The ease of mental calculations in metric got me into trouble in this case. Somehow a "k" slipped out, so I was off by a factor of 1000. The easier way (which I still messed up) is E=mgh. For m = 1000 kg, h=E/10000, or 170 m. Thanks for the direction. Bruce On 2001-07-25 18:55 -0400, James R. Frysinger wrote: > 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
