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

$$Ablockoficeslidesdowna45°$$$$inclineplaneintwicethetimeit$$$$takestoslidedowna45°frictionless$$$$inclineplane.Whatisthecoefficient$$$$ofkineticfrictionbetweenthe$$$$iceblockandtheinclineplqne.$$

Answered by mrW1 last updated on 12/Nov/17

$$\theta =angleofinclination$$$$s=distancetheblockslidesdown$$$$\mu =coefficientoffriction$$$$a=accelerationofblock$$$$t=timetheblocktakes$$$$$$$$Case1:withoutfriction$$$${ma}_1=mg\sin \theta$$$$\Rightarrow a_1=g\sin \theta$$$$s=\frac{1}{2}{a}_1{t}_1^2$$$$\Rightarrow t_1=\sqrt{\frac{2s}{a_1}}=\sqrt{\frac{2s}{g\sin \theta }}$$$$$$$$Case2:withfriction$$$${ma}_2=mg\sin \theta -\mu mg\cos \theta$$$$\Rightarrow a_2=g(\sin \theta -\mu \cos \theta )$$$$\Rightarrow t_2=\sqrt{\frac{2s}{g(\sin \theta -\mu \cos \theta )}}$$$$$$$$since{t}_2=2t_1$$$$\Rightarrow \sqrt{\frac{2s}{g(\sin \theta -\mu \cos \theta )}}=2\times \sqrt{\frac{2s}{g\sin \theta }}$$$$\Rightarrow 4(\sin \theta -\mu \cos \theta )=\sin \theta$$$$\Rightarrow \mu =\frac{3\sin \theta }{4\cos \theta }=\frac{3}{4}\times \tan \theta =\frac{3}{4}\times \tan 45°=\frac{3}{4}=0.75$$

Commented by NECx last updated on 12/Nov/17

$$thankssomuch.I^{\prime }vefinally$$$$seenmymistake.$$$$$$$$$$

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