Question

A helicopter lifts a 75.0 kg astronaut 14.2 m vertically from the ocean by means of a cable. The acceleration of the astronaut is g/10.
A) How much work is done on the astronaut by the force from the helicopter?
B)How much work is done on the astronaut by the gravitational force on her?
C)Just before she reaches the helicopter, what is her kinetic energy?
D) And just before she reaches the helicopter, what is her speed?

Answer 1

Intial astronaut speed u = 0 ,Hence its intial Kinetic Energy = 0

Astronaut height from ocean surface h_i = 0 , hence its intial Potential Energy = 0

Let us first calculate the final velocity v (upwards) of the astronaut (m=75 kg) who starts from rest (u=0) subject to an acceleration of a = g/10 and through a distance of h=14.2 m.
v²-u²=2ah
v² = u² + 2ah = 2 x g/10 x 14.2 = 27.832 (g = 9.8 m/s²)

v = 5.2756042308 m/sec

a. Work done on the astronaut by the force from the helicopter = change in potential energy + change in kinetic energy
Now change in Potential energy , E_p = mgh -0 = mgh = 10437 Joule
Change in Kinetic energy , E_k = (1/2)mv² -0 = (1/2)mv² = 1043.7 Joule
Work done = E_p + E_k = mgh + (1/2)mv² = 11480.7 Joule

b. Work done on the astronaut by the gravitational force = Change in Potential energy of astronaut =  mgh = 10437 Joule

c. Astronaut Kinetic energy at just before she reaches the helicopter  = (1/2)mv² = 1043.7 Joule

d. Astronaut speed at just before she reaches the helicopter = v = 5.2756042308 m/sec