Bruce Raup wrote: .... > Is that supposed to be kJ? Surely a hammer that delivers only 1700 kJ a > pop wouldn't be very useful. .... Well, let's see. Suppose we had a 1 t mass (1000 kg) and dropped it on top of the piling. It would have to be moving at some speed v to have 1700 kJ of energy. Since I picked a 1000 kg mass, I can divide 1700 kJ by 1000 to get rid of the "k" and I see that 0.5v^2 = 1700 m2/s2 or v^2 = 3400 m2/s2. That means that v would have to be just a bit under 60 m/s (square root of 3400). That means that it would have to fall for 6 s to attain that speed (v = gt, use 10 m/s2 for g). For it to fall for 6 s, we would have to drop it from a height of 180 m above the piling [= 0.5 gt2 = (0.5)(10 m/s2)(6 s)^2]. That's a pretty tall drop for a 1 t mass! Quadrupling the mass to 4 t would reduce the required speed by half and thus reduce the drop time by half. That would reduce the fall height by a factor of 4, making it about 45 m. this is a more reasonable value. (One can also see that quadrupling the mass reduces drop height by a factor of 4 from considering gravitational potential energy, mgh. If m goes up by a factor of 4 then h goes down by a factor of 4 to keep the same potential energy. One can use this to calculate the original drop height as well.) Gee, I didn't even look for my calculator. Thankfully I didn't have to do this in pounds, stones, feet, and cubits! Can a crane lift a BigBubbaSUV to a height of 45 m and drop it? Yep. Don't leave your lunch pail on top of the piling. Jim -- Metric Methods(SM) "Don't be late to metricate!" James R. Frysinger, CAMS http://www.metricmethods.com/ 10 Captiva Row e-mail: [EMAIL PROTECTED] Charleston, SC 29407 phone/FAX: 843.225.6789
