Vehicle with a weight 2.8 ton Slowing from 100 km/h to 50Kkm/h how much energy can was generate from this Slowing.
(if we Ignore form friction )
and to which Height this energy was can Climb up the Vehicle Up the hill
I convert the ton to Kg is gave me 2800Kg and the km/h to meter per second
I tried to use with the formula Ek=1/2mv^2 and U=mgh but I am not understand how to solve it.
so I need help.
thanks.

You're right so far. You can use your formula to compute the kinetic energy at the beginning (when the car was traveling 100 kph), and then compute after it slowed down to 50 kph. The difference in kinetic energy before and after is the amount of energy that was generated by this slowing down.

If the car climbs a hill, all of that kinetic energy gets converted to potential energy. Therefore, you can use



which means



You already computed in the previous step, so now just solve for .

hi thank for answer tell me if I correct till here:so as I told I convert the ton to Kg is gave me 2800Kg and the km/h to meter per second is gave me 100km/h =27.777 meter per second and 50 km/h= 13.888 meter pet second.

now I used the formula Ek=1/2mv^2 for Each velocity:1/2*2800*27.777^2/1000=1080
1/2*2800*13.888^2/1000=270
I am not sure if I need to divide in Thousand as I did but something here is not look me correct what you say?

thanks.

Well, you may or may not divide by 1000.

If you do however, you should be very careful.

The units are then kJ, and when you do the next part, you have to keep that in mind.

Yes I know, this why I am not sure.

Thank you.

Can you get the answer to the first part then?

'How much energy can be generated when the vehicle slows down?"

well 4.0*10^3 Kj
but I am not know how to find the way for this answer

Apply the law of conservation of energy (though I don't think that 4.0 x 10^3 kJ is the answer... )

The total initial energy is equal to the total final energy.

Initial KE = Final KE + 'Energy'

You are looking for 'Energy'

Quote:

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Originally Posted by pop000

hi thank for answer tell me if i correct till here:so as i told i convert the ton to Kg is gave me 2800Kg and the km/h to meter per second is gave me 100km/h =27.777 meter per second and 50 km/h= 13.888 meter pet second.

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Perfect so far.

Quote:

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now i used the formula Ek=1/2mv^2 for Each velocity:1/2*2800*27.777^2/1000=1080
1/2*2800*13.888^2/1000=270
i am not sure if i need to divide in Thousand as i did but something here is not look me correct what you say?

thanks.

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Again, you're correct. However, as unknown said, dividing by 1000 is okay but it changes the units from J to kJ. Personally, I would NOT divide by 1000. It just adds an extra step, adds confusion, and will have to be undone in order to calculate the height the car can go uphill. The standard unit for energy is J, so I'd just leave it that way. Thus, your answers would be 1080247 J before and 270062 J after. I know these values seem very large compared to the numbers we normally see in physics problems, but remember that a car is very heavy and moving very quickly.

Quote:

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well 4.0*10^3 Kj
but i am not know how to find the way for this answer

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The answer to this part of the question is simply the difference between the two numbers from above. I'm not sure where you got your answer of 4000000 J?

Finally, use what Unknown008 said, and apply the law of conservation of energy (my formula from the first answer is equivalent to that).

Originally the car had all kinetic energy (1080247 J of KE). After going up a hill, it sill has some KE (270062 J of KE) and some gravitational potential energy (GPE). How much GPE must it have if the total energy before and after is the same?

before = after
1080247 = 270062 + GPE

And since GPE = mgh, all you need to do is solve for h. What do you get?

hi.

well I got the answer from the book. :)

OK so I not dividing by 1000.
so the Ek that I can generate from this Slowing = 1080247-270062 ?
and the answer for the Height that this energy was can Climb up the Vehicle Up the hill is 150 meter (also I took it from the book.
GPE=810185

Unless you typed something incorrectly in the original question, I think your book is wrong! :)

The energy that you can generate from slowing is correct (and is, of course, the same as the amount of GPE that you can add, 810185).

Now if you take 810185 J = 2800 kg * 9.81 m/s^2 * h, you get h = 29.5 m.

If the amount of energy really was 4000 kJ, you'd get about h = 150m, so at least the book is consistent about THAT part. But I have no idea where they got 4000 kJ!

well my book is Chemical Principles the quest for insight.
maybe I typed something incorrectly in the original question but I don't think:)

about your last answer: "Now if you take 810185 J = 2800 kg * 9.81 m/s^2 * h, you get h = 29.5 m."
here you use in the Formula Ep=mgh? I am not work yet with GPE.
and what Value did you put in m/s^2 m=?
s=?
thank you really much.

Yes, this is the Gravitational potential energy formula for short distances :)

And

m is the mass of the vehicle, which is 2800 kg
g is the acceleration due to gravity, which is
h is the height the vehicle can reach.

m/s^2 is a unit, so, you don't really have to find what m and s are, just so you know :)

OK so to find h I need to do 810185 J/2800 kg * 9.81 m/s^2 = h?
I just divide the both side of the equation in 2800 kg * 9.81 m/s^2
if I correct is not gave me 29.5 m
thank you :)

This is what jcaron got :)

So I'll say you got it correct!

Sorry but I want to make sure I understood your English (which is quite good considering it's not your primary language). You did get 29.5 m? Or you did not?

You should put some parentheses in your equation to find h: h=810185 / (2800*9.81)= 29.5. Otherwise, as you wrote it, it's the same as (810185/2800)*9.81, which is incorrect.

I felt a lot more confident about it after you, too, said you thought 4000 kJ was incorrect. :)

No no I sorry I did some mistake in my calculation and yes I also get 29.5m
Thank you so much for helping also to you unknown008 :)

Just to be sure what we Calculate here on the first Section is the Difference between the Initial kinetic energy Minus the final kinetic energy?

Thanks.

Yes, it is :)