91Ó°ÊÓ

The mass is m = 800 kg.

The height is h = 14 m.

The velocity is v = 18 m/s.

When the external force is applied to an object then it gets moved by a distance in which transfer energy takes place. This is called work.

The expression to calculate the work done is given by,

W = mgh……………..(1)

Substitute 800 kg for m,$9.8\mathrm{m}/{\mathrm{s}}^2$ for g, and 14 m for h in equation (1), we get,

$$\begin{gathered} W=\left(800kg\right)\left(9.8m/s^2\right)\left(14m\right) \\ =109760J \\ =1.10×{10}^{5}J \end{gathered}$$

Therefore, the required work done is $1.10×{10}^{5}J$.

Find the kinetic energy the pump has when it has ejected the water as follows:

$KE=\frac{1}{2}m{v}^2$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦..(2)

Substitute 800 kg for m, and 18 m/s for v in equation (2).

role="math" localid="1663757519892" $$\begin{gathered} \mathrm{W}=\frac{1}{2}\left(800\mathrm{kg}\right){\left(18\mathrm{m}\right)}^2 \\ =129600\mathrm{J} \\ =1.30×{10}^5\mathrm{J} \end{gathered}$$

Therefore, the required work done is $1.30×{10}^5\mathrm{J}$.

The power can be calculated as follows.

$$\begin{gathered} P=\frac{W_{total}}{t} \\ =\frac{\left(KE+W\right)}{t} \end{gathered}$$…â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦â¶Ä¦..(3)

Substituterole="math" localid="1663757706553" $1.10×{10}^{5}J$ for W,$1.3×{10}^{5}J$ for KE, and 1min for t in equation (3), we get,

$$\begin{gathered} \mathrm{P}=\frac{\left(1.10×{10}^5\mathrm{J}+1.30×{10}^5\mathrm{J}\right)}{1\min \left({\displaystyle \frac{60\mathrm{s}}{1\min }}\right)} \\ =4000\mathrm{W}\left(\frac{1\mathrm{kW}}{1000\mathrm{W}}\right) \\ =4.0\mathrm{kW} \end{gathered}$$

Therefore, the required power is 4.0 kW.