Question
A helicopter lifts a $72kg$ astronaut $15m$ vertically from the ocean by means of a cable. The acceleration of the astronaut is $\frac{\text{g}}{10}$.

How much work is done on the astronaut by:
  1. The force from the helicopter and
  2. The gravitational force on him?
  3. What are the kinetic energy and
  4. The speed of the astronaut just before he reaches the helicopter?(Take, $g = 10ms^{‑2}$)

Answer

Here, mass of astronaut m = 72kg, vertical distance
h = 15m and acceleration of astronaut $\text{a}=\frac{\text{g}}{10}$
  1. Force ftom the helicopte
$\text{F}=\text{mg}+\text{ma}$
$=\text{mg}+\frac{\text{mg}}{10}=\frac{11\text{mg}}{10}$
$\therefore$ Work done by force from the helicopter
$\text{W}=\text{F}\times\text{h}=\frac{11\text{mg}}{10}\times\text{h}$
$\text{W}=\frac{11\times72\times10\times15}{10}$
$=11880\text{J}$
  1. Work done by gavitational force
$\text{W}'=\text{(mg)}(\text{h})(\cos180^\circ)$
$\text{W}'=-\text{mgh}=-72\times10\times15$
$=-10800\text{J}$
  1. KE of astronaut K = network done on him
= W + W'
= (11880 - 10800)J
= +1080J
  1. As $\text{KE}=\frac{1}{2}\text{mv}^2$, Hence
$\text{v}=\sqrt{\frac{2\text{K}}{\text{m}}}$
$=\sqrt{\frac{2\times1080}{72}}$
$=5.48\text{ms}^{-1}$

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