Question

A helicopter rescues a 105kg shipwreck survivor by lifting him straight up. Survivor moved 15 m up while his speed drops from 4m/s to 0. What is the work done on home by the applied force?

Answer 1

16275 Joule

Step-by-step explanation:

Work Done By A Constant Force

The work can be computed as

W=F.X, where F is the applied force (in this case, the force needed to overcome gravity)

Recall the second Newton's law which states that

`F_n=m.a`

Being
`F_n` the net force on a system of mass m and acceleration a

We know the helicopter is applying force against gravity because the speed of the shipwreck survivor changes from 4 m/s to 0 m/s in 15 meters

Let's calculate that acceleration by using the dynamics equation

`V_f^2=V_o^2+2aX`

Solving for a

`\displaystyle a=(V_f^2-V_o^2 )/(2X)`

a=-0.533\ m/sec^2

The negative sign indicates the body is braking, we'll later use the magnitude

The net force is, then

`F_n=(105\ kg)(0.533\ m/sec^2)`

`F_n=56\ Nw`

The only two forces acting upon the survivor is the lifting force of the helicopter and the weight of the man, so

`F_h-W=56\ Nw`

`F_h=mg+56\ Nw=105\ kg\ 9.8\ m/sec^2+56\ Nw`

`F_h=1085\ Nw`

Finally, the work done by the helicopter is

`W=(1085\ Nw)(15\ m)=16275 \Joule`

The work done by the helicopter is

`\boxed{16275\ Joule}`