# How are work and KE related?

• I
Hell Ok, I am in my thermos class and I ran into a question about work and energy. I figured out the correct answer based off of the formula given. But what MY question is, goes back to physics 1.

How are work and Ke related? IE, work is F*d and F is mg. But in Ke, I do not see the g. How did

Fd = (m(v2))/2 I know an integration is in the works because that is where the square and the half come from, but the rest, I do not remember

thanks.

I step by step would be nice, even if it is a vide Last edited by a moderator:

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kuruman
Homework Helper
Gold Member
Forces do work on objects. In so doing, the kinetic energy increases or decreases depending on whether the work done on the object is positive or negative. If you release an object from rest at height h, the work done by gravity is mgh and the kinetic energy change is ½mv2. The two are equal, so mgh = ½mv2.

• BeedS, LT72884 and davenn
Ok, I see the force in Pe, as in mass*gravity, and I see the distance h. So in that case work is being done because its a force*distance. that makes more sense. thank you.

Now, for the second half, where did the equation of 1/2mv2 come from again? I used to know it but I do not remember it anymore

Hell Ok, I am in my thermos class and I ran into a question about work and energy. I figured out the correct answer based off of the formula given. But what MY question is, goes back to physics 1.

How are work and Ke related? IE, work is F*d and F is mg. But in Ke, I do not see the g. How did

Fd = (m(v2))/2 I know an integration is in the works because that is where the square and the half come from, but the rest, I do not remember

thanks.

I step by step would be nice, even if it is a vide Hi LT72884,

Kinetic energy is just another way of representing work. I’ll show how you go from W=Fd to KE= (1/2)mv2. There is a equation which states that v2= u2 + 2as. If we leave s(distance) on one side of the equation, we are left with s= v2/(2a) ( let’s say that there is no initial velocity, which is written with the letter u in the equation ). Therefore, since s is distance and d is distance as well ( there are different letters but mean the same, it’s arbitrary), W=Fd , d= v2/(2a). So W= F(v2/(2a) ) .F=ma, so ma(v2/(2a) ). We cancel out a(acceleration), and we get W=(1/2)mv2. We represent this type of work as kinetic energy, but means the same thing just applied in a situation you might encounter in nature. Hopefully I was of help.

• LT72884
yes, Thank you:) that was interesting to see it done that way. After I continued to exploer a little bit, I did find this as well:

F=m(dv/dt) use chain rule end up with Fdx=mvdv then F = integration[mvdv] + c

so amazing how both of these equations take you to the same place. I love math!

sophiecentaur
Gold Member
Forces do work on objects. In so doing, the kinetic energy increases or decreases depending on whether the work done on the object is positive or negative. If you release an object from rest at height h, the work done by gravity is mgh and the kinetic energy change is ½mv2. The two are equal, so mgh = ½mv2.
But the change in KE can be as near zero as you like in many cases if the work is done slowly enough. No one seems to have taken this point which I have already made. KE change is only under certain circumstances.

kuruman
Homework Helper
Gold Member
But the change in KE can be as near zero as you like in many cases if the work is done slowly enough. No one seems to have taken this point which I have already made. KE change is only under certain circumstances.
I have not seen your earlier postings on this, so I am at a disadvantage. The understanding I have so far is that the work-energy theorem says that the change in kinetic energy of a mass is equal to the net work done on the mass by the external forces: ΔK = Wnet. It does not say anything about power, the rate at which work is done. It seems to me that if there is change in kinetic energy, there must be work done by the net force and vice-versa. The differential form of the work-energy theorem, dK = dWnet is also valid and that implies that the change in kinetic energy can be as small as one pleases. Am I missing something?

• LT72884
Acceleration or deceleration, it depends on the direction of the force applied relative to the direction of motion of the object being worked.

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• LT72884
But the change in KE can be as near zero as you like in many cases if the work is done slowly enough. No one seems to have taken this point which I have already made. KE change is only under certain circumstances.
As an example, Gravity constantly applies a force to the matter creating the gravity. There is no or little motion (KE) being transferred to the matter when the objects pressure to mass density is reached. Can I use this as an example of what you saying?

Is the pressure caused by the constant acceleration of the mass above it still considered (KE) that is instantly converted to pressure?

sophiecentaur
Gold Member
Am I missing something?
At some pointing history, there were images of a spring and someone lifting weights on the Earth in this thread. They are no longer there - weird (or was I hallucinating?). I was taking my context from those images and the post which contained them. My point was that Work, applied to a system from the outside, can change the KE and the PE of the system. That's pretty much obvious. The OP refers to "thermo" which implies that the gas laws were being discussed. For an ideal gas, the PE situation is ignored but real gases are all affected by PE between molecules. Adiabatic changes will always change the KE of a gas, unlike compressing a spring. All that is part of a normal thermodynamics course and it's not hard to understand in context.
My comments were pointing out that the KE change for a spring or a lifted mass will depend on the rate at which the Work is done. There is nothing contentious and the thread seems to sort of healed itself by losing the mechanical examples.

Lord Jestocost
Gold Member
.........and I ran into a question about work and energy.
"Energy is one measure of the state or condition of a system whereas work is not energy but a process that changes the energy of a system."
[PDF] Understanding energy

• LT72884 and sophiecentaur
sophiecentaur
Gold Member
As an example, Gravity constantly applies a force to the matter creating the gravity. There is no or little motion (KE) being transferred to the matter when the objects pressure to mass density is reached. Can I use this as an example of what you saying?
I can see what the problem could be, here. KE is frame dependent and you should consider the whole system. The Total Energy of the object w.r.t. to the Earth does not change if there are no outside forces. If an external force does work on the object then the Total Energy will change and the PE can change without affecting the KE.
If you choose to slow the fall of an object down so that its gain in KE is zero, then Energy will be transferred to the 'braking mechanism'.
Basically, the whole experiment needs to be described more fully in order to avoid our cross-purposes, I think.

• Lord Jestocost
"Energy is one measure of the state or condition of a system whereas work is not energy but a process that changes the energy of a system."
[PDF] Understanding energy
Thanks for the pdf. Thermo is a class that is different from all my other classes i have taken haha. Its a odd class

I can see what the problem could be, here. KE is frame dependent and you should consider the whole system. The Total Energy of the object w.r.t. to the Earth does not change if there are no outside forces. If an external force does work on the object then the Total Energy will change and the PE can change without affecting the KE.
I agree, only external energies can increase a systems total. Gravity is constantly providing the mass KE, but the systems total energy is not changing. Gravity is weird, it constantly supplies KE then steals/balances it back in another form.
If you choose to slow the fall of an object down so that its gain in KE is zero, then Energy will be transferred to the 'braking mechanism'.
In my example the matter being pressured below the object is the 'breaking mechanism' instantly decelerating the mass and causing/forcing pressure.

sophiecentaur
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