Chapter 2 - Energy, Energy Transfer, and General Energy Analysis - Problems - Page 107: 2-128

a) $56kW$

Here we are working with kinetic energy: The kinetic work is: $W_{k}=\frac{1}{2}m\Delta V^2$ And the kinetic power is: $P_{k}=\frac{W_{k}}{t}=\frac{\frac{1}{2}m\Delta V^2}{t}$ $P_{k}=\frac{\frac{1}{2}*900kg*[(100\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s})^2-(60\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s})^2]}{4s}=55.55kW$