Kinetic energy and momentum.

Ek = 0.5mv^2
P=mv

m=1, v=1
Ek = 0.5

m=1, v=2
Ek = 2

Accelerating a 1 kg object to 1m/s requires 0.5 joules whilst accelerating it a further 1m/s to 2m/s requires an additional 1.5 joules. Why is this?

I can't exactly explain it, but I could give you an analogy that might help.

replace "m" with dicks, and "v" with niggers, while "Ek" is france.

So, if we have 1 m and 2 v, that means we'd have exactly one dick inside of two niggers who are both in france times two!

>>1
Energy is acceleration times mass times distance. Suppose that over some distance you accelerate this mass from a m/s to b m/s. Then b^2=a^2+2*acceleration*distance. Multiply it by mass and you get mass*b^2=mass*a^2+2*acceleration*mass*distance=mass*a^2+2*energy.
Therefore, energy=(1/2)mass*(b^2-a^2)

>>3
How does that explain why in one instance it takes 0.5 joules to accelerate 1m/s whilst in another it takes 1.5 joules? If you are in a rocket in a vacuum with an engine that uses the same amount of energy per second, does this mean that acceleration decreases as you go faster?

>>4
This is because when you accelerate something over a certain distance, the acceleration*distance is directly proportional to the differences of the SQUARES of the velocities:

b^2=a^2+2*acceleration*distance

The reason why the above equation is so is simply by considering the fact that for any function f such that f''(x)=k where k is a constant, f'(b)^2-f'(a)^2=(f'(a)+(b-a)k)^2-f'(a)^2=f'(a)^2-f'(a)^2+2f'(a)(b-a)k+(k^2)(b-a)^2=2f'(a)bk-2f'(a)ak+(k^2)(b-a)^2=2k(f'(a)(b-a)+(1/2)k(b-a)^2)=2k(f(b)-f(a)).

>>4
For your second question, the answer is yes.

>>5
>>6
So why does the rocket engine apply less force even though it is expending the same amount of energy?

>>4
>>5
I know how v=u+at, v^2=u^2+2as and s=ut+0.5at^2 are derived. I just don't see how energy relates to force, I understand relativism, but the fact that you need 3 times as much energy to accelerate relative to an observer from a mere 1m/s to 2m/s compared to the acceleration from 0 m/s to 1m/s doesn't make sense. Even with air resistance and friction a car expending the same amount of energy will not experience a drop in acceleration by that amount.

Let's say a space maglev of 1 kilogram which experiences no resistance converts 1 joule of electrical energy into 1 joule of kinetic energy per second.

power=1, M=1
At first it accelerates enormously since the energy needed to accelerate to get from standing to 0.01 m/s is...
E = 0.5v^2 = 0.00005
and because 1 joule of kinetic energy is provided per second it takes 0.00005 seconds to get to 0.01 m/s
This the equivalent of hitting concrete after jumping off a 30 storey building.

When it reaches 1 m/s it's mean acceleration is at the reasonable rate of 0.5 m/s.

However it will take 3 times as much energy to reach 2 m/s as I explained. This means it's acceleration has dropped by a third even though it has used the same amount of energy to propel itself throughout it's acceleration.

>>8
Energy=force*distance. Thus the faster it goes, the more distance is covered, so there is less force. Do you understand?

A newton is the force required to accelerate 1 kg by 1 m/s^2.

The work done (ie the energy) is measured in newtons, and how far the object is moved. (W=F*d)

How far will a still object move if you apply an even 1 N force over the course of 1 second?

How far will that same moving object travel under the same conditions during the next second? That's the difference.

Kinetic energy depends on your frame of reference.

>>9
>>8 proves I understand, you don't understand my questions.

>>10
Why did he choose force*distance? Why not force*time? Shouldn't it require a constant amount of energy to generate a constant amount of force? Why do my calculations for kinetic energy prove that this isn't the case? How does kinetic energy relate to chemical energy when apparently you need more and more energy to accelerate at the same rate?

>>11
Why? I would prefer to be hit by a car going 1 mph than a bullet going 2000 mph even though they have the same momentum, but this is because of lower shock and higher surface area not kinetic energy.

>>10
momentum = force* time

Newton could have used the momentum needed to propel an object into the air instead of kinetic energy.

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