Kinetic Energy and height unknown

  • Thread starter future_vet
  • Start date
  • #1
169
0

Homework Statement


Determine the KE of a 2000 kg car with speed = 22.3 m/s.

Homework Equations


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

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!
 

Answers and Replies

  • #2
2,063
2
Hint: Potential energy
 
  • #3
236
0
Another hint:Think of total energy, what can you say about the total energy of the car?
 
  • #4
169
0
Is this for the first or the second question?
 
  • #5
236
0
both hints , are for the second part, where you need to know the height.
 
  • #6
169
0
PE = m x g x h
m = 2000Kg
g = 9.8 m/s^2
h = the unknown
PE = ? is it related to v?
 
  • #7
236
0
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
169
0
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
2,063
2
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:
  • #10
169
0
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
2,063
2
Do we still use: PE = m x g x h and Et=Ek+Ep ?
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
169
0
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
2,063
2
That's right.
 
  • #14
169
0
Thank you!
 

Related Threads on Kinetic Energy and height unknown

Replies
3
Views
1K
Replies
6
Views
3K
Replies
7
Views
4K
Replies
12
Views
687
Replies
1
Views
3K
Replies
6
Views
3K
Replies
2
Views
780
  • Last Post
Replies
4
Views
613
  • Last Post
Replies
6
Views
2K
Top