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=μmg sin θ ⇒(dm/dx) =−(f/(sΔT_0 ))=−((μgsin θ)/(s△T_0 )) m=−km ⇒∫_m_0 ^m (dm/m) =−k∫_0 ^x x ⇒ln (m/m_0 ) =−kx ⇒m=m_0 e^(−kx) F=mg(cos θ−μ sin θ)=m(dv/dt)+v(dm/dt) mg′=m(dv/dt)+v(dm/dt) mg′=m(dv/dt)×(dx/dx)+v(dm/dt)×(dx/dx) mg′=mv(dv/dx)+v^2 (dm/dx) mg′=mv(dv/dx)−v^2 km v(dv/dx)=g^′ +kv^2 ((vdv)/(((g′)/k)+v^2 ))=kdx (dv^2 /(((g′)/k)+v^2 ))=2kdx ln ((((g′)/k)+v^2 )/((g′)/k))=2kx ln (1+(k/(g′))v^2 )=2kx ⇒v=(√(((g′)/k)(e^(2kx) −1))) (dx/dt)=(√(((g′)/k)(e^(2kx) −1))) ⇒(dx/( (√(e^(2kx) −1))))=(√((g′)/k)) dt ⇒∫_0 ^( x) ((d(2kx))/( (√(e^(2kx) −1))))=2(√(kg′)) ∫_0 ^( t) dt ⇒2 tan^(−1) (√(e^(2kx) −1))=2(√(kg′)) t ⇒tan^(−1) (√(e^(2kx) −1))=(√(kg′)) t ⇒(√(e^(2kx) −1))=tan ((√(kg′)) t) ⇒e^(2kx) =1+tan^2 ((√(kg′)) t)=(1/(cos^2 ((√(kg′)) t))) ⇒x=(1/(2k))ln [1+tan^2 ((√(kg′)) t)]=(1/k)ln (1/(cos ((√(kg′)) t))) ⇒v=(√(k/(g′)))×tan ((√(kg′)) t) ⇒m=m_0 e^(−kx) =(m_0 /( (√e^(2kx) )))=m_0 cos ((√(kg′)) t) m=0⇒(√(kg′)) t=(π/2)⇒t=(π/(2(√(kg′)))) ⇒0≤t≤(π/(2(√(kg′)))) =======END========== e^(2kx) −1=2kx+(((2kx)^2 )/2)+(((2kx)^3 )/6)+... ⇒v=(√(((g′)/k)(e^(2kx) −1))) =(√(2g′x(1+kx+(2/3)k^2 x^2 +...)))>(√(2g′x)) Comparation with the case without mass loss: a=g(cos θ−μ sin θ)=g′ ⇒v=(√(2g′x)) ⇒A melting ice block moves faster than a normal ice block.](https://www.tinkutara.com/question/Q29292.png)
$${f}=\mu{mg}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow\frac{{dm}}{{dx}}\:=−\frac{{f}}{{s}\Delta{T}_{\mathrm{0}} }=−\frac{\mu{g}\mathrm{sin}\:\theta}{{s}\bigtriangleup{T}_{\mathrm{0}} }\:{m}=−{km} \\ $$$$\Rightarrow\int_{{m}_{\mathrm{0}} } ^{{m}} \frac{{dm}}{{m}}\:=−{k}\int_{\mathrm{0}} ^{{x}} {x} \\ $$$$\Rightarrow\mathrm{ln}\:\frac{{m}}{{m}_{\mathrm{0}} }\:=−{kx} \\ $$$$\Rightarrow{m}={m}_{\mathrm{0}} {e}^{−{kx}} \\ $$$$ \\ $$$${F}={mg}\left(\mathrm{cos}\:\theta−\mu\:\mathrm{sin}\:\theta\right)={m}\frac{{dv}}{{dt}}+{v}\frac{{dm}}{{dt}} \\ $$$${mg}'={m}\frac{{dv}}{{dt}}+{v}\frac{{dm}}{{dt}} \\ $$$${mg}'={m}\frac{{dv}}{{dt}}×\frac{{dx}}{{dx}}+{v}\frac{{dm}}{{dt}}×\frac{{dx}}{{dx}} \\ $$$${mg}'={mv}\frac{{dv}}{{dx}}+{v}^{\mathrm{2}} \frac{{dm}}{{dx}} \\ $$$${mg}'={mv}\frac{{dv}}{{dx}}−{v}^{\mathrm{2}} {km} \\ $$$${v}\frac{{dv}}{{dx}}={g}^{'} +{kv}^{\mathrm{2}} \\ $$$$\frac{{vdv}}{\frac{{g}'}{{k}}+{v}^{\mathrm{2}} }={kdx} \\ $$$$\frac{{dv}^{\mathrm{2}} }{\frac{{g}'}{{k}}+{v}^{\mathrm{2}} }=\mathrm{2}{kdx} \\ $$$$\mathrm{ln}\:\frac{\frac{{g}'}{{k}}+{v}^{\mathrm{2}} }{\frac{{g}'}{{k}}}=\mathrm{2}{kx} \\ $$$$\mathrm{ln}\:\left(\mathrm{1}+\frac{{k}}{{g}'}{v}^{\mathrm{2}} \right)=\mathrm{2}{kx} \\ $$$$\Rightarrow{v}=\sqrt{\frac{{g}'}{{k}}\left({e}^{\mathrm{2}{kx}} −\mathrm{1}\right)} \\ $$$$ \\ $$$$\frac{{dx}}{{dt}}=\sqrt{\frac{{g}'}{{k}}\left({e}^{\mathrm{2}{kx}} −\mathrm{1}\right)} \\ $$$$\Rightarrow\frac{{dx}}{\:\sqrt{{e}^{\mathrm{2}{kx}} −\mathrm{1}}}=\sqrt{\frac{{g}'}{{k}}}\:{dt} \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\:{x}} \frac{{d}\left(\mathrm{2}{kx}\right)}{\:\sqrt{{e}^{\mathrm{2}{kx}} −\mathrm{1}}}=\mathrm{2}\sqrt{{kg}'}\:\int_{\mathrm{0}} ^{\:{t}} {dt} \\ $$$$\Rightarrow\mathrm{2}\:\mathrm{tan}^{−\mathrm{1}} \sqrt{{e}^{\mathrm{2}{kx}} −\mathrm{1}}=\mathrm{2}\sqrt{{kg}'}\:{t} \\ $$$$\Rightarrow\mathrm{tan}^{−\mathrm{1}} \sqrt{{e}^{\mathrm{2}{kx}} −\mathrm{1}}=\sqrt{{kg}'}\:{t} \\ $$$$\Rightarrow\sqrt{{e}^{\mathrm{2}{kx}} −\mathrm{1}}=\mathrm{tan}\:\left(\sqrt{{kg}'}\:{t}\right) \\ $$$$\Rightarrow{e}^{\mathrm{2}{kx}} =\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\left(\sqrt{{kg}'}\:{t}\right)=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} \:\left(\sqrt{{kg}'}\:{t}\right)} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}{k}}\mathrm{ln}\:\left[\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \:\left(\sqrt{{kg}'}\:{t}\right)\right]=\frac{\mathrm{1}}{{k}}\mathrm{ln}\:\frac{\mathrm{1}}{\mathrm{cos}\:\left(\sqrt{{kg}'}\:{t}\right)} \\ $$$$\Rightarrow{v}=\sqrt{\frac{{k}}{{g}'}}×\mathrm{tan}\:\left(\sqrt{{kg}'}\:{t}\right) \\ $$$$\Rightarrow{m}={m}_{\mathrm{0}} {e}^{−{kx}} =\frac{{m}_{\mathrm{0}} }{\:\sqrt{{e}^{\mathrm{2}{kx}} }}={m}_{\mathrm{0}} \mathrm{cos}\:\left(\sqrt{{kg}'}\:{t}\right) \\ $$$${m}=\mathrm{0}\Rightarrow\sqrt{{kg}'}\:{t}=\frac{\pi}{\mathrm{2}}\Rightarrow{t}=\frac{\pi}{\mathrm{2}\sqrt{{kg}'}} \\ $$$$\Rightarrow\mathrm{0}\leqslant{t}\leqslant\frac{\pi}{\mathrm{2}\sqrt{{kg}'}} \\ $$$$======={END}========== \\ $$$${e}^{\mathrm{2}{kx}} −\mathrm{1}=\mathrm{2}{kx}+\frac{\left(\mathrm{2}{kx}\right)^{\mathrm{2}} }{\mathrm{2}}+\frac{\left(\mathrm{2}{kx}\right)^{\mathrm{3}} }{\mathrm{6}}+… \\ $$$$\Rightarrow{v}=\sqrt{\frac{{g}'}{{k}}\left({e}^{\mathrm{2}{kx}} −\mathrm{1}\right)}\:=\sqrt{\mathrm{2}{g}'{x}\left(\mathrm{1}+{kx}+\frac{\mathrm{2}}{\mathrm{3}}{k}^{\mathrm{2}} {x}^{\mathrm{2}} +…\right)}>\sqrt{\mathrm{2}{g}'{x}} \\ $$$$ \\ $$$${Comparation}\:{with}\:{the}\:{case}\:{without} \\ $$$${mass}\:{loss}: \\ $$$${a}={g}\left(\mathrm{cos}\:\theta−\mu\:\mathrm{sin}\:\theta\right)={g}' \\ $$$$\Rightarrow{v}=\sqrt{\mathrm{2}{g}'{x}} \\ $$$$ \\ $$$$\Rightarrow{A}\:{melting}\:{ice}\:{block}\:{moves}\:{faster} \\ $$$${than}\:{a}\:{normal}\:{ice}\:{block}. \\ $$
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
$${omg}!\:{what}\:{a}\:{solutionn}….. \\ $$