speed of light?

Whatever units you use to express the physical properties, the equations will work as long as you don't change units in the middle of the problem. That kind of mistake caused the failures of the tower of Babel and a probe to Mars; so it is best to forget about stone-age units like feet and British thermal units. We really should also forget about light-years, parsecs and cubits; it is best to stick to the MKS system of metric units.

Mass times acceleration is force; Ma = f.

force times velocity is power; fv = P.

power times time is energy; Pt = E.

So E = Mavt.

(Actually, the average velocity is only half of the final velocity, so the kinetic energy of a mass moving at velocity v is ½ Mavt.)

But at = v, so Mavt = Mv^2. So the kinetic energy of mass M at velocity v is ½ Mv^2.

The energy equivalent of a mass is equal to twice the kinetic energy of the mass moving at the speed of light. But you can never make a mass go that fast because the more kinetic energy you add to it the more massive it gets. As the velocity approaches the speed of light, the mass approaches infinity; so it takes more and more force to get the same acceleration, and you can never make it go as fast as light.

Mc^2 is the kinetic energy already stored in the mass when it is still. It is the energy of all those tiny subatomic particles spinning around each other—mostly inside the protons and neutrons.

P.S.: I like to think of the speed of light as 0.3 meter per nanosecond. My computer’s clock runs at 3.2 Ghz; so if two chips on my mother board are 0.1 meter apart, they are about one clock tick out of sync. Computer designers have to compensate for that so all parts of the mother board know what time it is at the microprocessor and how long it takes for messages to go back and forth.

n SI units : Joules, kilograms, metres per second.

So 1kg converts to (3x10^8)^2 joules. = 9x10^16 Joules.

Thats roughly the power output for a nuclear power station on full load for 1 y

You are not stupid. The speed of light is given in as 186212 MPH. The unit of energy is probably the Joule.

in SI units : Joules, kilograms, metres per second.

So 1kg converts to (3x10^8)^2 joules. = 9x10^16 Joules.

Thats roughly the power output for a nuclear power station on full load for 1 year.