A helicopter lifts a 72kg 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 (g=9.8 m//s^(2)) (a) what is the kinetic energy of the block as it passese through x=2.0m? (b) What is the maximum kinetic energy of the block between x=0 and x=2.0m?

A helicopter lifts a 72kg astronaut 15 m vertically from the ocean by

1.	A helicopter lifts a 72kg 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 (g=9.8 m//s^(2)) (a) what is the kinetic energy of the block as it passese through x=2.0m? (b) What is the maximum kinetic energy of the block between x=0 and x=2.0m?
Answer» Solution :`T-mg =ma` `:. T =m(g + a)=72(9.8 + 0.98)` `=776.16N` (a) `W_(T) =TS cos 0^@` `=(776.16)(15)` `=11642 J` (b) `W_(mg) =(mg)(S) cos 180^@` `=(72 xx 9.8 xx 15)(-1)` `=-10584 J` (c) `K=W("TOTAL") =11642 - 10584` `=1058 J` (d) `K =1/2mv^(2)` `:. v=SQRT((2K)/(m)) =sqrt((2 xx 1058)/(72))` `=5.42m//s`.