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A helicopter lifts a 66 kg astronaut 15 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/11. 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?

Answer :

Answer

given,

mass of the astronaut = 66 kg

height of the lift = 15 m

acceleration = g/11 = 0.89 m/s²

a) the work done by helicopter

W = F× h

    = m (a+g)h

    = 66 × (9.8+0.89)15

  W =10583.1 J

b) work done by the gravity

W= mgh

W = - 66 × 9.8 × 15

W = -9702 J

c) Δ KE = net work done

  Δ KE = 10583.1 - 9702

  Δ KE = 881.1 J

d) speed

[tex]\dfrac{1}{2}mv^2 = 881.1[/tex]

[tex]v = \sqrt{\dfrac{2 \times 881.1}{66}}[/tex]

v =5.17 m/s

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