# GPE and KE

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?

Related Other Physics Topics News on Phys.org
sophiecentaur
Gold Member
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.

Fabian901
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?

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

Fabian901
Assuming the ball is at rest when you kick it

Fabian901
sophiecentaur
Gold Member
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.

Fabian901
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

Fabian901
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?

sophiecentaur
Gold Member
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 out 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.

Fabian901
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!