Question

A helicopter lifts a $72\text{kg}$ astronaut $15\text{m}$ vertically from the ocean by means of a cable. The acceleration of the astronaut is $\frac{g}{10}$. How much work is done on the astronaut by the force from the helicopter ?

• A. $1188\text{J}$
• B. $11880\text{J}$
• C. $118.8\text{J}$
• D. $9720\text{J}$

Answer 1

The correct option is B $11880\text{J}$

Given, mass of an astronaut $\left(m\right)=72\text{kg}$
Displacement of astronaut $\left(s\right)=15\text{m}$
Acceleration of astronaut $\left(a\right)=\frac{g}{10}{\text{m/s}}^2$
By the FBD of the astronaut:
$\Rightarrow T-mg=ma$
$\therefore T=m(g+a)$
$T=72\left(\frac{11g}{10}\right)\text{N}$
Work done by the force from the helicopter
$W=Fs\cos \theta$
$\because$ Tension and displacement are along same direction $\theta =0^{\circ }$
$\therefore W=Ts\cos 0^{\circ }=Ts$
$W=72\times \frac{11\times 10}{10}\times 15$
$W=11880\text{J}$

Hence option B is the correct answer