Why Does Kinetic Energy Depend on the Square of Speed?

AI Thread Summary
Kinetic energy depends on the square of speed due to its definition, which aligns with dimensional analysis showing energy units as kg*m^2/s^2. The relationship can be derived from Newton's second law, where the work-energy theorem connects force, distance, and velocity, leading to the formula E = m*v^2/2. This squared dependency also distinguishes kinetic energy from momentum, which is linearly related to speed. The physical interpretation suggests that kinetic energy reflects the energy of motion without directional dependence. Understanding this relationship is crucial for grasping fundamental physics concepts.
lella
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Why is it that kinetic energy depends on the square of speed instead of simply speed? Is there an example that shows why?
 
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lella said:
Why is it that kinetic energy depends on the square of speed instead of simply speed? Is there an example that shows why?


Its really just because that's the definition of kinetic energy. Momentum is defined as the quantity that depends on the speed linearly.
 
That's a very interesting question - sure its defined that way, but why?

I suppose this isn't truly a satisfatory answer, but we could use dimensional analysis to show that it indeed must be squared:

In SI, we're looking for energy in kg*m^2/s^2

Potentional energy: m*g*h so (kg)(m/s^2)(m) looks good

Kinetic Energy: m*v^2 (kg)(m/s)^2 looks good too!

As far as a "deeper" reason, I don't think I can add anything off hand.
 
The guy above is cute. But I have another explanation.
If the mass is M, and at the beginning it rested on the non-fiction ground.
Then we give it a force F.
We know that after S meters' movement, the kientic of the mass will be E=F*S ;
but now we will express it in another way, with the V and M.
we know the acceleration of the mass is a = F/M
and there is the relation S = V^2/(2*a) = V^2/(2*F/M)=MV^2/(2F)
So E = F*S = MV^2/2
it is the kinetic.:approve:
 
You can get it from the second Newton's law
F=m*dv/dt

According to Work(W)-Energy(E) theorem we have the definition of kinetic energy
dE=dW=F*dx=F*v*dt=m*v*dv=d(m*v^2/2)

From which we get the kinetic energy
E = m*v^2/2

Hope this helps

Prof. Nikitin,
My Quick Online Physics Reference Guide and Help:
http://physics-help.info
 
From a pure foundation and without the benefit of hindsight, the answer of course, is that it is not necessary that kinetic energy be dependant on the square of velocity. This could only have been (and was, by Joule) determined empiracally. The other answer is because momentum, which is the quantity of motion, already does.

Kinetic energy is the energy of motion, it [energy] allows us to note a change of some system's state with respect to time. It also makes sense physically that kinetic energy would depend on the square of velocity since its value does not depend on direction.
 
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Hello everyone, Consider the problem in which a car is told to travel at 30 km/h for L kilometers and then at 60 km/h for another L kilometers. Next, you are asked to determine the average speed. My question is: although we know that the average speed in this case is the harmonic mean of the two speeds, is it also possible to state that the average speed over this 2L-kilometer stretch can be obtained as a weighted average of the two speeds? Best regards, DaTario
I know that mass does not affect the acceleration in a simple pendulum undergoing SHM, but how does the mass on the spring that makes up the elastic pendulum affect its acceleration? Certainly, there must be a change due to the displacement from equilibrium caused by each differing mass? I am talking about finding the acceleration at a specific time on each trial with different masses and comparing them. How would they compare and why?
This has been discussed many times on PF, and will likely come up again, so the video might come handy. Previous threads: https://www.physicsforums.com/threads/is-a-treadmill-incline-just-a-marketing-gimmick.937725/ https://www.physicsforums.com/threads/work-done-running-on-an-inclined-treadmill.927825/ https://www.physicsforums.com/threads/how-do-we-calculate-the-energy-we-used-to-do-something.1052162/

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