Register to reply

Kinetic/potential energies for a ball thrown down

Share this thread:
Kork
#1
Mar14-12, 02:20 PM
P: 33
Hi again!

So, this time I have a ball with a mass m that I throw down from some point with a height h.

I want to write the kinetic and potentiel energies that the ball has in the start and in the end for this ball that is falling to h=0.

What I have thought of is that the ball must have a potential energy, u = mgh before it is thrown and when it hits the grown, and when it's moving down the energy is kinetic. I dont understand completely why and when it's undergoing a certain form of energy.

How does mgh=1/2mv^2 relate to this situation?

In my notes I also have that E = mgh + 1/2mv2 , which for some reason is the start energy?
Phys.Org News Partner Physics news on Phys.org
Step lightly: All-optical transistor triggered by single photon promises advances in quantum applications
The unifying framework of symmetry reveals properties of a broad range of physical systems
What time is it in the universe?
Kork
#2
Mar14-12, 02:41 PM
P: 33
Or is this correct:

In the start I have that:
E = mgh + 1/2mv^2

What about when it hits the ground?
Doc Al
#3
Mar14-12, 02:53 PM
Mentor
Doc Al's Avatar
P: 41,475
Quote Quote by Kork View Post
Or is this correct:

In the start I have that:
E = mgh + 1/2mv^2
Yes, where h is the initial height above the ground and v the initial speed at which you throw the ball.
What about when it hits the ground?
Energy is conserved, so that potential energy gets transformed to additional kinetic energy.

The same equation applies, only now the height is the final height and the velocity is the final velocity. The total energy remains the same as it was at the start.

emailanmol
#4
Mar14-12, 03:05 PM
P: 297
Kinetic/potential energies for a ball thrown down

The basic concept is that mechanical energy of a body is constant if no external force does work on it.

In the start just before you throw the ball, the ball was at height h above the ground and had speed 0.
When you throw it you apply a force which does work on it and instantaneously makes the ball have speed v.

From here, till the ball reaches the ground the only external force acting on it is the gravity.

Now consider the ball and earth as a system so that gravity is now an internal conservative force and regard the energy due to it as potential energy.
Thus no external force acts on this system (till it hits the ground)and its total mechanical energy i.e sum of potential and kinetic energies is constant.

Initially the ball was at height h above the ground(where we regard potential energy to be 0).
So its Total E was mv^2/2 + mgh.

Later when it reached the ground (just before touching) suppose its speed becomes x.here height is 0 and body has only kinetic energy . Thus mx^2/2 = E.


Register to reply

Related Discussions
Potential and kinetic energies in Quantum Oscillator Advanced Physics Homework 0
Potential and Kinetic Energy with Time Thrown In Introductory Physics Homework 6
Potential/Kinetic Energies Introductory Physics Homework 7
Kinetic/Potential energy and thrown rocks Introductory Physics Homework 3