A-100kg-man-lowers-himself-to-the-ground-from-a-height-of-10m-by-means-of-a-rope-passed-over-a-frictionless-pulley-and-the-other-end-attached-to-a-70kg-sandbag-a-with-what-speed-does-the-man-hit-the

Question Number 25253 by NECx last updated on 07/Dec/17

A 100 k g m a n l o w e r s h i m s e l f t o

t h e g r o u n d f r o m a h e i g h t o f 10 m

b y m e a n s o f a r o p e p a s s e d o v e r a

f r i c t i o n l e s s p u l l e y a n d t h e o t h e r

e n d a t t a c h e d t o a 70 k g s a n d b a g .

a) w i t h w h a t s p e e d d o e s t h e m a n

h i t t h e g r o u n d ?

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

r e d u c e t h e s p e e d ?

Commented by NECx last updated on 07/Dec/17

p l e a s e d o w e l l t o i n t e r p r e t e w i t h a

d i a g r a m . T h a t w a s 1 o f m y

c h a l l e n g e w h i l e I t r i e d s o l v i n g i t .

T h a n k s

Answered by mrW1 last updated on 07/Dec/17

a)

a c c e l e r a t i o n o f m a n a n d s a n d b a g i s

a = \frac{(m_{M} - m_{S}) g}{m_{M} + m_{S}} = \frac{100 - 70}{100 + 70} \times 10 = \frac{30}{17} m / s^{2}

s p e e d o f t h e m i s

v = \sqrt{2 a h} = \sqrt{2 \times \frac{30}{17} \times 10} = 5.96 m / s

b)

t h e m a n c a n r e d u c e h i s s p e e d i f h e a l s o

h o l d s t h e r o p e o n t h e s i d e o f s a n d b a y

a n d r e l e a s e s t h i s r o p e g r a d u a l l y .

Commented by NECx last updated on 07/Dec/17

h o w d i d y o u d e r i v e t h e f o r m u l a

f o r t h e a c c e l e r a t i o n ?

Commented by mrW1 last updated on 07/Dec/17

m a n a n d s a n d b a g m o v e t o g e t h e r

a s a s i n g l e o b j e c t . t o t a l m a s s o f t h e m

i s M = m_{M a n} + m_{S a n d} . T h e t o t a l f o r c e w h i c h

a c t s o n t h e m i s F = m_{M a n} g - m_{S a n d} g = (m_{M a n} - m_{S a n d}) g,

t h e r e f o r e t h e a c c e l e r a t i o n o f t h e m

i s a = \frac{F}{M} = \frac{(m_{M} - m_{S}) g}{m_{M} + m_{S}} .

A n o t h e r w a y i s :

l e t T = t e n s i o n i n r o p e

m_{M a n} a = m_{M a n} g - T \dots (i)

m_{S a n d} a = T - m_{S a n d} g \dots (i i)

(i) + (i i) :

(m_{M} + m_{S}) a = (m_{M} - m_{S}) g

\Rightarrow a = \frac{(m_{M} - m_{S}) g}{m_{M} + m_{S}}

Commented by NECx last updated on 08/Dec/17

t h a n k s b o s s . I n o w u n d e r s t a n d i t .

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