(a) 0 W
In this situation, the car is moving on a level road at constant velocity. This means that its acceleration is zero:
a = 0
But this also means that the net force acting on the car is zero, according to Newton's second law:
F = ma = 0
The power required to move the car is
P = Fv
where F is the force applied to the car and v the velocity. Since there are no air drag/frictional forces, the force applied to the car is exactly zero: so, the extra power needed to keep the car in motion at constant velocity is also zero.
(b) 81815 W
In this situation, the car is moving on an uphill road, so there is a component of the weight parallel to the incline acting agains the direction of motion of the car. This component of the weight is given by
[tex]W=mg sin \theta[/tex]
where
m = 1200 kg is the mass of the car
g = 9.81 m/s^2 is the acceleration of gravity
[tex]\theta = 30^{\circ} [/tex] is the angle of the incline
Substituting,
[tex]W=(1200 kg)(9.81 m/s^2)(sin 30^{\circ} )=5886 N[/tex]
So the truck must exerts an additional force of F = 5886 N to keep the car in motion at constant velocity.
The velocity of the car is
v = 50 km/h = 13.9 m/s
So the extra power needed is
P = Fv = (5886 N)(13.9 m/s) = 81,815 W
(c) 31250 W
Here the car accelerates to a final velocity of
v = 90 km/h = 25 m/s
from an initial velocity of
u = 0
So the work done on the car is equal to its change in kinetic energy, so:
[tex]W=K_f - K_i = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 = \frac{1}{2}(1200 kg)(25 m/s)^2 -0 =3.75\cdot 10^5 J[/tex]
The extra power erogated by the truck therefore must be
[tex]P=\frac{W}{t}[/tex]
where
t = 12 s
is the time taken
Substituting,
[tex]P=\frac{3.75\cdot 10^5 J}{12 s}=31250 W[/tex]
