Problem 7-49 (b) Energy cons. - Part 2 - C

A river is flowing with a speed of \(3.74\; m/s\) just upstream from a waterfall of vertical height \(8.74\; m\). During each second \(7.12\times10^4\; kg\) of water pass over the fall.

(a) [We will ignore this part]
(b) If there is complete conversion of gravitational potential energy to kinetic energy, what is the speed of the water at the bottom of the fall?

(Note: This is actually an unrealistic assumption; some of the energy in the fall will be converted into heat)

Correct!

Mechanical energy is conserved

\((E_K + E_P)_{top} = (E_K + E_P)_{bottom} \\ ½ mv{^2}_{top}+ mgh = ½ \;mv{^2}_{bot}+ 0 \\ v{^2}_{bot} = v{^2}_{top} + 2gh = (3.74 \;m/s)^2 + 2(9.8\; m/s^2)(8.74 \;m) = 185.3 \\ v_{bot} = 14.4 \;m/s\)
 

Notice that the mass cancelled out. The result would be the same for a river of molten iron or a river of water. This is another aspect of what you already know-in the absence of friction all objects fall the same distance in the same time. 

You have completed this problem.