Question

A helicopter lifts a 76 kg astronaut 14 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/14. 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

Part-a

We use $\vec F$ to denote the upward force exerted by the cable on the astronaut. The force of the cable is upward and the force of gravity is mg downward. Furthermore, the acceleration of the astronaut is g/14 upward. According to Newton’s second law,

$F-mg = \frac{mg}{14}$

so   $F= \frac{15mg}{14}$

Since the force $\vec F$   and the displacement $\vec d$ are in the same direction, the work done by $\vec F$ is

$W_F = Fd = \frac{15}{14}mgd$

by plugging all the values in above equation, we get

$W_F = Fd = \frac{15}{14}(76)(9.8)(14) = 11172J$

Part-b

The force of gravity has magnitude mg and is opposite in direction to the displacement. Thus the work done by gravity is

Part-c

The total work done is

Since the astronaut started from rest, the work-kinetic energy theorem tells us that this is her final kinetic energy.

Part-d

Since   $K.E. = \frac{1}{2}mv^2$ , her final speed is

$v = \sqrt{\frac{2K.E.}{m}}$

$v = \sqrt{\frac{2 \times 744.8}{76}}$