Question

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?

Answer 1

Answer:

$E_{o} = 47489.4\,J$

Explanation:

The helicopter is described by means of the Principle of Energy Conservation and the Work-Energy Theorem. Let assume that final height is equal to zero:

$U_{g,A} + K_{A} = U_{g,B} + K_{g,B} + W_{thrust}$

The work done by the increase on upward thrust is:

$W_{thrust} = (K_{A} - K_{B}) + (U_{g,A}-U_{g,B})$

$W_{thrust} = (1.40\times 10^{3}\,kg)\cdot\left[ \frac{1}{2}\cdot [ (3\,\frac{m}{s} )^{2}-(0.45\,\frac{m}{s} )^{2}]+(9.807\,\frac{m}{s^{2}} )\cdot (3\,m)\right]$

$W_{thrust} = 47347.65\,J$

The initial energy of the helicopter is:

$E_{o} = \frac{1}{2}\cdot (1.40\times 10^{3}\,kg)\cdot (0.45\,\frac{m}{s} )^{2} +47347.65\,J$

$E_{o} = 47489.4\,J$