7-49(b).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 7.12X10^{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}
+ 2*gh* = (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.

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