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

Respuesta :

Answer:

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

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

Explanation:

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

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

                                               [tex]F_{h}-F_{g}=m.a[/tex]

                                              [tex]F_{h}-mg=m.\frac{g}{15}[/tex]

                                              [tex]F_{h}=mg(1+\frac{1}{15})[/tex]

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

We define work as the product of the force, the displacement of the body and the cosine of the angle [tex]\theta[/tex] between the direction of the force and the displacement of the body:

                                                 [tex]W=F.\Delta x.\ cos(\theta)[/tex]

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

                                                  [tex]W_{h}=F_{h}.\Delta x[/tex]

                                             [tex]W_{h}=846.72\ N.\ 19\ m[/tex]

                                                  [tex]W_{h}=16087.68\ J[/tex]

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

                                                   [tex]W_{g}=-F_{g}.\Delta x[/tex]

                                                    [tex]W_{g}=-mg\Delta x[/tex]

                                            [tex]W_{g}=-81\ kg.\ 9.8\ \frac{m}{s^{2}}.\ 19\ m[/tex]

                                                  [tex]W_{g}=-15082.2\ J[/tex]

                                             

                                                 

                                               

                                       

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

[tex]W = -f_{g} h[/tex]

[tex]f_{g} = mg = 81 * 9.8\\f_{g} = 793.8 N[/tex]

[tex]W = -793.8 * 19\\W =- 15082.2 J[/tex]

W = -15.082 kJ