Physics Help Forum [SOLVED] Falling water (power problem)

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 Sep 5th 2010, 03:20 AM #1 Member   Join Date: Jul 2010 Posts: 70 [SOLVED] Falling water (power problem) After a particularly wet winter, a weir is overflowing at the rate of 800 litres of water every second (1 litre of water has a mass of 1 kg). The water takes 1.3 s to fall to the river below. With what vertical velocity does the water hit the river below? 1.3 * 10 m/s down What height does the water fall through to the river below? 8.3 m What weight of water falls over the weir every 10 s? 8000 kg Calculate the work that has been done on this weight of water by gravitational forces by the time it reaches the river. 6.5 * 10^5 J Calculate the power developed by the falling water at the instant before it hits the river. [All of the above answers are correct.] P = 6.5 * 10^5 / 10 = 6.5 * 10^4 W yet the answers say 1 * 10^6 W. Where have I gone wrong? TIA, Fred
 Sep 5th 2010, 04:37 AM #2 Senior Member   Join Date: Dec 2008 Location: Udupi, Karnataka, India Posts: 289 Power = mgv P = 8000*10*13 = ...?
 Sep 9th 2010, 09:01 AM #3 Physics Team     Join Date: Jun 2010 Location: Morristown, NJ USA Posts: 2,272 xtheunknown0: I believe you have the correct answer, and the book is wrong. Power is energy per unit time. Each second you have 800 Kg of water falling through a distance of 13 meters. Hence the energy per second is: mgh = 800 Kg x 10 m/s^2 x 8.45m = 6.76 x 10^4 J . Since this energy is expended every second, you have 6.76 x 10^5 J/s = 6.76 x 10^5 W An alternative method: use the velocity of the water to calculate its kinetic energy of 800 Kg of water moving at 13 m/s: KE = 1/2 m v^2 = 1/2 x 800 Kg x (13 m/s)^2 = 6.76 x 10^4 J So power = energy/time = 6.7 x 10^4 W

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