Question

A helicopter lifts a 72 kg astronaut 15 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/10. 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 1

Answer:

a) $F_H=776.952\ N$

b) $F_g=706.32\ N$

c) $v=5.4249\ m.s^{-1}$

d) $KE=1059.48\ J$

Explanation:

Given:

• mass of the astronaut, $m=72\ kg$

• vertical displacement of the astronaut, $h=15\ m$

• acceleration of the astronaut while the lift, $a=\frac{g}{10} =0.981\ m.s^{-2}$

a)

Now the force of lift by the helicopter:

Here the lift force is the resultant of the force of gravity being overcome by the force of helicopter.

$F_H-F_g=m.a$

where:

• $F_H=$ force by the helicopter

• $F_g=$ force of gravity

$F_H=72\times 0.981+72\times9.81$

$F_H=776.952\ N$

b)

The gravitational force on the astronaut:

$F_g=m.g$

$F_g=72\times 9.81$

$F_g=706.32\ N$

d)

Since the astronaut has been picked from an ocean we assume her initial velocity to be zero, $u=0\ m.s^{-1}$

using equation of motion:

$v^2=u^2+2a.h$

$v^2=0^2+2\times 0.981\times 15$

$v=5.4249\ m.s^{-1}$

c)

Hence the kinetic energy:

$KE=\frac{1}{2} m.v^2$

$KE=0.5\times 72\times 5.4249^2$

$KE=1059.48\ J$

Answer 2

Answer:

Explanation:

mass of helicopter, m = 72 kg

height, h = 15 m

acceleration, a = g/10

(a) Work done by the force

Work, W = force due to helicopter x distance

W = m x ( g + a) x h

W = 72 ( 9.8 + 0.98) x 15

W = 11642.4 J

(b) Work done by the gravitational force

W = - m x g x h

W = - 72 x 9.8 x 15

W = - 10584 J

(c) Kinetic energy = total Work done

K = 11642.4 - 10584

K = 1058.4 J

(d) Let the speed is v.

K = 0.5 x m v²

1058.4 = 0.5 x 72 x v²

v = 5.42 m/s

Now the force of lift by the helicopter:

where:

force by the helicopter

force of gravity

b)

The gravitational force on the astronaut:

d)

Since the astronaut has been picked from an ocean we assume her initial velocity to be zero,

using equation of motion:

c)

Hence the kinetic energy:

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