Requesting help with Potential and Kinetic energy

AI Thread Summary
The discussion centers on the concepts of potential and kinetic energy using the example of a ball on a ramp. When a force is applied to the ball, it gains kinetic energy, which is converted to gravitational potential energy as it ascends the ramp. At the top, the ball has only potential energy, and as it descends, that potential energy is converted back into kinetic energy. Before any force is applied, the ball has gravitational potential energy relative to its height, even if it is at ground level. The conversation emphasizes the conservation of energy principle, stating that energy is always conserved regardless of the starting height.
soccer5454
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Hello PF. I have a question on a concept of Potential and Kinetic energy. So for this let's take a ramp and a ball at the bottom. If I apply a force on the ball to make it go up the ramp, when it starts to move it will gain kinetic and potential gravitational ( I think ) but what about at the very top. Will it just be gravitational-potential? Also at the ground before a force is applied, does it have any energy, thanks!
 

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If you first hit the ball, it will gain kinetic energy. As it goes up the ramp, the kinetic energy will gradually be converted to gravitational potential energy. When it stops, it has indeed only potential energy. As it rolls back down, the process is reversed.

Before the force is applied, it has gravitational potential energy: if you were to make a hole under the ball, it would fall. But it of course has less gravitational potential energy than when it is up the ramp.
 
But I thought that it can only have gravitational potential energy if it has height..also I don't understand how it is able to start off that that much kinetic energy right from the beginning :(
 
soccer5454 said:
But I thought that it can only have gravitational potential energy if it has height..also I don't understand how it is able to start off that that much kinetic energy right from the beginning :(
Unless it is at the center of the Earth, it has height :smile:

The amount of kinetic energy it will have depends on how much force is applied for how long of a time. You must know by experience how much of a push you need to give to a ball to get it up a ramp.
 
DrClaude said:
Unless it is at the center of the Earth, it has height :smile:

The amount of kinetic energy it will have depends on how much force is applied for how long of a time. You must know by experience how much of a push you need to give to a ball to get it up a ramp.
Ohhhh I get it, so if I just put a pencil on the table and don't touch it, it still has gravitational potential energy right?
 
soccer5454 said:
Ohhhh I get it, so if I just put a pencil on the table and don't touch it, it still has gravitational potential energy right?
Right. Just remove the table and you'll see.
 
DrClaude said:
Right. Just remove the table and you'll see.
Awesome, thanks! :)
 
I should probably clarify that for what is important for the vast majority of problems is the change in gravitational potential energy. Therefore, for convenience, you could take the table to be at the zero of energy (height = 0), and work from that, not having to consider the additional potential energy from the height of the table, which would cancel out in the end. Likewise, for the setup you posted, one would most often take the ball at the base of the ramp to have no gravitational potential energy for the purpose of studying its motion up and down the ramp.
 
Don't worry so much about the initial force applied, only consider the ball after the force is applied. Considering Newton's First law of motion, an object at rest will stay at rest or in motion until a force is acted upon it. When you apply a push to the ball to go up the ramp, you apply a force to give it an initial velocity, but once you let go of the ball, the force become irrelevant because you are no longer applying a force on the ball, thus we only need to consider the ball after you push it.

As the ball goes up the ramp, it has some initial velocity and is deaccelerating because there is a net force of gravity acting on it, causing the ball to slow down until it eventually stops and falls back to Earth.

Consider the conservation of energy here:
When you first push the ball, it is has kinetic energy, and as it travels up, some height h, it converts that kinetic energy into potential energy, and when it reaches the top it has all of its energy stored in potential, but as it falls back towards the Earth, it gains speed due to the acceleration of gravity and that potential energy is converted back into Kinetic energy. So it doesn't matter what height you start above the Earth, energy is always conserved.

Remember that:
\begin{equation}
\frac{1}{2}mv^2 = \Delta mgh
\end{equation}

\begin{equation}
\Delta KE = \Delta PE
\end{equation}
 
  • #10
Dopplershift said:
Don't worry so much about the initial force applied, only consider the ball after the force is applied. Considering Newton's First law of motion, an object at rest will stay at rest or in motion until a force is acted upon it. When you apply a push to the ball to go up the ramp, you apply a force to give it an initial velocity, but once you let go of the ball, the force become irrelevant because you are no longer applying a force on the ball, thus we only need to consider the ball after you push it.

As the ball goes up the ramp, it has some initial velocity and is deaccelerating because there is a net force of gravity acting on it, causing the ball to slow down until it eventually stops and falls back to Earth.

Consider the conservation of energy here:
When you first push the ball, it is has kinetic energy, and as it travels up, some height h, it converts that kinetic energy into potential energy, and when it reaches the top it has all of its energy stored in potential, but as it falls back towards the Earth, it gains speed due to the acceleration of gravity and that potential energy is converted back into Kinetic energy. So it doesn't matter what height you start above the Earth, energy is always conserved.

Remember that:
\begin{equation}
\frac{1}{2}mv^2 = \Delta mgh
\end{equation}

\begin{equation}
\Delta KE = \Delta PE
\end{equation}
Thank you so much :))
 
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