(a) 147,500 J
Newton's second law, applied to the helicopter, states that
[tex]F-mg=ma[/tex]
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:
[tex]F=m(g+a)=(500 kg)(9.8 m/s^2+2.0 m/s^2)=5900 N[/tex]
The vertical distance covered by the helicopter is given by
[tex]S=\frac{1}{2}at^2=\frac{1}{2}(2.0 m/s^2)(5.0 s)^2=25 m[/tex]
So, the work done by the lifting force F is
[tex]W_1=FS=(5900 N)(25 m)=147,500[/tex]
2) -122,500 J
The magnitude of the gravitational force acting on the helicopter is
[tex]mg=(500 kg)(9.8 m/s^2)=4900 N[/tex]
And the work done by this force on the helicopter is
[tex]W_2 = -(mg)S=-(4900 N)(25 m)=-122,500 J[/tex]
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:
[tex]W=W_1+W_2 =147500 J-122500 J=25500 J[/tex]
