1-The-gravitational-force-in-lb-of-attraction-between-two-objects-is-given-by-F-k-x-2-where-x-is-the-distance-between-the-objects-If-the-objects-are-10-ft-apart-find-the-work-required-to-se

Question Number 124976 by liberty last updated on 07/Dec/20

(1) T h e g r a v i t a t i o n a l f o r c e (i n l b) o f

a t t r a c t i o n b e t w e e n t w o o b j e c t s i s g i v e n

b y F = \frac{k}{x^{2}}, w h e r e x i s t h e d i s t a n c e

b e t w e e n t h e o b j e c t s . I f t h e o b j e c t s a r e

10 f t a p a r t, f i n d t h e w o r k r e q u i r e d t o

s e p a r a t e t h e m u n t i l t h e y a r e 50 f t a p a r t . E x p r e s s

t h e r e s u l t i n t e r m s o f k .

(a) \frac{k}{500} (b) \frac{2 k}{25} (c) \frac{k}{5} (d) \frac{k}{40}

(2) O n e e n d o f a p o o l i s v e r t i c a l w a l l 15 f t

w i d e . W h a t i s t h e f o r c e e x e r t e d o n t h i s

w a l l b y t h e w a t e r i f i t i s 6 f t d e e p ?

T h e d e n s i t y o f w a t e r i s 62.4 l b / {f t}^{3}

(a) 8420 l b (b) 33, 700 l b (c) 2810 l b (d) 16, 800 l b

(3) F i n d t h e a r e a o f t h e s u r f a c e g e n e r a t e d

b y r e v o l v i n g t h e c u r v e a b o u t t h a t

i n d i c a t e d a x i s . x = 3 \sqrt{4 - y}, 0 ⩽ y ⩽ \frac{15}{4}, y - a x i s

(a) (\frac{125}{2} + 5 \sqrt{10}) π (b) (\frac{125}{2} - 5 \sqrt{10}) π

(c) \frac{125}{2} π (d) 5 π \sqrt{10}

Commented by Dwaipayan Shikari last updated on 07/Dec/20

W o r k d o n e = \int_{10}^{50} \frac{k}{x^{2}} d x = {[- \frac{k}{x}]}_{10}^{50} = \frac{2 k}{25}

Answered by bemath last updated on 07/Dec/20

(3) S = \underset{0}{\int^{15 / 4}} 2 π x \sqrt{1 + {(\frac{d x}{d y})}^{2}} d y

x = 3 \sqrt{4 - y}; \frac{d x}{d y} = \frac{- 3}{2 \sqrt{4 - y}}

t h e n 1 + {(\frac{d x}{d y})}^{2} = 1 + \frac{9}{4 (4 - y)} = \frac{25 - 4 y}{4 (4 - y)}

S = \underset{0}{\int^{\frac{15}{4}}} 6 π \sqrt{4 - y} \sqrt{\frac{25 - 4 y}{4 (4 - y)}} d y

S = 3 π \underset{0}{\int^{\frac{15}{4}}} \sqrt{25 - 4 y} d y = {[\frac{1}{\frac{3}{2} (- 4)} \times 3 π \times \sqrt{{(25 - 4 y)}^{3}}]}_{0}^{\frac{15}{4}}

= - \frac{π}{2} {\sqrt{1000} - 125}

= \frac{125 π}{2} - 5 π \sqrt{10} = (\frac{125}{2} - 5 \sqrt{10}) π

Answered by bemath last updated on 07/Dec/20

(2) F o r c e = w h A = 62.4 l b / {f t}^{3} \times 6 f t \times 15^{2} {f t}^{2}

= 84, 240 l b

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