00:01
In this exercise, we have a car that is coasting in neutral, meaning that no power is being given by the engine to the car, at a speed of 20 meters per second, and down a hill that has an inclination of 2 degrees with the horizontal.
00:22
The car has a mass of 1 ,500 kilograms.
00:27
In question a, we have to suppose that the car is now moving in a level road, and we want to calculate what is the power that the engine must give to the car such that it travels at the same speed of 20 meters per second.
00:50
Well, first of all, what we need to consider in this exercise is that when the car is coasting down the road, the forces that act upon it are the gravitational force that causes an acceleration parallel to the inclined plane of m .g.
01:15
Sine theta.
01:19
And we also have an air resistance force that i'm going to call f air that points upward.
01:30
Now, if the car is moving with a constant speed, that is the car reached a terminal, speed, then the sum of the forces must be zero.
01:44
So you know that f air, the force that the air exerts over the car minus m g sine theta is equal to zero.
01:57
So from here we can obtain that the force that the air exerts over the car is mg times sine theta and theta here is just two.
02:08
Okay.
02:10
Now when when the car is moving in the level road, we also have the force of air, the air resistance, but we do not have the parallel tangential component of the gravitational force.
02:31
The gravitational force is just pointing downwards and not causing any acceleration on the car.
02:37
This means that the force that is making the car go forward is the force.
02:46
That the engine of the car is providing it.
02:52
And if the car is moving with a constant speed of 20 meters per second, then the force that the car, the engine of the car, exerts over the car is equal to, i'm sorry, minus the force of resistance of air is equal to zero.
03:12
This means that the force, the engine exerts, is equal to the force, the air exerts.
03:21
And as we saw before, this is equal to m g times the sign of two.
03:29
Now we want to calculate what is the average power that's given by the engine.
03:37
And we know that the power is equal to the energy that's given divided by the time.
03:48
Now the energy is just the work that the engine does, and this is equal to the force of the engine.
03:55
Times the distance d divided by t.
04:00
Now notice that the car is moving at a constant speed, so d divided by t is just the speed.
04:06
So we have the force of the engine times the speed.
04:11
So this is m -g sine of theta times the speed v.
04:17
So now we can calculate it explicitly.
04:20
M is 48 kilograms.
04:24
G is 9 .8 meters per second squared.
04:27
Then we have to multiply it by sign of 2 degrees, then by 20 meters per second.
04:37
So the power is equal to 10 to the 4 watts or 10 kilowatts.
04:45
This is the power that must be provided by the engine.
04:51
In question b, we have to suppose that the power that the engine gives to the car is equal to 40 kilowatts...
