Target880 t1_j5wncaw wrote
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
ThrowawayHomesch OP t1_j5x79rh wrote
Thanks, I think this is starting to make sense. So when the car changes speed, it's applying work against the ground/earth and that's the frame of reference? I think that's what I was missing
If I'm understanding it right, this should not apply to a spacecraft right? I'm assuming it has some kind of rocket/thruster for it to change speed. In that case, the work is being done against the exhaust gasses exiting the nozzle? So it shouldn't matter at what speed the spacecraft is traveling since the thing it's doing work against (propellant) is always traveling at the same speed as the spacecraft?
UntangledQubit t1_j5xj958 wrote
> So it shouldn't matter at what speed the spacecraft is traveling since the thing it's doing work against (propellant) is always traveling at the same speed as the spacecraft?
That is exactly right.
From the spaceship, when I expel some exhaust, I gain some constant amount of energy. So, it makes sense to me that me acceleration is steady.
From the ground, for a spaceship to have steady acceleration, it must be gaining more and more energy per second. The faster I am going, the more energy I need to gain a little bit more speed. This extra energy comes from the fact that the fuel is moving along with the spaceship - from the ground, the fact that it's moving means it has extra kinetic energy to expend on propelling the spaceship. If you do the algebra, you'll find that the extra quadratic terms on the spaceship's energy and the fuel's energy exactly cancel out.
So the actual effect (constant acceleration of the spaceship) looks the same, but the accounting of which object has how much kinetic energy looks completely different.
Target880 t1_j5yvdhl wrote
>Thanks, I think this is starting to make sense. So when the car changes speed, it's applying work against the ground/earth and that's the frame of reference? I think that's what I was missing
The frame of reference is just what you define as being stationary for your calculations. It is something that just exists in the model you use for the calculation.
The speedometer of the car measures the speed of the car relative to what is rolling on. Let's assume that it is a day with no wind. let's say the speedometer shows 120, the unit does not matter, that is the relative speed between the car and the ground.
So you can use the ground as the frame of reference then the car moves forward at 120 or you can use the car then the ground and the air move backward at 120. So any calculation you do need to include both parts
So the situation is like if you drive a remote-controlled toy car on a treadmill and add a fan to get the air moving. Compare that car to a remote-controlled car on the floor beside it. From the behavior, it is quite clear that the moment of the treadmill has an effect on the behavior of the car.
So if you use a car that is more relative to the earth as the frame of reference you need to include that earth is moving backward just like if you drive on a treadmill and the frame of reference is the ground you need to include the motion of the band on the treadmill
For spacecraft or any other rocket engine, the propellant is moving at the same speed as it. They will both have the same speed regardless of what frame of reference is used. What speed at both moves depends on the frame of reference you use but it will always be the same.
The result is it works out the same regardless of what framer of reference you use, what can change it how hard the calculation is
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