(a) 11500 W
The power required for the car's engine is given by
P = Fv
where
F is the force that the engine must apply
v = 23 m/s is the velocity of the car
In this situation,the car is moving at constant velocity: this means that its acceleration is zero, so the net force on the car must be zero. Since there is a drag force of 500 N against the motion of the car, this means that the force applied by the engine in the forward direction must also be 500 N:
F = 500 N
So the power erogated by the engine is
P = (500 N)(23 m/s)= 11500 W
(b) 17084 W
In this situation, there is not only the drag force opposing the motion of the car, but also the component of the weight which is parallel to the incline.
This component is given by
[tex]W_p = mg sin \theta[/tex]
where
m = 710 kg is the mass
g = 9.8 m/s^2 is the acceleration of gravity
[tex]\theta=2^{\circ}[/tex] is the slope of the incline
Substituting,
[tex]W_p = (710 kg)(9.8 m/s^2)sin 2^{\circ} =242.8 N[/tex]
So now the total backward force against the motion of the car is the sum of the drag force (500 N) and this force:
F = 500 N + 242.8 N = 742.8 N
And so the force applied by the engine must be the same; so the power erogated will be
P = (742.8 N)(23 m/s)= 17,084 W
