A block of ice slides down a 45° incline plane in twice the time it takes to slide down a 45° frictionless incline plane.What is the coefficient of kinetic friction between the ice block and the incline plqne.


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Question Number 24057 by NECx last updated on 12/Nov/17

$A b l o c k o f i c e s l i d e s d o w n a 45 °$ $i n c l i n e p l a n e i n t w i c e t h e t i m e i t$ $t a k e s t o s l i d e d o w n a 45 ° f r i c t i o n l e s s$ $i n c l i n e p l a n e . W h a t i s t h e c o e f f i c i e n t$ $o f k i n e t i c f r i c t i o n b e t w e e n t h e$ $i c e b l o c k a n d t h e i n c l i n e p l q n e .$
	Answered by mrW1 last updated on 12/Nov/17

	$θ = a n g l e o f i n c l i n a t i o n$ $s = d i s t a n c e t h e b l o c k s l i d e s d o w n$ $μ = c o e f f i c i e n t o f f r i c t i o n$ $a = a c c e l e r a t i o n o f b l o c k$ $t = t i m e t h e b l o c k t a k e s$ $C a s e 1 : w i t h o u t f r i c t i o n$ ${m a}_{1} = m g \sin θ$ $\Rightarrow a_{1} = g \sin θ$ $s = \frac{1}{2} a_{1} t_{1}^{2}$ $\Rightarrow t_{1} = \sqrt{\frac{2 s}{a_{1}}} = \sqrt{\frac{2 s}{g \sin θ}}$ $C a s e 2 : w i t h f r i c t i o n$ ${m a}_{2} = m g \sin θ - μ m g \cos θ$ $\Rightarrow a_{2} = g (\sin θ - μ \cos θ)$ $\Rightarrow t_{2} = \sqrt{\frac{2 s}{g (\sin θ - μ \cos θ)}}$ $s i n c e t_{2} = 2 t_{1}$ $\Rightarrow \sqrt{\frac{2 s}{g (\sin θ - μ \cos θ)}} = 2 \times \sqrt{\frac{2 s}{g \sin θ}}$ $\Rightarrow 4 (\sin θ - μ \cos θ) = \sin θ$ $\Rightarrow μ = \frac{3 \sin θ}{4 \cos θ} = \frac{3}{4} \times \tan θ = \frac{3}{4} \times \tan 45 ° = \frac{3}{4} = 0.75$
	Commented by NECx last updated on 12/Nov/17

	$t h a n k s s o m u c h . I^{'} v e f i n a l l y$ $s e e n m y m i s t a k e .$
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