A car with a mass of 900 kg climbs a 20 degree incline at a steady speed of 60 km/hr. If the total resistive forces acting on the car add to 500 N, what is the power output of the car in watts? In horsepower?
As usual, draw a free-body diagram.
Using a coordinate system with +x up the slope and +y perpendicular to the slope, the gravitational force is the only force that needs to be split into components.
mg sin(20) acts down the slope
mg cos(20) acts into the slope
Fr represents the resistive forces.
The power output by the car's engine goes into the force directed up the slope. This force is actually static friction exerted on the drive wheels by the road - the road exerts this force F because the engine causes the drive wheels to rotate. With no friction they would spin on the road - friction opposes this tendency.
The velocity is constant, so the forces must balance. Applying Newton's second law in the x-direction gives:
F - Fr- mg sin(20) = 0
The force up the slope is then
F = Fr + mg sin(20) = 500 + 3017 = 3517 N
Converting the car's speed to m/s gives 16.67 m/s. The power output can then be found from
P = Fv = (3517) (16.67) = 58620 W.
This can be converted to horsepower, using 1 hp = 746 W. This gives a power output of 78.6 hp.
Most cars have engines with maximum power outputs of about 100 hp, so this is a reasonable value.