Question

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 1

Correct Answer - A::B::C::D
`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`.