I'm getting the KE of a car to be roughly double your estimate: assuming ~4000lb[1] car, 140mph ~= 62.6m/s, 0.5*m*v*v = 0.5 * (4000*0.454) * 62.6^2 =

I'm getting the KE of a car to be roughly double your estimate: assuming ~4000lb[1] car, 140mph ~= 62.6m/s, 0.5*m*v*v = 0.5 * (4000*0.454) * 62.6^2 = 3.5E6Joules[2] ~= 830kcal Did you miss the 0.5, or use a different Spherical Co^Wstandard automobile mass? Which is still only 2 kitkats :) The other potentially confusing thing (judging by the sibling comments here) is that this isn't the energy required to accelerate an actual car from 0 to 140mph. It's the kinetic energy embodied in a car traveling at that speed. Wind & tyre resistance losses, engine/transmission, etc, etc. It's really the (upper bound of the) amount of energy you could expect it to impart if it hit something, or the amount of heat the brakes would need to dissipate to stop it. Finally, for all those commenting on how much energy there appears to be in a kitkat, consider the famous E=MC^2, and just how staggeringly huge that C^2 term really is (9E16J/kg, if you're wondering) Getting all that out in a usable form, however, is left as an exercise for the reader. [1] http://www.nytimes.com/2004/05/05/business/05weight.html http://www.nytimes.com/2004/05/05/business/05weight.html http://hypertextbook.com/facts/2000/YanaZorina.shtml http://hypertextbook.com/facts/2000/YanaZorina.shtml [2] http://www.wolframalpha.com/input/?i=0.5+*+%284000+*+0.454%29+*+%2862.6^2%29 http://www.wolframalpha.com/input/?i=0.5+*+%284000+*+0.454%2...