# Kinetic Energy and height unknown

1. Mar 8, 2007

### future_vet

1. The problem statement, all variables and given/known data
Determine the KE of a 2000 kg car with speed = 22.3 m/s.

2. Relevant equations
E = 0.5 x 2000 x 22.3^2= 497290 Joules or 497 KJ.

3. The attempt at a solution
Above.

The car is lifted vertically and dropped from rest. Find the height from which it was released if v before hitting the ground = 22.3m/s.

Here I don't know what to do...

Thanks!

2. Mar 8, 2007

### neutrino

Hint: Potential energy

3. Mar 8, 2007

### ziad1985

Another hint:Think of total energy, what can you say about the total energy of the car?

4. Mar 8, 2007

### future_vet

Is this for the first or the second question?

5. Mar 8, 2007

### ziad1985

both hints , are for the second part, where you need to know the height.

6. Mar 8, 2007

### future_vet

PE = m x g x h
m = 2000Kg
g = 9.8 m/s^2
h = the unknown
PE = ? is it related to v?

7. Mar 8, 2007

### ziad1985

the potential enegry at the height h, should be equale to the potential energy of the car as it reach the ground..
Et=Ek+Ep
Et is conserved..
can you continue now ?

8. Mar 8, 2007

### future_vet

Is the total energy the value I found in the first part? 497 KJ?
So I divide 497 by 2 to find Ep?
And then I can find h...

Is this correct?

9. Mar 9, 2007

### neutrino

I think ziad meant to say, "the potential enegry at the height h, should be equal to the kinetic energy of the car as it reach the ground."

In general, for a particle moving in a gravitational field not affected by forces of friction and such, the total energy will remain a constant, and it will be equal to the sum of its (gravitational) potential and kinetic energies.

Last edited: Mar 9, 2007
10. Mar 9, 2007

### future_vet

I am completely confused... Could you show me what you would do so that I can understand the process?
Do we still use: PE = m x g x h and Et=Ek+Ep ?

Thank you...

11. Mar 9, 2007

### neutrino

Yes.

Et, at all times, is a constant.

When the car is at a height h (which you need to find out), its potential energy(Ep) is mgh, but it is at rest, so Ek = ?

But when the car hits the ground, it does so with some velocity, and you already know what it's kinetic energy is at the velocity. If you take the ground to be the 0 of the height, Ep at the bottom = ?

Et(at h) = Et(at 0), solve for h.

12. Mar 9, 2007

### future_vet

Here's what I did, if it's not right, then I have no clue...

h = 1/2mv^2/mg=1/2v^2/g
= (22.3^2 x 0.5)/9.8 = 25.4 meters.

13. Mar 9, 2007

### neutrino

That's right.

14. Mar 9, 2007

### future_vet

Thank you!

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