91Ó°ÊÓ

i) The height at which the river descends, h = 15m

ii) Initial velocity of the river,$v_i=3.2m/s$

iii) Final velocity of the river,$v_f=13m/s$

iv) Mass of water,$m=10kg$

We use the concept of kinetic energy and gravitational potential energy. We find the gravitational potential energy and kinetic energy, and then we can take its ratio to find the percentage.

Formulae:

The potential energy at a height, PE = mgh (i)

The kinetic energy of the body, $KE=\frac{1}{2}m{v}^2$ (ii)

Gravitational potential energy of the water is given using equation (i) as follows:

$$\begin{gathered} P.E=10×9.8×15 \\ =1470J \\ =1.47×{10}^{3}J \end{gathered}$$

Change in kinetic energy is given using equation (ii) as follows:

$$\begin{gathered} K.E=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2 \\ =\frac{1}{2}\left(10\right){\left(13\right)}^2-\frac{1}{2}\left(10\right){\left(3.2\right)}^2 \\ =0.79×{10}^{3}J \end{gathered}$$

So, the percentage of gravitational potential energy transferred to the kinetic energy is given as:

$$\begin{gathered} percentage=\frac{K.E}{P.E}×100 \\ =\frac{0.79×{10}^3}{1.47×{10}^3}×100 \\ =54\% \end{gathered}$$

Hence, the value of the required percentage is 54% .