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Question Number 24057 by NECx last updated on 12/Nov/17
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.
$${A}\:{block}\:{of}\:{ice}\:{slides}\:{down}\:{a}\:\mathrm{45}° \\ $$$${incline}\:{plane}\:{in}\:{twice}\:{the}\:{time}\:{it} \\ $$$${takes}\:{to}\:{slide}\:{down}\:{a}\:\mathrm{45}°\:{frictionless} \\ $$$${incline}\:{plane}.{What}\:{is}\:{the}\:{coefficient} \\ $$$${of}\:{kinetic}\:{friction}\:{between}\:{the} \\ $$$${ice}\:{block}\:{and}\:{the}\:{incline}\:{plqne}. \\ $$
Answered by mrW1 last updated on 12/Nov/17
θ=angle of inclination s=distance the block slides down μ=coefficient of friction a=acceleration of block t=time the block takes Case 1: without friction ma_1 =mgsin θ ⇒a_1 =gsin θ s=(1/2)a_1 t_1 ^2 ⇒t_1 =(√((2s)/a_1 ))=(√((2s)/(gsin θ))) Case 2: with friction ma_2 =mgsin θ−μmgcos θ ⇒a_2 =g(sin θ−μcos θ) ⇒t_2 =(√((2s)/(g(sin θ−μcos θ)))) since t_2 =2t_1 ⇒(√((2s)/(g(sin θ−μcos θ))))=2×(√((2s)/(gsin θ))) ⇒4(sin θ−μcos θ)=sin θ ⇒μ=((3sin θ)/(4cos θ))=(3/4)×tan θ=(3/4)×tan 45°=(3/4)=0.75
$$\theta={angle}\:{of}\:{inclination} \\ $$$${s}={distance}\:{the}\:{block}\:{slides}\:{down} \\ $$$$\mu={coefficient}\:{of}\:{friction} \\ $$$${a}={acceleration}\:{of}\:{block} \\ $$$${t}={time}\:{the}\:{block}\:{takes} \\ $$$$ \\ $$$${Case}\:\mathrm{1}:\:{without}\:{friction} \\ $$$${ma}_{\mathrm{1}} ={mg}\mathrm{sin}\:\theta \\ $$$$\Rightarrow{a}_{\mathrm{1}} ={g}\mathrm{sin}\:\theta \\ $$$${s}=\frac{\mathrm{1}}{\mathrm{2}}{a}_{\mathrm{1}} {t}_{\mathrm{1}} ^{\mathrm{2}} \\ $$$$\Rightarrow{t}_{\mathrm{1}} =\sqrt{\frac{\mathrm{2}{s}}{{a}_{\mathrm{1}} }}=\sqrt{\frac{\mathrm{2}{s}}{{g}\mathrm{sin}\:\theta}} \\ $$$$ \\ $$$${Case}\:\mathrm{2}:\:{with}\:{friction} \\ $$$${ma}_{\mathrm{2}} ={mg}\mathrm{sin}\:\theta−\mu{mg}\mathrm{cos}\:\theta \\ $$$$\Rightarrow{a}_{\mathrm{2}} ={g}\left(\mathrm{sin}\:\theta−\mu\mathrm{cos}\:\theta\right) \\ $$$$\Rightarrow{t}_{\mathrm{2}} =\sqrt{\frac{\mathrm{2}{s}}{{g}\left(\mathrm{sin}\:\theta−\mu\mathrm{cos}\:\theta\right)}} \\ $$$$ \\ $$$${since}\:{t}_{\mathrm{2}} =\mathrm{2}{t}_{\mathrm{1}} \\ $$$$\Rightarrow\sqrt{\frac{\mathrm{2}{s}}{{g}\left(\mathrm{sin}\:\theta−\mu\mathrm{cos}\:\theta\right)}}=\mathrm{2}×\sqrt{\frac{\mathrm{2}{s}}{{g}\mathrm{sin}\:\theta}} \\ $$$$\Rightarrow\mathrm{4}\left(\mathrm{sin}\:\theta−\mu\mathrm{cos}\:\theta\right)=\mathrm{sin}\:\theta \\ $$$$\Rightarrow\mu=\frac{\mathrm{3sin}\:\theta}{\mathrm{4cos}\:\theta}=\frac{\mathrm{3}}{\mathrm{4}}×\mathrm{tan}\:\theta=\frac{\mathrm{3}}{\mathrm{4}}×\mathrm{tan}\:\mathrm{45}°=\frac{\mathrm{3}}{\mathrm{4}}=\mathrm{0}.\mathrm{75} \\ $$
Commented by NECx last updated on 12/Nov/17
thanks so much. I′ve finally seen my mistake.
$${thanks}\:{so}\:{much}.\:{I}'{ve}\:{finally} \\ $$$${seen}\:{my}\:{mistake}. \\ $$$$ \\ $$$$ \\ $$

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