Question

A helicopter lifts a 81 kg astronaut 19 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/15. How much work is done on the astronaut by (a) the force from the helicopter and (b) the gravitational force on her

Answer 1

Answer:

a) Work done on the astronaut by the force from the helicopter = 16.104 kJ

b) Work done on the astronaut by the gravitational force = -15.082 kJ

Explanation:

mass of the astronaut, m = 81 kg

height, h = 19 m

acceleration of the astronaut, a = g/15

Since the astronaut is lifted up, using the third law of motion:

T - mg = ma

T = mg + ma

T = (81*9.81) + 81*(9.81/15)

T = 847.584 N

Work done on the astronaut by the helicopter

Work done = Tension * height

W = T* h

W = 847.584 * 19

Work done, W = 16104.096 Joules

W = 16.104 kJ

b) Work done on the astronaut by the gravitational force on her

$W = -f_{g} h$

$f_{g} = mg = 81 * 9.8\\f_{g} = 793.8 N$

$W = -793.8 * 19\\W =- 15082.2 J$

W = -15.082 kJ

Answer 2

Answer:

a) The work done on the astronaut by the force from the helicopter is $W_{h}=16087.68\ J$.

b) The work done on the astronaut by the gravitational force is  $W_{g}=-15082.2\ J$ .

Explanation:

We are told that the mass of the astronaut is $m=81\ kg$, the displacement is $\Delta x=19\ m$, the acceleration of the astronaut is  $|\vec{a}|=\frac{g}{15}$  and the acceleration of gravity is $g=9.8\ \frac{m}{s^{2}}$ .

We suppose that in the vertical direction the force from the helicopter $F_{h}$ is upwards and the gravitational force $F_{g}$ is downwards. From the sum of forces we can get the value of $F_{h}$:

                                               $F_{h}-F_{g}=m.a$

                                              $F_{h}-mg=m.\frac{g}{15}$

                                              $F_{h}=mg(1+\frac{1}{15})$

                              $F_{h}=(\frac{16}{15}).81\ kg.\ 9.8\ \frac{m}{s^{2}}\ \Longrightarrow\ F_{h}=846.72\ N$

We define work as the product of the force, the displacement of the body and the cosine of the angle $\theta$ between the direction of the force and the displacement of the body:

                                                 $W=F.\Delta x.\ cos(\theta)$

a) The work done on the astronaut by the force from the helicopter

                                                  $W_{h}=F_{h}.\Delta x$

                                             $W_{h}=846.72\ N.\ 19\ m$

                                                  $W_{h}=16087.68\ J$

b) The work done on the astronaut by the gravitational force

                                                   $W_{g}=-F_{g}.\Delta x$

                                                    $W_{g}=-mg\Delta x$

                                            $W_{g}=-81\ kg.\ 9.8\ \frac{m}{s^{2}}.\ 19\ m$

                                                  $W_{g}=-15082.2\ J$