Answer
(a) FBD for the car is shown in the figure
(b) Weight of the car $= 11 kN$
(c) Net force acting on the car$=f_{cr}-f_{ca}=3.3kN-1.2kN=2.1kN northward$
Work Step by Step
(a) Essentially there are four forces acting on the car: the gravitational force due to Earth on the car ($F_{cE}$) acting downward, the normal force on the car due to road ($N_{cr}$) acting upward, drag force exerted by the air on the car ($f_{ca}$) towards south and the frictional force on the car due to road ($f_{cr}$) towards north.
(b) Since the car speeds up while moving north, there is a net force component in this direction. whereas, the net force is zero in the vertical direction. Therefore,
$F_{cE}+N_{cr}=0$
we know that the weight of an object is the force of attraction of Earth on that object.
hence, the weight of the car, $W_{c} =F_{cE}$
Therefore we can write,
$W_{c}+N_{cr}=0$
or
$W_{c}=-N_{cr}$
by considering the magnitude only, we can say
$W_{c}=N_{cr}=11 kN$
(c) To speed up northward, the friction on the car due to road should dominate the drag force of air acting southward. Therefore,
the net force acting on the car$=f_{cr}-f_{ca}=3.3kN-1.2kN=2.1kN northward$
