How to intuitively see the v^2 relation to kinetic energy?

[ForTheGreater]

KE is proportional to v^2. In a gravitational field KE=1/2 m*v^2.
It's easy to find mathematically Work=Fd=mad=m(v/t)(v*t)=m*v^2.

But how to visualize it or get an intuitively "feel" for this v^2 relationship?

 

[Mastermind01]

Perhaps you can check out this stackexchange answer: http://physics.stackexchange.com/qu...ncrease-quadratically-not-linearly-with-speed

 

[mrspeedybob]

The amount of power it takes to push something is equal to how hard you push times how fast you push. So far that's pretty intuitive right?

Accelerating something that's moving fast requires me to push fast.
Accelerating something that's moving slow requires that I push slow.

If I push equally hard on the fast thing and the slow thing, then I'll put more power into the fast thing, so, it will be accumulating more kinetic energy while accelerating at the same rate.

 

[ForTheGreater]

If I push equally hard on the fast thing and the slow thing, then I'll put more power into the fast thing, so, it will be accumulating more kinetic energy while accelerating at the same rate.

Yes, thank you.

Funny you should respond today. I was thinking about this again today and I reason as such that momentum is m*v and the rate at which momentum is traveling is (m*v)*v which is kinetic energy. I like the way you put it very much too.

 

[anorlunda]

There is a completely different way to approach a semi-intuitive understanding that agrees with everyday observations.

Consider two cars colliding while traveling 60 mph. If they were traveling in the same direction, the consequences are very different than if they were traveling in opposite directions. Thus, momentum has a +- sign. $v^1$ has a +- sign.

Consider the energy needed to accelerate a car to 60 mph. You could measure that by the fuel consumed during acceleration. The fuel needed is always positive regardless of the +- orientation of the drag strip. $v^2$ is always positive.

Using these arguments, momentum must be proportional to an odd power of v and energy proportional to an even power of v. 1 and 2 are the smallest (simplest) nonzero powers that fit. If you always guess the simplest possible solution, you will nearly always be correct.

Therefore, based only what you observe at the drag strip, your intuition should make you guess $v^2$ for K.E. and $v^1$ for momentum. Further search of your intuitive knowledge of car collisions (based on watching Mythbusters on TV) should lead you to conclude that both momentum and K.E. must also be proportional to mass.

$p=mv$ and $K.E.=mv^2$ can be completely intuitive without studying physics. Do you agree?

 

[robphy]

Good Definitions are a good place to start.
$Work=\int v dp$ is a very general definition.
Now use p=mv. Then work = the change in kinetic energy. The important factor of 1/2 shows up.

The same definition works in special relativity... but you don't get a quadratic relation.