First, I’ve never taken a physics class or attended highschool before, hence the ELI5. I’ve read many explanations but it doesn’t really make intuitive sense to me. For example (assuming there’s no air resistance / drag), let’s say I was traveling in a car going 120 mph and I wanted to decelerate to 90 mph. This would take four times as much energy than going from 30 mph to 0.
But let’s say there were two cars traveling at 120 mph. The car next to me decelerates to 90 mph, but I’m still going 120. From my point of view, the car next to me just started going 30 mph in the opposite direction. Why would this require 4 times as much energy than if both cars were just stationary, and the car next to me actually started going 30 mph in the opposite direction?
And, let’s say we’re both standing on earth. One person at the north pole and one at the equator. Both of us throw a ball, but the ball at the equator is already traveling at something like 1,000 mph due to the earth’s rotation. Shouldn’t throwing a ball eastward then require way more energy to go from 1,000 to say 1,020 mph, than the person throwing the ball at the north pole who just has to accelerate it from 0 mph to 20 mph?
In: 3
Other have explained the frame of reference part but let’s look at the square part.
The reason is the square of the velocity is at high speed you need to apply the force for a longer distance.
(1) W = F *s
where W =Work (transferred enerergy), F = force and s= applied distance
(2) s = v *t
Where v is the speed and t is the time.
(3) a= F/m => F =m *a
where a is the acceleration and m is the mass.
(4) v= a * t => t= v/a
Consider if you an object with a mass of 1 kg that you accelerate with a force of 1N that means the acceleration is 1m/S
If you start at a speed of 0 the speed after 1 second is 1m/s so the average speed for that second is(0+1)/2= 0.5 m/s and you only travel 0.5m That is the distance you apply the force. So the work is 1 N * 0.5m = 0.5 Joules.
If you start at 10m/s the speed after 1 second is 11m/s so the average speed is (10+11)/2=10.5m/s . So you traveled 11.5m. That means the work is 1 N * 11.5m = 11.5 joule
If we just have constant acceleration 0 to v the average speed is v/2.
Combine 1 and 3 and we get (5) W= m *a *s
If we use 2 with the average speed you get (6) s= v/2 *t
Combine 5 and 6 and we get (7) W= m *a * v/2 *t
Combine 7 with 4 and we get W= m *a * v/2 * v/a = m * v/2 *v * a/A = m * v/2 *v = (m * v^2)/2
So the square is a result of that you need to apply the force over a longer and longer distance the higher the speed is and the work gets larger and larger.
It can also explain what we see in a different frame of reference. When a car slowed down from 120 to 90 by applying breaks the force is applied from the road. So from your point of view, the car change is speed but pushing on something that moved backward at a speed of 120 relatives to you. You need to include that the ground is not stationary compared to you.
For it to do less work whilst slowing down it needs to apply force to something that moves at the same speed as you.
You can compare this to you walkin’ in a train where you start waking forward in the direction of the moment. Your feet apply the force relative to the train so the distance is short. But at the same time, the train needs to apply the same force relative to the ground, if it did not it would slow down.
Compare it to walking in a canoe on the water next to the dock. It has very little friction from the water and the mass close to you. If you stand up and walk forward you push the canoe backward. When you reach the front of the canoe you have walked a short distance relative to the dock. So if you use the canoe as the frame of reference you need to consider how it moves relative to the dock. You moving in a train change it a moment too but it will be a lot less compared to you because its mass is many times higher than you
So if you jump off the train and hit the ground, the difference in energy from you walking forward versus stationary when you hit the ground from the work the train needed to do to keep its speed constant.
So it all works out regardless of the frame of reference. You just need to consider everything like what you apply a force relative to and what your action has on the frame of reference
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