W

= 22.38 kW

Velocity, v = 140 mph

1 mile = 1609 m

Converting mph to m/s,

140 mile/hour × 1609 m/1 mile × 1 hour/3600 s

= 62.572 m/s

Summation of forces = 0

Power, p = energy/time

Energy = force × distance;

Velocity = distance/time;

Power = force × velocity

Force, fr = 22380/62.572

= 357.67 N

Frictional force, fr = 0.358 kN.

Answer 2

Answer:

The total frictional force is 358.0 newtons

Explanation:

Power is the amount of average work (W) an object does on a period of time (Δt):

$P=\frac{W}{\Delta t}$

Remember average work is average force (F) times displacement (Δs):

$P=\frac{F \Delta s}{\Delta t}$

but displacement over time is average speed $v=\frac{\Delta s}{\Delta t}$, then:

$P=Fv$ (1)

That is, the power of the car is the force the engine does times the speed of the car. As the question states, if the car is at constant velocity then the power developed is used to overcome the frictional forces exerted by the air and the road, that is by Newton's first law, the force the motor of the car does is equal the force of frictional forces. So, to find the frictional forces we only have to solve (1) for F:

$F=\frac{P}{v}$

Knowing that 1hp is 746W then 30hp=22380W and 1 mile = 1609m then 140 mph = 225308$\frac{m}{h}$ = $62.5 \frac{m}{s}$, then:

$F=\frac{22380}{62.5}=358.0 N$