A rescue cable attached to a helicopter′s weighs 2 lb/ft. A man 180−lb grabs the end of the rope and his pulled from the ocean into the helicopter. How much work is done in lifting the man if the helicopter is 40 ft above the water ? (a) 8800 lb−ft (b) 1780 lb−ft (c) 7280 lb−ft (d) 10,400 lb−ft


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Question Number 125052 by bramlexs22 last updated on 08/Dec/20

$A r e s c u e c a b l e a t t a c h e d t o a$ ${h e l i c o p t e r}^{'} s w e i g h s 2 l b / f t .$ $A m a n 180 - l b g r a b s t h e e n d$ $o f t h e r o p e a n d h i s p u l l e d$ $f r o m t h e o c e a n i n t o t h e h e l i c o p t e r .$ $H o w m u c h w o r k i s d o n e i n$ $l i f t i n g t h e m a n i f t h e h e l i c o p t e r$ $i s 40 f t a b o v e t h e w a t e r ?$ $(a) 8800 l b - f t$ $(b) 1780 l b - f t$ $(c) 7280 l b - f t$ $(d) 10, 400 l b - f t$
Commented by mr W last updated on 08/Dec/20

$W = 180 \times 40 + 2 \times 40 \times \frac{40}{2} = 8800 l b - f t$ $\Rightarrow a n s w e r (a)$
Commented by benjo_mathlover last updated on 08/Dec/20

$s i r p l e a s e w h a t y o u r f o r m u l a$
Commented by mr W last updated on 08/Dec/20

$m a n :$ $w e i g h t = 180 l b$ $d i s t a n c e m o v e d = 40 f t$ $w o r k d o n e = 180 \times 40 l b - f t$ $r o p e :$ $l e n g t h = 40 f t$ $w e i g h t = 2 \times 40 l b$ $d i s t a n c e m o v e d = \frac{40}{2} f t$ $(c e n t e r o f m a s s o f r o p e i s \frac{40}{2} b e n e a t h$ $t h e h e l i c o p t e r)$ $w o r k d o n e = 2 \times 40 \times \frac{40}{2} l b - f t$
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