Question

A helicopter of mass 1.40×103kg1.40×103kg is descending vertically at 3.00 m/sm/s. The pilot increases the upward thrust provided by the main rotor so that the vertical speed decreases to 0.450 m/sm/s as the helicopter descends 3.00 mm. What is the initial energy of the helicopter?

Answer 1

Answer:

KE=6300 J

Explanation:

Given that

mass of the helicopter , m = 1.4 x 10³ kg

Initial velocity , u = 3 m/s

Velocity after increasing thrust , v = 0.45 m/s

Distance , d= 3 mm

The initial kinetic energy KE is given as follows

$KE=\dfrac{1}{2}mu^2$

Now by putting the values in the above equation we get

$KE=\dfrac{1}{2}\times 1.4\times 10^3\times 3^2\ J$

KE=6300 J

Therefore the initial energy of the helicopter will be 6300  J.