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Question Number 27899 by Tinkutara last updated on 16/Jan/18

	Answered by mrW2 last updated on 17/Jan/18

	$l e t e = m a x . e l o n g a t i o n o f s p r i n g a f t e r r e l e a s e$ $k e = μ m_{A} g . . . (i)$ $\frac{1}{2} {k d}^{2} - μ m_{B} g (d + e) = \frac{1}{2} {k e}^{2} . . . (i i)$ ${k d}^{2} - 2 μ m_{B} g d - (2 μ m_{B} g + k e) e = 0$ ${k d}^{2} - 2 μ m_{B} g d - \frac{m_{A} (2 m_{B} + m_{A}) {(μ g)}^{2}}{k} = 0$ $\Rightarrow d = \frac{2 μ m_{B} g + \sqrt{{(2 μ {g m}_{B})}^{2} + 4 m_{A} (2 m_{B} + m_{A}) {(μ g)}^{2}}}{2 k}$ $\Rightarrow d = \frac{μ g (m_{A} + 2 m_{B})}{k}$ $\Rightarrow d = \frac{0.5 \times 10 (2 + 2 \times 4)}{50} = 1 m$
	Commented by Tinkutara last updated on 16/Jan/18

	Commented by Tinkutara last updated on 16/Jan/18
	This is the solution in book. Can you please explain this? And why your method is wrong? This question was integer type so they approximated it to 1 m.
	Commented by mrW2 last updated on 17/Jan/18

	$W h e n b l o c k A i s p r e s s e d a g a i n s t t h e$ $w a l l, t h e c o n t a c t f o r c e b e t w e e n b l o c k$ $A a n d t h e w a l l i s t h e p u s h f o r c e i n$ $s p r i n g, w h i c h i s k d .$ $W h e n b l o c k B i s r e l e a s e d, t h e c o n t a c t$ $f o r c e b t w . A a n d w a l l w i l l b e r e d u c e d$ $t i l l z e r o (s a y a t i n s t a n t t_{1}), t h e n d u e t o$ $t e n s i o n f o r c e i n s p r i n g t h e f r i c t i o n f o r c e$ $b t w . A a n d f l o o r i n c r e a s e s f r o m z e r o t i l l μ m_{A} g$ $(s a y a t i n s t a n t t_{2}), a f t e r t h a t t h e b l o c k A$ $b e g i n s t o m o v e .$ $T h e a n s w e r d e p e n d s v e r y m u c h o n h o w$ $y o u u n d e r s t a n d ‘ ‘ b l o c k A l o s e s c o n t a c t$ $t o {w a l l}^{″} .$ $M y a n s w e r i s c o r r e c t, i f m a n$ $u n d e r s t a n d s t h e l o s e o f c o n t a c t a s t h e i n s t a n t w h e n$ $b l o c k A i s a b o u t t o m o v e a w a y f r o m w a l l,$ $i . e . a t i n s t a n t t_{2}, w h e n t h e t e n s i o n f o r c e$ $i n s p r i n g j u s t o v e r c o m e s t h e s t a t i c f r i c t i o n$ $b e t w e e n b l o c k A a n d f l o o r, i . e .$ $k e = μ m_{A} g .$ $T h e a n s w e r i n b o o k i s c o r r e c t, i f m a n$ $u n d e r s t a n d s t h e l o s e o f c o n t a c t a s t h e i n s t a n t w h e n$ $t h e c o n t a c t f o r c e t o t h e w a l l b e c o m e s z e r o,$ $i . e . a t i n s t a n t t_{1} . I n t h i s c a s e t h e b l o c k$ $B m a y m o v e m a x . t o a p o s i t i o n w h e r e$ $t h e s p r i n g h a s n o e l o n g a t i o n (e = 0) s o t h a t$ $t h e s p r i n g f o r c e (a n d t h e r e f o r e t h e c o n t a c t f o r c e t o$ $w a l l) i s z e r o . T h e e n e r g y s t o r e d i n t h e$ $c o m p r e s s e d s p r i n g i s j u s t e n o u g h t o c o m p e n s a t e$ $t h e l o s e d u e t o f r i c t i o n d u r i n g t h e m o v e m e n t o f b l o c k B i n a d i s t a n c e d, i . e .$ $\frac{1}{2} {k d}^{2} = μ m_{B} g d$ $T h i s i s t h e s o l u t i o n g i v e n i n b o o k .$ $Y o u c a n a l s o g e t t h i s s o l u t i o n i f y o u p u t$ $k e = 0 i n s t e a d o f k e = μ m_{A} g i n m y e q n . (i) .$ $I {d o n}^{'} t a g r e e t h a t t h e {b o o k}^{'} s$ $a n s w e r i s t h e o n l y c o r r e c t a n s w e r .$ $B e c a u s e i t i s n o t c l e a r l y d e f i n e d w h a t$ $i s m e a n t w i t h ‘ ‘ b l o c k A l o s e s c o n t a c t$ $t o {w a l l}^{″} .$
	Commented by Tinkutara last updated on 17/Jan/18
	Thank you very much Sir! I got the answer.
	Answered by ajfour last updated on 17/Jan/18

	$k e = μ m_{A} g . . . . (i)$ $\frac{1}{2} {k d}^{2} = \frac{1}{2} {k e}^{2} + μ m_{B} g (d + e) . . . (i i)$ $f r o m (i) : e = \frac{0.5 \times 2 \times 10}{50} = \frac{1}{5}$ $r e p l a c i n g i n (i i) : e = \frac{1}{5}, k = 50$ $m_{A} = 2, m_{B} = 4$ $25 d^{2} = 1 + 20 (d + \frac{1}{5})$ $\Rightarrow 5 d^{2} - 4 d - 1 = 0$ $5 {(d - \frac{2}{5})}^{2} = \frac{9}{5}$ $d = \frac{2}{5} + \frac{3}{5} = 1 .$
	Commented by Tinkutara last updated on 17/Jan/18
	Thank you very much Sir! I got the answer.
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