(a) 11,500 W
The car is driving at constant speed on a horizontal surface. Constant speed means that the acceleration of the car is zero, so according to Newton's second law (F=ma), the net force must be zero.
We have two forces acting on the car: the driving force F, forward, and the drag force, D, acting backward. The net force must be zero, so:
[tex]F-D=0[/tex]
We know that D=500 N, so
[tex]F=D\\F=500 N[/tex]
The power used by the engine is equal to the product between the force generated by the engine (F) and the velocity of the car (v=23 m/s), therefore:
[tex]P=Fv=(500 N)(23 m/s)=11,500 W[/tex]
(b) 17,084 W
In this case, the car is moving on a hill with slope of [tex]2.0^{\circ}[/tex]. Therefore, in this case, there is another force acting along the direction parallel to the surface: the component of the weight of the car parallel to the slope, which acts against the motion of the car. Its magnitude is given by:
[tex]W_x = (mg) sin \theta = (710 kg)(9.8 m/s^2)(sin 2.0^{\circ})=242.8 N[/tex]
As before, the net force along the slope must be zero, since the car is moving at constant speed; therefore, the equation for the forces is
[tex]F-D-W_x =0\\F=D+W_x = 500 N+242.8 N=742.8 N[/tex]
So, this time the power used by the engine is
[tex]P=Fv=(742.8 N)(23 m/s)=17,084 W[/tex]
