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igomit [66]
3 years ago
7

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?
Physics
1 answer:
artcher [175]3 years ago
4 0

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

\dfrac{1}{2}mv^2 = 881.1

v = \sqrt{\dfrac{2 \times 881.1}{66}}

v =5.17 m/s

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The motion of the diver is a free-fall motion, so it is a uniform accelerated motion. Choosing downward as positive direction, the final velocity can be found by using the following suvat equation:

v^2-u^2=2as

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For the diver jumping from 5 m, s = 5 m, so

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For the diver jumping from 10 m, s = 10 m, so

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The time of flight of each diver can be found by using the other suvat equation

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For the diver jumping from 10 m, s = 10 m, so we find

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(10)}{9.8}}=1.43 s

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