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 - QAmmunity.org

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

asked Feb 15, 2021 23.1k views

2 Answers

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
$W_(g)=-15082.2\ J$ .

Step-by-step 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
$|\vec{a}|=(g)/(15)$ and the acceleration of gravity is
$g=9.8\ (m)/(s^(2))$ .

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.(g)/(15)

F_(h)=mg(1+(1)/(15))

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

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

$W_(g)=-F_(g).\Delta x$

$W_(g)=-mg\Delta x$

$W_(g)=-81\ kg.\ 9.8\ (m)/(s^(2)).\ 19\ m$

$W_(g)=-15082.2\ J$

Niels Robben answered Feb 22, 2021

by Niels Robben

5.0k points

Search QAmmunity.org