Water is filled to a depth of 18 meters above a small hole in a large container. Using PE and KE changes on a mass of 8 grams, determine the velocity of 8 grams of water as it flows out through the hole, provided there are no dissipative losses.
a=fnf(2,20,1) b=fnf(3,9,1) c=b/1000 d=c*9.8 e=d*a f=fng(2*e/c)
The PE of the system changes as the 8 grams of water at the surface are replaced by 8 grams of water flowing out the hole. The change in the PE is equal to the work required to raise 8 grams of water the distance 18 meters through which it effectively falls.
The associated work is equal to the product of the weight of the water and the distance through which it is raised. The weight of the water is
The work to raise the water is therefore
which is equal to the PE loss. We thus have PE change
By conservation of energy, if there is no dissipative loss, `dPE + `dKE = 0 so
Since the hole is small the water at the surface is effectively at rest, with no KE. Thus `dKE = KE at exit = .5 m v^2, where v is exit velocity. We therefore have
We easily solve for v, obtaining v = `sqrt(2 * 1.4112 Joules / m) = `sqrt(2 * 1.4112 Joules / .008 kg) = 18.78 m/s.
When mass m at altitude y is replaced by mass m at altitude 0 there is a PE change of
If there are no dissipative losses, we have `dKE + `dPE = 0 so
Because the hole is small and the container large, we assume that the velocity of the water at the surface is negligible, so that the mass m had negligible KE at the surface. Thus `dKE = KE at exit, or
Thus we have
which we easily solve for v to obtain
We note that this is the same velocity as would be attained by an object freely falling through vertical distance y from rest.
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