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Question Number 218221 by Tawa11 last updated on 01/Apr/25

Commented by Tawa11 last updated on 01/Apr/25

Ground is rough But resistance not given how come sir.
Commented by mr W last updated on 02/Apr/25

$s i m i l a r t o t h e q u e s t i o n i a s k e d y o u$ $i n Q 218030$
Commented by mr W last updated on 02/Apr/25

$a c c . t o t h a t s o l v e d q u e s t i o n {w h a t}^{'} s$ $y o u r a n s w e r t o t h i s q u e s t i o n ?$
Commented by Tawa11 last updated on 02/Apr/25

$Yes sir .$ $That is why I posted this sir .$ $For help .$
Commented by Tawa11 last updated on 02/Apr/25

$Tension = \frac{2 mg}{0.5} \times 0.1 = 0 . 4 mg$ $Friction = 0 . 8 mg$ $Permanent .$
Commented by mr W last updated on 02/Apr/25

$w h y n o t$ $T e n s i o n = \frac{2 m g}{0.5} \times 0.2 = 0.8 m g$ $F r i c t i o n = 0.09 m g$ $t h e n i n s t a n t a n e o u s ?$
Commented by mr W last updated on 03/Apr/25

$i t h i n k t h e q u e s t i o n {d o e s n}^{'} t h a v e a n$ $u n i q u e a n s w e r .$
	Answered by mr W last updated on 02/Apr/25

	Commented by mr W last updated on 04/Apr/25

	$s a y ∣ B C ∣= x$ $f = f r i c t i o n f o r c e (= μ m g)$ $k = \frac{λ}{l_{0}} = \frac{2 m g}{l_{0}}$ $v = 1.4 m / s$ ${K E}_{B} - {P E}_{C} = w o r k d o n e b y f r i c t i o n$ $\frac{{m v}^{2}}{2} - \frac{{k x}^{2}}{2} = f x = μ m g x$ $\frac{{m v}^{2}}{2} - \frac{2 {m g x}^{2}}{2 l_{0}} = μ m g x$ $\frac{x^{2}}{l_{0}} + μ x - \frac{v^{2}}{2 g} = 0$ $\Rightarrow x = (- μ + \sqrt{μ^{2} + \frac{2 v^{2}}{{g l}_{0}}}) \frac{l_{0}}{2}$ $t e n s i o n i n s t r i n g$ $T = k x = \frac{2 m g}{l_{0}} \times (- μ + \sqrt{μ^{2} + \frac{2 v^{2}}{{g l}_{0}}}) \frac{l_{0}}{2}$ $= (- μ + \sqrt{μ^{2} + \frac{2 v^{2}}{{g l}_{0}}}) m g$ $i f T > f, p a r t i c l e w i l l m o v e f u r t h e r .$ $i f T ⩽ f, p a r t i c l e r e s t s p e r m a n e n t l y .$ $T - f = (- 2 μ + \sqrt{μ^{2} + \frac{2 v^{2}}{{g l}_{0}}}) m g$ $- 2 μ + \sqrt{μ^{2} + \frac{2 v^{2}}{{g l}_{0}}} = 0$ $\Rightarrow μ = \sqrt{\frac{2 v^{2}}{3 {g l}_{0}}} = \sqrt{\frac{2 \times 1 . 4^{2}}{3 \times 9.81 \times 0.5}} \approx 0.516$ $i f μ < 0.516, p o s i t i o n o f r e s t i s$ $i n s t a n t a n e o u s . i f μ ⩾ 0.516, p o s i t i o n$ $o f r e s t i s p e r m a n e n t .$ $s i n c e μ i s u n k n o w n i n q u e s t i o n,$ $w e c a n n o t u n i q u e l y d e t e r m i n e i f$ $t h e p o s i t i o n o f r e s t i s p e r m a n e n t$ $o r i n s t a n t a n e o u s .$
	Commented by Tawa11 last updated on 02/Apr/25

	Commented by Tawa11 last updated on 02/Apr/25

	$Sir, see ∣ BC ∣ = 0.1 .$ $Is this approach correct ? .$ $This will tally with your working sir .$ $If ∣ BC ∣ ⩽ 0 . 13 m it is permanent .$
	Commented by mr W last updated on 03/Apr/25

	$i a s k e d y o u w h a t y o u g o t, b u t y o u$ $a n s w e r e d m e w h a t a n o t h e r p e r s o n$ $d i d a n d y o u a r e n o t s u r e i f h e$ $o r s h e i s r i g h t . i h a v e s h o w e d m y$ $c o m p l e t e w o r k i n g w i t h a l l s t e p s$ $w h i c h y o u c a n f o l l o w a n d c h e c k t o$ $m a k e y o u r o w n j u d g e .$ $a s i h a v e s a i d w i t h g i v e n c o n d i t i o n s$ $t h e q u e s t i o n {d o e s n}^{'} t h a v e a n u n i q u e$ $a n s w e r . i h a v e s h o w e d w h y .$ $i n t h e s o l u t i o n y o u p r e s e n t e d a b o v e$ $a f r i c t i o n c o e f f i c i e n t μ = 0.8 i s$ $t a k e n, b u t \underset{―}{t h i s i s n o t g i v e n i n t h e}$ $\underset{―}{o r i g i n a l q u e s t i o n} . i n t h i s c a s e h o w$ $c a n i a n s w e r y o u i f t h e a p p r o a c h$ $i s c o r r e c t ? w h y {d i d n}^{'} t h e / s h e t a k e$ $μ = 0.1 ? t h e n h e / s h e w o u l d h a v e$ $g o t a t o t a l l y d i f f e r e n t r e s u l t a n d$ $s a y t h a t t h e s t o p i s i n s t a n t a n e o u s .$
	Commented by Tawa11 last updated on 03/Apr/25

	$Sir,$ $I am only showing you sir .$ $After I see your solution .$ $I see the mistake I also made .$ $I followed their concept before .$ $But I understand your point sir .$ $And I asked the question because$ $I doubt their workings .$ $I just need to show you that you are$ $right since I have their duplicate workings .$ $I only doubt their workings .$ $Thanks for your time sir$
	Commented by mr W last updated on 03/Apr/25

	$o k . n o w y o u k n o w t h a t t h e p r o b l e m$ $l i e s i n t h e q u e s t i o n i t s e l f .$
	Commented by Tawa11 last updated on 03/Apr/25

	$Yes sir . Clearly seen .$
	Commented by Tawa11 last updated on 03/Apr/25

	$God bless you more sir .$
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