If you look up the formula for the kinetic energy of a body in motion, you’ll see: KE = ½ mv^2, where ‘m’ is ‘mass’ and ‘v’ is ‘velocity’.  What is the derivation of this formula? 

Let’s start with the concept of uniform acceleration, ‘a’.  If a body is uniformly accelerated from a resting position, for example, at the rate of one meter per second per second, it means that at the end of one second, it will be moving at a velocity of one meter per second.  At the end of two seconds, it will be two meters a second.  At the end of three, three, and so forth.  So it immediately follows that velocity, ‘v’, in this example, has the following formula: v = at, where ‘t’ is ‘time’. Velocity equals acceleration times time.  Now the distance, ‘d’, traveled by a body going at a uniform velocity, ‘v’, is merely as follows: d = vt.  If the body is going 3 meters a second and it goes for 2 seconds, it will go 6 meters.  But in our example, the acceleration, not the velocity, is uniform.  Our velocity keeps increasing uniformly.  Our initial velocity was 0 and our final velocity was ‘at’.  Since we have uniform acceleration, we can get our average velocity, by averaging inital and final velocity: v_average = (0 + at) / 2 = ½ at.  Multiplying our average velocity by the total time, we get the total distance covered: d = ½ at^2.  If we imagine a force, F, acting over a distance, we say that the it does work and imparts energy.  Force is equal to mass times acceleration: F = ma.  Obviously, for a prescribed acceleration, the force necessary to propel the body will be proportional to the mass.  The energy received by the propelled body is equal to the work done, which is the product of force times distance, providing the force applied is uniform.  So we have KE = Fd.  But F = ma and d = ½ at^2.  So Fd = ½ m(at)^2.  But we know that v = at, so a = v / t. Therefore KE = Fd = ½ m (v/t x t)^2 = ½ mv^2.

Some things are dependent on the path.  Depending on the route I take, I could walk one, two, or three miles and still be only a mile from where I began.  Thus my effort was dependent on the path.

But kinetic energy is said to be independent of the path.  In other words, given two equally heavy cannonballs traveling side by side at the same velocity, it will not make any difference how they attained the velocity.  They have the same kinetic energy.  Thus, it is immaterial whether one was accelerated uniformly and the other desultorily.  So I can use the formula derived from the uniform-acceleration case and apply it to all bodies of equal mass and velocity.  Thus, the formula “KE = ½ mv^2” is universal.

When you drive your car, once you attain the speed you want, you have a given KE, and your car would keep advancing at the same speed, without the introduction of more fuel, except that you are slowed down by aerodynamic drag, which is the wind blowing back against your car, and friction, which is the retardation caused by the surface properties of your tires and the pavement, as well as the internal friction of your engine, as with pistons in cylinders.  All the fuel you consume once you attain your speed goes to overcoming drag and friction.  None of it goes to KE, unless you accelerate again.

For a spaceship in outer space, aerodynamic drag and friction  are negligible.  This means that, once it has attained its cruising speed, the spaceship will hurtle forward at the same speed without the introduction of more fuel.  However, the downside is that, when you want to stop, you cannot rely upon drag and friction to help you.  What you have to do is apply a force equal to the force that put the ship in motion in the first place, and acting over the same distance, but opposite in direction.  In other words, you have to launch it backwards to stop.

In an earlier article, I quoted NASA  with regard to the three breakthroughs one or another of which they say must be made in order for space travel to be feasible:

“To overcome this difficulty, we need either a breakthrough where we can take advantage of the energy in the space vacuum, a breakthrough in energy production physics, or a breakthrough where the laws of kinetic energy don’t apply.”

The third breakthrough they mention seems to be an allusion to the need of finding a way to brake a high-velocity spaceship, with an enormous KE, in near-vacuum conditions, that is, in the absence of drag and friction.

Can KE be shut off when it gets in the way?  When the spaceship is ready to stop, can’t we just turn off KE?    It looks pretty futile to me.  Can you abolish mass when you wish to be massless?  Can you abolish velocity when you wish to stand still?  If so, then you can also abolish KE.  Just set the condition: m = 0 or v = 0.  Then, since KE = ½ mv^2, KE = O.  Presto! We’ve stopped the ship.

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