Posted by **jessi** on Sunday, March 22, 2009 at 11:16pm.

Water falls without splashing at a rate of 0.220 L/s from a height of 2.20 m into a 0.720 kg bucket on a scale. If the bucket is originally empty, what does the scale read 3.20 s after water starts to accumulate in it?

The answer is supposed to be in Newtons but I don't see how that would work. I know that with water one L is about one kg, so that's no problem converting. Does gravity factor in somehow because it's up at that height?

- physics -
**drwls**, Sunday, March 22, 2009 at 11:28pm
That's a neat problem. The falling water applies a force to the bucket that must be added to the weight to get the scale reading. The added force equals the force that the bucket applies to the water to stop its fall. The bucket reduces the momentum of the water at a rate of dM/dt * V, where V is the velocity when it hits the bucket,

V = sqrt(2gH) = 6.57 m/s.

The water flow rate dM/dt is 0.22 kg/s

Total force = (0.720)*g + 6.57*0.22 + (3.2)(0.22)*g = 7.06 + 1.45 + 6.90 N

The last term added is the weight of water already collected at t = 3.2 s

- physics -
**jessi**, Sunday, March 22, 2009 at 11:34pm
how did you get that the velocity is sqrt(2gH)?

- physics -
**jessi**, Sunday, March 22, 2009 at 11:36pm
nevermind i figured it out

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