Question

A helicopter lifts a 48 kg astronaut 11 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) Wf = 5808 J

b) Wg = -5280 J

c) Kf = 528 J

d) Vf = 4.7 m/s

Explanation:

The force exerted by the helicopter is:

F - m*g = m*g/10

F = 11 / 10 * m * g = 528 N

So, the work done by the helicopter was:

Wf = F*h = 528 * 11 = 5808 J

The work done by gravitational force is:

Wg = -ΔUg = - m*g*h = -5280J

Kinetic energy is:

Kf + ΔUg = Wf

Kf = Wf - ΔUg = 5808 - 5280 = 528 J

And her speed is:

$Vf = \sqrt{2*Kf/m}$

Vf = 4.7 m/s