GPE & KE: Understanding the Fundamentals

In summary, to determine the velocity of a ball as it leaves your foot and the KE and GPE at the time of release, you need to know the formulae for GPE and KE and the quantities involved. To calculate how high the ball will go, you can use the principle of Energy Conservation, assuming no energy is lost. In this conversation, the participants also discuss how to measure the velocity at the time of release, the effects of force and distance on motion, and the impact of air resistance on objects.
  • #1
Fabian901
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I’m trying to get my head around the concepts of GPE and KE.

Imagine I kick a ball vertically. What information do I need to determine the velocity of the ball as it leaves my foot, the KE and the GPE at the time the ball is released?
 
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  • #2
Do you know the formulae for GPE and KE? They contain the quantities that you need to know about. What do you need to calculate? You haven't actually asked a practical question yet. To find how high a ball will go (or a number of other similar questions, you can rely on the fact that (ideally) no energy is lost so that the GPE at the maximum height will equal the initial KE + GPE in the system. This is referred to as Energy Conservation.
 
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  • #3
sophiecentaur said:
Do you know the formulae for GPE and KE? They contain the quantities that you need to know about. What do you need to calculate? You haven't actually asked a practical question yet. To find how high a ball will go (or a number of other similar questions, you can rely on the fact that (ideally) no energy is lost so that the GPE at the maximum height will equal the initial KE + GPE in the system. This is referred to as Energy Conservation.
Okay, say I have the following problem and I want to calculate the velocity of the ball as soon as it is released and the maximum height it reaches.

If I kick a ball of mass 0.5Kg vertically with a forcé of 30N what will be the ball’s velocity as soon as the ball is released? Assuming the ball leaves my foot at a height of 0.3m from the ground and my foot makes contact with the ball for a distance of 0.05m before releasing it. Gravity = 9.8m/s^2.

Once I have the velocity it should be straight forward to calculate maximum height with V^2=u^2 + 2as, but how do I measure the velocity at the time of release?
 
  • #4
Calculating velocity isn't too hard. If your foot produces a force of 30 N, the net force on the ball is 30-4.9= 25.1 N
So the acceleration of the ball upwards is 25.1/0.5= 50.2 m/s^2. Use v^2 = u^2 + 2as to calculate the velocity at release. So v^2 = 0+2*50.2*0.05
 
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  • #5
Assuming the ball is at rest when you kick it
 
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  • #6
UncertaintyAjay said:
Calculating velocity isn't too hard. If your foot produces a force of 30 N, the net force on the ball is 30-4.9= 25.1 N
So the acceleration of the ball upwards is 25.1/0.5= 50.2 m/s^2. Use v^2 = u^2 + 2as to calculate the velocity at release. So v^2 = 0+2*50.2*0.05
The 'Force' of the kick is not enough to predict how fast the ball will travel. You need to specify either the distance over which the kick is applied (giving the work done) or the time (the Impulse). This is a collision problem (which is what the kick is, basically). Either way, it is not easy to specify those initial conditions because a kick does not apply a constant force during the contact. Text book problems usually specify some very simplified conditions in their set questions but, to apply this to a practical problem is a bit of a nightmare - as are all mechanical calculations involving muscles and skeletons. In many ways it's easier to work backwards from the height reached and to work out either the KE or the Impulse imparted to the ball.
 
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  • #7
True, in a practical situation you can't get the data. But Fabian specified a figure for the distance ( don't know how he got it) . Working from that its pretty simple
 
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  • #8
Thanks a lot for the answers! Obviously in practical situations the force would not be constant across that small distance, but if we knew how the force varied and plotted it on a force displacement graph would we still be able to calculate KE (the area under the curve) and therefore the balls velocity when released?
Also, now that we know all the variables, would GPE and KE (at the time the ball is released) have the following values:

GPE = mgh = 0.5 x 9.81 x 0.3 = 1.47 J
KE = 0.5mv^2 = 0.5 x 0.5 x 5.02 = 1.255 J
Would all that KE be converted to GPE as the ball reaches maximum height?

One more question. Are we exerting a force through a distance every time we are giving motion to an object? Kicking a ball would be a clear example but if we are playing ping pong for instance, are we exerting a force through a distance every time we hit the ball with our table tennis racket? Obviously that distance would be excessively small but it would still not equal 0 would it?
 
  • #9
Fabian901 said:
Thanks a lot for the answers! Obviously in practical situations the force would not be constant across that small distance, but if we knew how the force varied and plotted it on a force displacement graph would we still be able to calculate KE (the area under the curve) and therefore the balls velocity when released?
Also, now that we know all the variables, would GPE and KE (at the time the ball is released) have the following values:

GPE = mgh = 0.5 x 9.81 x 0.3 = 1.47 J
KE = 0.5mv^2 = 0.5 x 0.5 x 5.02 = 1.255 J
Would all that KE be converted to GPE as the ball reaches maximum height?

One more question. Are we exerting a force through a distance every time we are giving motion to an object? Kicking a ball would be a clear example but if we are playing ping pong for instance, are we exerting a force through a distance every time we hit the ball without table tennis racket? Obviously that distance would be excessively small but it would still not equal 0 would it?

Air resistance will have an effect on a relative low mass / large area object. You can work out the drag factor for an spherical object in air. Incorporating that would give a better idea of final height. Low velocities will produce less effects from drag, of course.

Even a table tennis ball impact takes a finite time (so does the impact between the steel balls in a Newton's Cradle). In many calculations about collisions, they use a quantity called Coefficient of Restitution which is the ratio of the separation speed to the approach speed. It's 1 (100%) for a perfect elastic collision and can be anything down to Zero (for two lumps of putty). That is a simplified figure which assumes that the same proportion is lost for all speeds but is pretty successful in practice.
In many collision calculations, the actual time or distance during impact can be ignored and you can just work on the assumption that Momentum is Conserved and deal with the initial and final situations. COR comes in very handy for this, of course. I remember doing endless collision calculations in A Level Applied Maths.
 
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  • #10
sophiecentaur said:
Air resistance will have an effect on a relative low mass / large area object. You can work out the drag factor for an spherical object in air. Incorporating that would give a better idea of final height. Low velocities will produce less effects from drag, of course.

Even a table tennis ball impact takes a finite time (so does the impact between the steel balls in a Newton's Cradle). In many calculations about collisions, they use a quantity called Coefficient of Restitution which is the ratio of the separation speed to the approach speed. It's 1 (100%) for a perfect elastic collision and can be anything down to Zero (for two lumps of putty). That is a simplified figure which assumes that the same proportion is lost for all speeds but is pretty successful in practice.
In many collision calculations, the actual time or distance during impact can be ignored and you can just work on the assumption that Momentum is Conserved and deal with the initial and final situations. COR comes in very handy for this, of course. I remember doing endless collision calculations in A Level Applied Maths.
I see! Thanks a lot for the answer!
 

What is the difference between GPE and KE?

GPE stands for gravitational potential energy, which is the energy an object has due to its position in a gravitational field. KE stands for kinetic energy, which is the energy an object has due to its motion. The main difference between the two is that GPE is related to an object's position, while KE is related to an object's motion.

How are GPE and KE related?

GPE and KE are related through the Law of Conservation of Energy. This law states that energy cannot be created or destroyed, only transformed from one form to another. In the case of GPE and KE, when an object falls, its GPE is converted into KE. Similarly, when an object is lifted, its KE is converted into GPE.

What factors affect an object's GPE and KE?

The main factor that affects an object's GPE is its height above the ground. The higher the object is, the greater its GPE. On the other hand, an object's KE is affected by its mass and velocity. The greater the mass and velocity of an object, the greater its KE.

Can GPE and KE be negative?

Yes, both GPE and KE can be negative. This usually occurs when the reference point for measuring GPE is below the ground level, or when an object is moving in the opposite direction of the chosen positive direction for measuring KE. Negative values indicate that the object has less energy than the reference point.

How are GPE and KE used in real-life applications?

GPE and KE are used in various real-life applications, such as roller coasters, water dams, and bungee jumping. In roller coasters, the potential energy is converted into kinetic energy as the cars go down a hill, providing the thrill of the ride. Water dams use the potential energy of water stored at a height to generate electricity. Bungee jumping relies on the conversion of GPE into KE as the person jumps off a tall platform.

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