00:01
In this question, you are given that there's a lead car that pulls a train.
00:08
Okay, so a lead car, it pulls a train, okay? and then, so this, the train is 9 .1 times 10 to the 5 kg, and then it's traveling at 45 meters per second.
00:32
There are three parts in this question.
00:34
So it's about calculating power and additional power needed.
00:40
So in part a, you want to find the power provided to the lead car.
00:46
In the case when the car is pulling the chain horizontally.
00:52
So the solution in part a will be using the formula power equals to f times v and the force is given to be 53, kilo nuken so 53 ,000 newtons times 45.
01:12
And calculate this you get 2 .4 times 10 to the 6, what? okay, so this is the answer for part a.
01:23
And then in part b, we want to calculate the additional power needed if the lead card were to accelerate the train at 1 .5 meters per second.
01:38
Square.
01:38
Okay.
01:39
So what happens is that at constant speed on the train itself, there's a friction, which is actually 53 kiloons.
01:56
Then when the when the lead car is supplying for the 53 kiloons and the train is able to travel at 45 meters per second.
02:06
Okay...