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Question Number 29273 by ajfour last updated on 06/Feb/18

Commented by ajfour last updated on 06/Feb/18

$I c e c u b e m e l t s d u e t o f r i c t i o n,$ $a s i t s l i d e s d o w n a r o u g h i n c l i n e .$ $\frac{d m}{d x} = - \frac{f}{s △ T_{0}} (f i s f o r c e o f f r i c t i o n)$ $s t h e s p e c i f i c h e a t o f i c e a n d$ $△ T_{0} = c o n s t a n t (t e m p . d i f f e r e n c e)$ $I f w o r k d o n e b y f r i c t i o n g o e s a s$ $h e a t t o t h e i c e, t h e n$ $F i n d v e l o c i t y a s a f u n c t i o n o f$ $x, t h e d i s t a n c e a l o n g i n c l i n e$ $t h a t i t s l i d e s .$
Commented by mrW2 last updated on 07/Feb/18

$f = μ m g \sin θ$ $\Rightarrow \frac{d m}{d x} = - \frac{f}{s Δ T_{0}} = - \frac{μ g \sin θ}{s △ T_{0}} m = - k m$ $\Rightarrow \int_{m_{0}}^{m} \frac{d m}{m} = - k \int_{0}^{x} x$ $\Rightarrow \ln \frac{m}{m_{0}} = - k x$ $\Rightarrow m = m_{0} e^{- k x}$ $F = m g (\cos θ - μ \sin θ) = m \frac{d v}{d t} + v \frac{d m}{d t}$ ${m g}^{'} = m \frac{d v}{d t} + v \frac{d m}{d t}$ ${m g}^{'} = m \frac{d v}{d t} \times \frac{d x}{d x} + v \frac{d m}{d t} \times \frac{d x}{d x}$ ${m g}^{'} = m v \frac{d v}{d x} + v^{2} \frac{d m}{d x}$ ${m g}^{'} = m v \frac{d v}{d x} - v^{2} k m$ $v \frac{d v}{d x} = g^{^{'}} + {k v}^{2}$ $\frac{v d v}{\frac{g^{'}}{k} + v^{2}} = k d x$ $\frac{{d v}^{2}}{\frac{g^{'}}{k} + v^{2}} = 2 k d x$ $\ln \frac{\frac{g^{'}}{k} + v^{2}}{\frac{g^{'}}{k}} = 2 k x$ $\ln (1 + \frac{k}{g^{'}} v^{2}) = 2 k x$ $\Rightarrow v = \sqrt{\frac{g^{'}}{k} (e^{2 k x} - 1)}$ $\frac{d x}{d t} = \sqrt{\frac{g^{'}}{k} (e^{2 k x} - 1)}$ $\Rightarrow \frac{d x}{\sqrt{e^{2 k x} - 1}} = \sqrt{\frac{g^{'}}{k}} d t$ $\Rightarrow \int_{0}^{x} \frac{d (2 k x)}{\sqrt{e^{2 k x} - 1}} = 2 \sqrt{{k g}^{'}} \int_{0}^{t} d t$ $\Rightarrow 2 \tan^{- 1} \sqrt{e^{2 k x} - 1} = 2 \sqrt{{k g}^{'}} t$ $\Rightarrow \tan^{- 1} \sqrt{e^{2 k x} - 1} = \sqrt{{k g}^{'}} t$ $\Rightarrow \sqrt{e^{2 k x} - 1} = \tan (\sqrt{{k g}^{'}} t)$ $\Rightarrow e^{2 k x} = 1 + \tan^{2} (\sqrt{{k g}^{'}} t) = \frac{1}{\cos^{2} (\sqrt{{k g}^{'}} t)}$ $\Rightarrow x = \frac{1}{2 k} \ln [1 + \tan^{2} (\sqrt{{k g}^{'}} t)] = \frac{1}{k} \ln \frac{1}{\cos (\sqrt{{k g}^{'}} t)}$ $\Rightarrow v = \sqrt{\frac{k}{g^{'}}} \times \tan (\sqrt{{k g}^{'}} t)$ $\Rightarrow m = m_{0} e^{- k x} = \frac{m_{0}}{\sqrt{e^{2 k x}}} = m_{0} \cos (\sqrt{{k g}^{'}} t)$ $m = 0 \Rightarrow \sqrt{{k g}^{'}} t = \frac{π}{2} \Rightarrow t = \frac{π}{2 \sqrt{{k g}^{'}}}$ $\Rightarrow 0 ⩽ t ⩽ \frac{π}{2 \sqrt{{k g}^{'}}}$ $======= E N D ==========$ $e^{2 k x} - 1 = 2 k x + \frac{{(2 k x)}^{2}}{2} + \frac{{(2 k x)}^{3}}{6} + . . .$ $\Rightarrow v = \sqrt{\frac{g^{'}}{k} (e^{2 k x} - 1)} = \sqrt{2 g^{'} x (1 + k x + \frac{2}{3} k^{2} x^{2} + . . .)} > \sqrt{2 g^{'} x}$ $C o m p a r a t i o n w i t h t h e c a s e w i t h o u t$ $m a s s l o s s :$ $a = g (\cos θ - μ \sin θ) = g^{'}$ $\Rightarrow v = \sqrt{2 g^{'} x}$ $\Rightarrow A m e l t i n g i c e b l o c k m o v e s f a s t e r$ $t h a n a n o r m a l i c e b l o c k .$
Commented by ajfour last updated on 06/Feb/18

$W o n d e r F u l S i r! T h a n k s a l o t .$
Commented by math solver last updated on 08/Feb/18

$o m g! w h a t a s o l u t i o n n . . . . .$
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