**what is the cars speed after coasting down the other side?** This is a topic that many people are looking for. **newyorkcityvoices.org** is a channel providing useful information about learning, life, digital marketing and online courses …. it will help you have an overview and solid multi-faceted knowledge . Today, ** newyorkcityvoices.org ** would like to introduce to you **Law of Conservation of Energy – Mastering Physics A 1500 kg car is approaching the hill shown at**. Following along are instructions in the video below:

1500 kilogram car is approaching the hill shown at 15 meters per second when it it suddenly runs out of gas and neglect. Any friction can the car make it the other or to the top of the hill by coasting and what is the car speed after coasting down the other side of the hill all right. So lets just think about this for a second.

How do we know if the car is gonna be able to make it to the top of the hill. We need to know the car has to go how how much energy does it need to get to the top so that is going to be its gravitational potential energy. Which is going to be mg y and we have everything for that we have the mass.

We have gravity of course and we have the delta y to get to the top is going to be 5. Meters. So if we plug that in the.

Mass is 1500 kilograms. Times. 98.

Times 5 meters. And that is equal to 68 thousand six hundred joules for the car to get it to the top of the hill. But how much energy does the car have well as traveling at 15 meters per second.

We know kinetic energy is one half mv squared. So lets plug that in so we have one half of the mass of the car. Which is 1500 kilograms times its velocity of 15 meters per second squared.

And that is equal to one hundred and fifty seven thousand five hundred and joules so yes we have more than enough energy to get to the top of the hill by coasting and in fact. Theres going to be energy left over so for part a on mastering physics. The answer is yes.

It will pass the top and keep moving down the other side of the hill now they say whats the speed. When we get to the bottom of the hill. Yes.

We goes yes pass top and move down. Okay. So now for part b.

What is the final velocity at the bottom. After its coasted down the bottom of the hill. So how much so we know it makes the top.

But how much energy does it have left over so we know were gonna have lets just write out the whole equation and talk about which one to cancel out so we have 1 2 m. V. Initial.

Squared plus mg y. We neglect friction. So that is equal to initial one half mv squared final plus mg y final ok.

So were at the top of the hill. Here. And or whoops sorry let back up a second we need to figure out how fast its going at the top of the hill.

So we know what its going initially at the bottom of the hill. But initially lets say that this right here where the car started is y equals zero. So since thats y equals zero.

It started at zero so mg y. All goes away to zero. And then were trying to find what the speed of the car is at the top of the hill.

So then we can figure out how fast itll go at the bottom of the hill. So its delta y. Final will be positive five.

So. Lets solve for this v. Final.

Right here. So lets move that over to the other side and divide by one half m. So we get 1 2.

And the initial squared minus mgy final all over 1 2. The mass of the car. And then square root.

That puppy equal to the final so this top part thats pretty logical. Were gonna say well whatever energy. It has whatever energy.

It took to get to the top and then we just isolated the rest of software v final. It makes sense right so v final. When we plug all that in so 1 2.

The mass of 1500 times its. Initial velocity 15 meters per second squared minus mgy. 1500 times.

98. Times 5 meters for delta y. That is all over 1 2.

Of 1500 and then we take the square root of that answer so at the top of the hill at the top. The car is going it is going eleven point two six nine meters per second. But we dont want the top of course.

We know we want the bottom so now lets do another one of these equations right here to solve for the bottom of the hill. So if we do that were going to say you have one half. And.

The initial squared plus mg y. Initial equals 1 2 m. Naught v m.

V. Final. Squared plus mg y final okay so now music whoops sorry so now.

Were gonna start at the top of the hill here start right here and end up here. So were going to have an initial velocity. So this stays.

We are going to start out with a with a ye. Im actually going to change this a little bit no i wont i was just going to change where y equals zero was to make this simpler. But ill leave it so were gonna start out with a y initial of positive five.

So that stays now what you know i am going to change it lets now lets get rid of this and lets say down here is y equals zero. So where it ends up is y equals zero. So now our y final is zero.

And were looking for v final. Of course. So.

We have now one half and the initial squared. Plus. Mg.

Y. Initial. Which in this case.

Is 10. Because were saying the bottom of the hill is zero. And so hes five and five ten of course you get that and that is all over one half m square root that was terrible square root is equal to v final so it looks pretty similar right the only difference really is our y initial changed and were adding that to it instead of taking it away so.

We have one half of 1500 times. V. Initial.

Which now we are v. Initial at the top of the hill is eleven point two six nine meters per second. Were going to add the mass of the car times nine point eight times.

The delta y. Which we said was ten meters. And that is all over one half of the mass of the car.

Then we take the square root. That gives us an answer of seventeen point nine seven two meters per second at the bottom of the hill and rounded for mastering physics. It gives us 18 meters per second you .

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