Question

A 500-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?

Answer 1

(a) 147,500 J

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

$F-mg=ma$

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:

$F=m(g+a)=(500 kg)(9.8 m/s^2+2.0 m/s^2)=5900 N$

The vertical distance covered by the helicopter is given by

$S=\frac{1}{2}at^2=\frac{1}{2}(2.0 m/s^2)(5.0 s)^2=25 m$

So, the work done by the lifting force F is

$W_1=FS=(5900 N)(25 m)=147,500$

2) -122,500 J

The magnitude of the gravitational force acting on the helicopter is

$mg=(500 kg)(9.8 m/s^2)=4900 N$

And the work done by this force on the helicopter is

$W_2 = -(mg)S=-(4900 N)(25 m)=-122,500 J$

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:

$W=W_1+W_2 =147500 J-122500 J=25500 J$