A helicopter lifts a 66 kg astronaut 18 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? Just before she reaches the helicopter, what are her (c) kinetic energy and (d) speed?

The mass of astronaut m = 66kg

the height lifted h = 18m

The acceleration a = g/15 = 0.65m/s^2

(a) The work done by helicopter

W = F*h = m(g+a) h

= (66)(9.8 + 0.65) (18)

= 12414.6 J

(b) The work done by gravity

W = mgh = 66(9.8)(18) = -11642.4 J

(c) From work energy theorem

net work done = ?K

  12414.6-11642.4=772.2 = ?K

772.2J = ?K

(d) The speed

772.2 = 1/2 mv^2

v = v2(772.2) /66 = 4.83735 m/s^2

a.) is the correct answer.

explanation:

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