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

How much work is done on the astronaut by (a) the force from the helicopter and (b) the gravitational force on her? Just before she reaches the helicopter, what are her (c) kinetic energy and (d) speed?

1. Mass of the astronaut is, $\mathrm{m}=72\text{\hspace{0.17em}kg}$.
2. Vertical distance is, $\mathrm{d}=15\text{\hspace{0.17em}m}$.
3. Acceleration of the astronaut is, $\mathrm{a}=\frac{\mathrm{g}}{10}$.

By usingtheconcept of Newton’s third law, we can find the force oftheastronaut. Next, we can find the work done bytheastronaut, bythehelicopter as well as the gravity. We can use the work-energy theorem, which states that change in energy is equal tothework done.

The free body diagram can be drawn as,

From the above figure,

The forces in the vertical directions are balanced as,

$\begin{array}{c}\mathrm{F}-\mathrm{mg}=\mathrm{ma}\\ \mathrm{F}=\mathrm{m}(\mathrm{a}+\mathrm{g})\\ \mathrm{F}=\mathrm{m}\left(\frac{\mathrm{g}}{10}+\mathrm{g}\right)\\ \mathrm{F}=\frac{11}{10}\mathrm{mg}\end{array}$

Now, work done on the astronaut by the helicopter can be calculated as,

$\mathrm{W}=\mathrm{Fd}$ (1)

Substitute the values in the above expression, and we get,

$\begin{array}{c}{\mathrm{W}}_{\text{h}}=\frac{11}{10}\mathrm{mgd}\\ {\mathrm{W}}_{\text{h}}=\frac{11}{10}\left(72\right)\left(9.8\right)\left(15\right)\\ {\mathrm{W}}_{\text{h}}=11642.4\text{J}\\ \approx 1.2\times {10}^4\text{\hspace{0.17em}J}\end{array}$

Thus, work done on the astronaut by the helicopter is $1.2\times {10}^4\text{\hspace{0.17em}J}$.

The gravitational force on the astronaut can be calculated as,

${\mathrm{F}}_{\mathrm{g}}=-\mathrm{mg}$

As an astronaut is moving against gravity, the force will be negative.

From equation (2), the work done can be written as,

${\mathrm{W}}_{\text{g}}=-\mathrm{mgd}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}{\mathrm{W}}_{\text{g}}=-\left(72\right)\left(9.8\right)\left(15\right)\\ {\mathrm{W}}_{\text{g}}=-10584\text{J}\\ \approx -1.1\times {10}^4\text{\hspace{0.17em}J}\end{array}$

Thus, work done on the astronaut by the gravitational force is$-1.1\times {10}^4\text{\hspace{0.17em}J}$ .

Energy work done theory states that kinetic energy isequal to thetotal work done by the system.

${\mathrm{W}}_{\text{total}}=\mathrm{KE}$ (2)

Total work done can be calculated as,

${\mathrm{W}}_{\text{total}}={\mathrm{W}}_{\text{h}}+{\mathrm{W}}_{\text{g}}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}{\mathrm{W}}_{\text{total}}=11642.4\text{J}-10584\text{J}\\ =1058.4\text{\hspace{0.17em}J}\\ \approx 0.1\times {10}^4\text{\hspace{0.17em}J}\end{array}$

From equation 2, kinetic energy can be calculated as,

$\begin{array}{c}\mathrm{KE}={\mathrm{W}}_{\text{total}}\\ =0.1\times {10}^4\text{\hspace{0.17em}J}\end{array}$

Thus, the kinetic energy of the astronaut when she reaches the helicopter is $0.1\times {10}^4\text{\hspace{0.17em}J}$.

The expression for kinetic energy is,

$\mathrm{KE}=\frac{1}{2}\times {\mathrm{mv}}^2$

Here $\mathrm{v}$is the speed.

From equation (2), we can calculate the velocity as,

$\begin{array}{c}{\mathrm{W}}_{\text{total}}=\frac{1}{2}\times {\mathrm{mv}}^2\\ \mathrm{v}=\sqrt{\frac{2{\mathrm{W}}_{\text{total}}}{\mathrm{m}}}\end{array}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}\mathrm{v}=\sqrt{\frac{2\left(1058.4\right)}{72}}\\ \mathrm{v}=5.42\text{\hspace{0.17em}m/s}\end{array}$

Thus, the speed of an astronaut when she reaches the helicopter is $5.42\text{\hspace{0.17em}m/s}$.