A500-kg, light-weight helicopter ascends from the ground with an acceleration of 2.00 m/s2. over a 5.00-s interval, what is (a) the work done by the lifting force, (b) the work done by the gravitational force, and (c) the net work done on the helicopter?

(a) 147,500 J

Newton's second law, applied to the helicopter, states that

where

F is the lifting force

mg is the weight of the helicopter, with m=500 kg being the mass of the helicopter and g=9.8 m/s^2 the acceleration due to gravity

a=2.00 m/s^2 is the acceleration of the helicopter

From the equation, we can calculate the magnitude of the lifting force:

The vertical distance covered by the helicopter is given by

So, the work done by the lifting force F is

2) -122,500 J

The magnitude of the gravitational force acting on the helicopter is

And the work done by this force on the helicopter is

And the negative sign is due to the fact that the direction of the gravitational force is opposite to the displacement of the helicopter.

3) 25,500 J

The net work done on the helicopter is given by the sum of the work done by the two forces:

the bal will not change its velocity.