Question Number 29273 by ajfour last updated on 06/Feb/18
Commented by ajfour last updated on 06/Feb/18
$${Ice}\:{cube}\:{melts}\:{due}\:{to}\:{friction}, \\ $$$${as}\:{it}\:{slides}\:{down}\:{a}\:{rough}\:{incline}. \\ $$$$\frac{{dm}}{{dx}}\:=−\frac{{f}}{{s}\bigtriangleup{T}_{\mathrm{0}} }\:\:\:\:\left({f}\:{is}\:{force}\:{of}\:{friction}\right) \\ $$$${s}\:{the}\:{specific}\:{heat}\:{of}\:{ice}\:{and} \\ $$$$\bigtriangleup{T}_{\mathrm{0}} \:\:={constant}\:\left({temp}.\:{difference}\right) \\ $$$${If}\:{work}\:{done}\:{by}\:{friction}\:{goes}\:{as} \\ $$$${heat}\:{to}\:{the}\:{ice},\:{then} \\ $$$${Find}\:{velocity}\:{as}\:{a}\:{function}\:{of} \\ $$$${x},\:{the}\:{distance}\:{along}\:{incline} \\ $$$${that}\:{it}\:{slides}. \\ $$$$ \\ $$
Commented by mrW2 last updated on 07/Feb/18
$${f}=\mu{mg}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow\frac{{dm}}{{dx}}\:=−\frac{{f}}{{s}\Delta{T}_{\mathrm{0}} }=−\frac{\mu{g}\mathrm{sin}\:\theta}{{s}\bigtriangleup{T}_{\mathrm{0}} }\:{m}=−{km} \\ $$$$\Rightarrow\int_{{m}_{\mathrm{0}} } ^{{m}} \frac{{dm}}{{m}}\:=−{k}\int_{\mathrm{0}} ^{{x}} {x} \\ $$$$\Rightarrow\mathrm{ln}\:\frac{{m}}{{m}_{\mathrm{0}} }\:=−{kx} \\ $$$$\Rightarrow{m}={m}_{\mathrm{0}} {e}^{−{kx}} \\ $$$$ \\ $$$${F}={mg}\left(\mathrm{cos}\:\theta−\mu\:\mathrm{sin}\:\theta\right)={m}\frac{{dv}}{{dt}}+{v}\frac{{dm}}{{dt}} \\ $$$${mg}'={m}\frac{{dv}}{{dt}}+{v}\frac{{dm}}{{dt}} \\ $$$${mg}'={m}\frac{{dv}}{{dt}}×\frac{{dx}}{{dx}}+{v}\frac{{dm}}{{dt}}×\frac{{dx}}{{dx}} \\ $$$${mg}'={mv}\frac{{dv}}{{dx}}+{v}^{\mathrm{2}} \frac{{dm}}{{dx}} \\ $$$${mg}'={mv}\frac{{dv}}{{dx}}−{v}^{\mathrm{2}} {km} \\ $$$${v}\frac{{dv}}{{dx}}={g}^{'} +{kv}^{\mathrm{2}} \\ $$$$\frac{{vdv}}{\frac{{g}'}{{k}}+{v}^{\mathrm{2}} }={kdx} \\ $$$$\frac{{dv}^{\mathrm{2}} }{\frac{{g}'}{{k}}+{v}^{\mathrm{2}} }=\mathrm{2}{kdx} \\ $$$$\mathrm{ln}\:\frac{\frac{{g}'}{{k}}+{v}^{\mathrm{2}} }{\frac{{g}'}{{k}}}=\mathrm{2}{kx} \\ $$$$\mathrm{ln}\:\left(\mathrm{1}+\frac{{k}}{{g}'}{v}^{\mathrm{2}} \right)=\mathrm{2}{kx} \\ $$$$\Rightarrow{v}=\sqrt{\frac{{g}'}{{k}}\left({e}^{\mathrm{2}{kx}} −\mathrm{1}\right)} \\ $$$$ \\ $$$$\frac{{dx}}{{dt}}=\sqrt{\frac{{g}'}{{k}}\left({e}^{\mathrm{2}{kx}} −\mathrm{1}\right)} \\ $$$$\Rightarrow\frac{{dx}}{\:\sqrt{{e}^{\mathrm{2}{kx}} −\mathrm{1}}}=\sqrt{\frac{{g}'}{{k}}}\:{dt} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\:{x}} \frac{{d}\left(\mathrm{2}{kx}\right)}{\:\sqrt{{e}^{\mathrm{2}{kx}} −\mathrm{1}}}=\mathrm{2}\sqrt{{kg}'}\:\int_{\mathrm{0}} ^{\:{t}} {dt} \\ $$$$\Rightarrow\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \sqrt{{e}^{\mathrm{2}{kx}} −\mathrm{1}}=\mathrm{2}\sqrt{{kg}'}\:{t} \\ $$$$\Rightarrow\mathrm{tan}^{−\mathrm{1}} \sqrt{{e}^{\mathrm{2}{kx}} −\mathrm{1}}=\sqrt{{kg}'}\:{t} \\ $$$$\Rightarrow\sqrt{{e}^{\mathrm{2}{kx}} −\mathrm{1}}=\mathrm{tan}\:\left(\sqrt{{kg}'}\:{t}\right) \\ $$$$\Rightarrow{e}^{\mathrm{2}{kx}} =\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\left(\sqrt{{kg}'}\:{t}\right)=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\left(\sqrt{{kg}'}\:{t}\right)} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}{k}}\mathrm{ln}\:\left[\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\left(\sqrt{{kg}'}\:{t}\right)\right]=\frac{\mathrm{1}}{{k}}\mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{cos}\:\left(\sqrt{{kg}'}\:{t}\right)} \\ $$$$\Rightarrow{v}=\sqrt{\frac{{k}}{{g}'}}×\mathrm{tan}\:\left(\sqrt{{kg}'}\:{t}\right) \\ $$$$\Rightarrow{m}={m}_{\mathrm{0}} {e}^{−{kx}} =\frac{{m}_{\mathrm{0}} }{\:\sqrt{{e}^{\mathrm{2}{kx}} }}={m}_{\mathrm{0}} \mathrm{cos}\:\left(\sqrt{{kg}'}\:{t}\right) \\ $$$${m}=\mathrm{0}\Rightarrow\sqrt{{kg}'}\:{t}=\frac{\pi}{\mathrm{2}}\Rightarrow{t}=\frac{\pi}{\mathrm{2}\sqrt{{kg}'}} \\ $$$$\Rightarrow\mathrm{0}\leqslant{t}\leqslant\frac{\pi}{\mathrm{2}\sqrt{{kg}'}} \\ $$$$======={END}========== \\ $$$${e}^{\mathrm{2}{kx}} −\mathrm{1}=\mathrm{2}{kx}+\frac{\left(\mathrm{2}{kx}\right)^{\mathrm{2}} }{\mathrm{2}}+\frac{\left(\mathrm{2}{kx}\right)^{\mathrm{3}} }{\mathrm{6}}+… \\ $$$$\Rightarrow{v}=\sqrt{\frac{{g}'}{{k}}\left({e}^{\mathrm{2}{kx}} −\mathrm{1}\right)}\:=\sqrt{\mathrm{2}{g}'{x}\left(\mathrm{1}+{kx}+\frac{\mathrm{2}}{\mathrm{3}}{k}^{\mathrm{2}} {x}^{\mathrm{2}} +…\right)}>\sqrt{\mathrm{2}{g}'{x}} \\ $$$$ \\ $$$${Comparation}\:{with}\:{the}\:{case}\:{without} \\ $$$${mass}\:{loss}: \\ $$$${a}={g}\left(\mathrm{cos}\:\theta−\mu\:\mathrm{sin}\:\theta\right)={g}' \\ $$$$\Rightarrow{v}=\sqrt{\mathrm{2}{g}'{x}} \\ $$$$ \\ $$$$\Rightarrow{A}\:{melting}\:{ice}\:{block}\:{moves}\:{faster} \\ $$$${than}\:{a}\:{normal}\:{ice}\:{block}. \\ $$
Commented by ajfour last updated on 06/Feb/18
$$\mathscr{W}{onder}\mathcal{F}{ul}\:\mathcal{S}{ir}\:!\:{Thanks}\:{a}\:{lot}. \\ $$
Commented by math solver last updated on 08/Feb/18
$${omg}!\:{what}\:{a}\:{solutionn}….. \\ $$