A helicopter hoists a 72kg astronaut 15m vertically in the air by means of a cable. The acceleration of the astronaut is g/10. How much work is done on the astronaut by the force from the helicopter?In showing the steps for the answer, please explain how to handle the g/10. I think it means to use g=1/10 in the F=-mg equation, but I do not get the correct answer by doing this.
you are almost correct. Very often acceleration is spoken of in terms of fraction of 1g, so g/10 is exactly that 9.8/10 or 0.98 m/s^2. Since you used 1/10 = 0.1 you are probably out by factor of 10. Forces on the man Tension up, gravity down. Net resulty is acceleration up of g/10 T - mg = m(g/10) T = mg + mg/10 = 1.1mg So the Force from the helicopter on the man is 1.1mg, so the work is 1.1mg x 15 = 16.5mg = 16.5 x 72 x 9.8 hope that helps.
