A helicopter of mass 1.40×10^3 kg is descending vertically at 3.00 m/s. The pilot increases the upward thrust provided by the main rotor so that the vertical speed decreases to 0.450 m/s as the helicopter descends 3.00 m. What is the initial energy of the helicopter?
Answers: 1
![W_{thrust} = (1.40\times 10^{3}\,kg)\cdot\left[ \frac{1}{2}\cdot [ (3\,\frac{m}{s} )^{2}-(0.45\,\frac{m}{s} )^{2}]+(9.807\,\frac{m}{s^{2}} )\cdot (3\,m)\right]\\](/tpl/images/0555/1712/6e3fe.png)
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A helicopter of mass 1.40×10^3 kg is descending vertically at 3.00 m/s. The pilot increases the upwa...
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