The problem is as follows:
A car whose has a mass of $1500\,kg$ is cruising with a constant speed in 3 hours a distance of $120\,km$ over a road which makes a slope with the horizontal. The car reaches a height of $400\,m$ at the end. The measured external resistance to the advancing of the vehicle are $\frac{200}{1000}\frac{N}{kg}$. Using the given information, find the power of the engine in the car if the manufacturer indicates that its efficiency is $80\%$.
My book list the answer to be: $4861\,W$
But to get to there?. This problem did not come with a drawing, so the sketch from below is my interpretation of what it is meant to said in the problem. As it can be seen in the figure the car is ascending over the ramp.
Because the problem indicates that the external forces causing friction is $f=\frac{200}{1000}\frac{N}{kg}$
then for the given $1500\,kg$ the mass of the car then the frictional force is equal to:
$f=\frac{200}{1000}\frac{N}{kg}\cdot 1500\,kg=300\,N$
Because it says that the car is advancing at a constant rate then:
$\sum_{i=1}^{n}F_{i}=ma = 0$
$F-f=0$
$F=300\,N$
Thus the force involved to calculate the work and then the power is $300\,N$
But the source of confusion is, does the net work here accounts for zero?. How do I find the power?.
$P=\frac{W}{t}$
The time here is $3(3600\,s)$
The only thing which I can recall is that the work, thus to find the power of the engine, would be the sum of the work in the $\textrm{x-direction}$ and in the $\textrm{y-direction}$.
$w_{net}=w_{x}+w_{y}$
But how do I account for this?. Should it account for the work done by lifting the object, I mean for the gravity?. How do I account for this?. All and all, can someone help me with this problem?.
Speed of the car = $\frac{120000}{3*3600} = \frac{100}{9} \frac{m}{s}$
Angle of slope = $\arcsin{\frac{400}{120000}} = \arcsin \frac{1}{300}$
Now, for the car, the net force on it is zero as it moves with constant velocity. Let $F_e$ be the force provided by the engine
Hence $$F_{e, act} = F_{air} + F_{g, slope}$$
$$\implies F_{e,act} = 300 + 1500(10)\frac{1}{300} = 350 N$$
Note here the answer assumes $g = 10$
Now this is at 80% efficiency. Hence the real engine force is
$$F_{e,real} = \frac{F_{e,act}}{0.8} = 437.5 N$$
Hence power of the engine is $Fv = \frac{437.5 \times 100}{9} = 4861 W$