Question Number 225146 by fantastic last updated on 17/Oct/25
Commented by fantastic last updated on 17/Oct/25
$${maybe}\:{it}\:{is}\:{an}\:{easy}\:{question} \\ $$$${but}\:{i}\:{feel}\:{very}\:{happy} \\ $$$${by}\:{solving}\:{this}\:{i}\:.{e}\:{irodov} \\ $$$${q}\:{by}\:{myself} \\ $$
Commented by mr W last updated on 18/Oct/25
$${x}_{{max}} =\frac{\mathrm{2}\:\mathrm{tan}\:\alpha}{{a}} \\ $$$${v}_{{max}} =\mathrm{sin}\:\alpha\:\sqrt{\frac{{g}}{{a}\:\mathrm{cos}\:\alpha}} \\ $$
Commented by ajfour last updated on 19/Oct/25
$$\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} ={mgx}\mathrm{sin}\:\alpha−{kmg}\mathrm{cos}\:\alpha\int_{\mathrm{0}} ^{\:{x}} {xdx} \\ $$$${v}^{\mathrm{2}} ={gx}\left(\mathrm{2sin}\:\alpha−{kx}\mathrm{cos}\:\alpha\right) \\ $$$${v}=\mathrm{0}\:{at}\:{x}=\mathrm{0},\:\frac{\mathrm{2tan}\:\alpha}{{k}} \\ $$$${v}^{\mathrm{2}} =\left({gk}\mathrm{cos}\:\alpha\right)\left({x}−\mathrm{0}\right)\left(\frac{\mathrm{2tan}\:\alpha}{{k}}−{x}\right) \\ $$$${parabolic}.\:{so}\:{v}={v}_{{max}} \:{when} \\ $$$${x}=\frac{\mathrm{0}+\frac{\mathrm{2tan}\:\alpha}{{k}}}{\mathrm{2}}=\frac{\mathrm{tan}\:\alpha}{{k}} \\ $$$${v}_{{max}} ^{\mathrm{2}} =\left({gk}\mathrm{cos}\:\alpha\right)\left(\frac{\mathrm{tan}\:\alpha}{{k}}\right)^{\mathrm{2}} \\ $$$${v}_{{max}} =\mathrm{sin}\:\alpha\sqrt{\frac{{g}}{{k}\mathrm{cos}\:\alpha}} \\ $$$$ \\ $$
Commented by fantastic last updated on 18/Oct/25
$${how} \\ $$
Commented by mr W last updated on 18/Oct/25
$${what}\:{did}\:{you}\:{get}? \\ $$
Commented by fantastic last updated on 18/Oct/25
$${same} \\ $$
Commented by fantastic last updated on 18/Oct/25
$${i}\:{didnt}\:{get}\:{the}\:{v}_{{max}} \\ $$
Commented by fantastic last updated on 20/Oct/25
$${thanks}\:{sir} \\ $$
Commented by fantastic last updated on 20/Oct/25
$${i}\:{am}\:{going}\:{to}\:{post}\:{another}\:{Q} \\ $$$${related}\:{to}\:{this} \\ $$
Answered by mr W last updated on 19/Oct/25
![μ=ξx (μ & ξ replace k & a in question) v=(dx/dt) a=(dv/dt)=v(dv/dx) f=μN=μmg cos α ma=mg sin α−μmg cos α v(dv/dx)=g(sin α−ξ cos α x) ∫_0 ^v vdv=g∫_0 ^x (sin α−ξ cos α x)dx (v^2 /2)=((g[sin^2 α−(sin α−ξ cos α x)^2 ])/(2ξ cos α)) v^2 =gx(2 sin α−ξ cos α x) ⇒v=(√(gx(2 sin α−ξ cos α x))) at x_(max) : v=0 gx(2 sin α−ξ cos α x)=0 ⇒x_(max) =((2 sin α)/(ξ cos α))=((2 tan α)/ξ) ✓ v^2 =gx(2 sin α−ξ cos α x) =(g/(ξ cos α))×ξ cos α x (2 sin α−ξ cos α x) ≤(g/(ξ cos α))×(((ξ cos α x+2 sin α−ξ cos α x)/2))^2 =((g sin^2 α)/(ξ cos α)) ⇒v_(max) =sin α (√(g/(ξ cos α))) ✓ when ξ cos α x=2 sin α−ξ cos α x, i.e. x=((tan α)/ξ)=(x_(max) /2). (dx/dt)=v=(√(gξ cos α(((2 tan α)/ξ)−x)x)) dt=(dx/( (√(gξ cos α (((2 tan α)/ξ)−x)x)))) ∫_0 ^t dt=(1/( (√(gξ cos α))))∫_0 ^x (dx/( (√((((2 tan α)/ξ)−x)x)))) ⇒t=(2/( (√(gξ cos α)))) sin^(−1) (√((ξx)/(2 tan α))) at x=((tan α)/ξ) for v_(max) : t=(π/( 2(√(gξ cos α)))) at x_(max) =((2 tan α)/ξ): t=(π/( (√(gξ cos α))))](https://www.tinkutara.com/question/Q225169.png)
$$\mu=\xi{x}\: \\ $$$$\left(\mu\:\&\:\xi\:{replace}\:{k}\:\&\:{a}\:{in}\:{question}\right) \\ $$$${v}=\frac{{dx}}{{dt}} \\ $$$${a}=\frac{{dv}}{{dt}}={v}\frac{{dv}}{{dx}} \\ $$$${f}=\mu{N}=\mu{mg}\:\mathrm{cos}\:\alpha \\ $$$${ma}={mg}\:\mathrm{sin}\:\alpha−\mu{mg}\:\mathrm{cos}\:\alpha \\ $$$${v}\frac{{dv}}{{dx}}={g}\left(\mathrm{sin}\:\alpha−\xi\:\mathrm{cos}\:\alpha\:{x}\right) \\ $$$$\int_{\mathrm{0}} ^{{v}} {vdv}={g}\int_{\mathrm{0}} ^{{x}} \left(\mathrm{sin}\:\alpha−\xi\:\mathrm{cos}\:\alpha\:{x}\right){dx} \\ $$$$\frac{{v}^{\mathrm{2}} }{\mathrm{2}}=\frac{{g}\left[\mathrm{sin}^{\mathrm{2}} \:\alpha−\left(\mathrm{sin}\:\alpha−\xi\:\mathrm{cos}\:\alpha\:{x}\right)^{\mathrm{2}} \right]}{\mathrm{2}\xi\:\mathrm{cos}\:\alpha} \\ $$$${v}^{\mathrm{2}} ={gx}\left(\mathrm{2}\:\mathrm{sin}\:\alpha−\xi\:\mathrm{cos}\:\alpha\:{x}\right) \\ $$$$\Rightarrow{v}=\sqrt{{gx}\left(\mathrm{2}\:\mathrm{sin}\:\alpha−\xi\:\mathrm{cos}\:\alpha\:{x}\right)} \\ $$$${at}\:{x}_{{max}} :\:{v}=\mathrm{0} \\ $$$${gx}\left(\mathrm{2}\:\mathrm{sin}\:\alpha−\xi\:\mathrm{cos}\:\alpha\:{x}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}_{{max}} =\frac{\mathrm{2}\:\mathrm{sin}\:\alpha}{\xi\:\mathrm{cos}\:\alpha}=\frac{\mathrm{2}\:\mathrm{tan}\:\alpha}{\xi}\:\checkmark \\ $$$${v}^{\mathrm{2}} ={gx}\left(\mathrm{2}\:\mathrm{sin}\:\alpha−\xi\:\mathrm{cos}\:\alpha\:{x}\right) \\ $$$$\:\:\:=\frac{{g}}{\xi\:\mathrm{cos}\:\alpha}×\xi\:\mathrm{cos}\:\alpha\:{x}\:\left(\mathrm{2}\:\mathrm{sin}\:\alpha−\xi\:\mathrm{cos}\:\alpha\:{x}\right) \\ $$$$\:\:\:\leqslant\frac{{g}}{\xi\:\mathrm{cos}\:\alpha}×\left(\frac{\xi\:\mathrm{cos}\:\alpha\:{x}+\mathrm{2}\:\mathrm{sin}\:\alpha−\xi\:\mathrm{cos}\:\alpha\:{x}}{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$$\:\:\:=\frac{{g}\:\mathrm{sin}^{\mathrm{2}} \:\alpha}{\xi\:\mathrm{cos}\:\alpha} \\ $$$$\Rightarrow{v}_{{max}} =\mathrm{sin}\:\alpha\:\sqrt{\frac{{g}}{\xi\:\mathrm{cos}\:\alpha}}\:\:\checkmark \\ $$$${when}\:\xi\:\mathrm{cos}\:\alpha\:{x}=\mathrm{2}\:\mathrm{sin}\:\alpha−\xi\:\mathrm{cos}\:\alpha\:{x},\: \\ $$$${i}.{e}.\:{x}=\frac{\mathrm{tan}\:\alpha}{\xi}=\frac{{x}_{{max}} }{\mathrm{2}}. \\ $$$$ \\ $$$$\frac{{dx}}{{dt}}={v}=\sqrt{{g}\xi\:\mathrm{cos}\:\alpha\left(\frac{\mathrm{2}\:\mathrm{tan}\:\alpha}{\xi}−{x}\right){x}} \\ $$$${dt}=\frac{{dx}}{\:\sqrt{{g}\xi\:\mathrm{cos}\:\alpha\:\left(\frac{\mathrm{2}\:\mathrm{tan}\:\alpha}{\xi}−{x}\right){x}}} \\ $$$$\int_{\mathrm{0}} ^{{t}} {dt}=\frac{\mathrm{1}}{\:\sqrt{{g}\xi\:\mathrm{cos}\:\alpha}}\int_{\mathrm{0}} ^{{x}} \frac{{dx}}{\:\sqrt{\left(\frac{\mathrm{2}\:\mathrm{tan}\:\alpha}{\xi}−{x}\right){x}}} \\ $$$$\Rightarrow{t}=\frac{\mathrm{2}}{\:\sqrt{{g}\xi\:\mathrm{cos}\:\alpha}}\:\mathrm{sin}^{−\mathrm{1}} \sqrt{\frac{\xi{x}}{\mathrm{2}\:\mathrm{tan}\:\alpha}} \\ $$$${at}\:{x}=\frac{\mathrm{tan}\:\alpha}{\xi}\:{for}\:{v}_{{max}} : \\ $$$${t}=\frac{\pi}{\:\mathrm{2}\sqrt{{g}\xi\:\mathrm{cos}\:\alpha}} \\ $$$${at}\:{x}_{{max}} =\frac{\mathrm{2}\:\mathrm{tan}\:\alpha}{\xi}: \\ $$$${t}=\frac{\pi}{\:\sqrt{{g}\xi\:\mathrm{cos}\:\alpha}} \\ $$
Commented by fantastic last updated on 18/Oct/25
$${v}_{{max}} =\sqrt{\frac{{g}}{{a}}\mathrm{tan}\:\alpha\mathrm{sin}\:\alpha} \\ $$
Commented by fantastic last updated on 18/Oct/25
$${when}\:{x}=\frac{\mathrm{1}}{{a}}\mathrm{tan}\:\alpha \\ $$$${v}\:{is}\:{max} \\ $$
Answered by fantastic last updated on 18/Oct/25
![a_r =gsin α−μgcos α=g(sin α−ϕxcos α)[μ=ϕx] v(dv/dx)=gsin α−μgcos α vdv=gsin αdx−μgcos αdx ∫_0 ^v vdv=∫_0 ^x gsin αdx−∫_0 ^x ϕxgcos αdx (v^2 /2)=gxsin α−ϕg(x^2 /2)cos α when the bar stops v=0 ⇒gxsin α=ϕg(x^2 /2)cos α x=((2tan α)/ϕ) we got v=(√(2gxsin α−gϕx^2 cos α))=(√u)[let] (dv/dx)=(dv/du)×(du/dx) (du/dx)=(d/dx)(2gxsin α−gϕx^2 cos α) =2gsin α−2gϕxcos α (dv/du)=(d/du)(√u)=(1/(2(√u)))=(1/(2(√(2gxsin α−gϕx^2 cos α)))) ∴(dv/dx)=(1/(2(√(2gxsin α−gϕx^2 cos α))))×2gsin α−2gϕxcos α =((g(sin α−ϕxcos α))/( (√(2gxsin α−gϕx^2 cos α)))) at v_(max) (dv/dx)=0 ⇒g(sin α−ϕxcos α)=0 ⇒x=(1/ϕ)tan α](https://www.tinkutara.com/question/Q225175.png)
$${a}_{{r}} ={g}\mathrm{sin}\:\alpha−\mu{g}\mathrm{cos}\:\alpha={g}\left(\mathrm{sin}\:\alpha−\varphi{x}\mathrm{cos}\:\alpha\right)\left[\mu=\varphi{x}\right] \\ $$$${v}\frac{{dv}}{{dx}}={g}\mathrm{sin}\:\alpha−\mu{g}\mathrm{cos}\:\alpha \\ $$$${vdv}={g}\mathrm{sin}\:\alpha{dx}−\mu{g}\mathrm{cos}\:\alpha{dx} \\ $$$$\underset{\mathrm{0}} {\overset{{v}} {\int}}{vdv}=\underset{\mathrm{0}} {\overset{{x}} {\int}}{g}\mathrm{sin}\:\alpha{dx}−\underset{\mathrm{0}} {\overset{{x}} {\int}}\varphi{xg}\mathrm{cos}\:\alpha{dx} \\ $$$$\frac{{v}^{\mathrm{2}} }{\mathrm{2}}={gx}\mathrm{sin}\:\alpha−\varphi{g}\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{cos}\:\alpha \\ $$$${when}\:{the}\:{bar}\:{stops}\:{v}=\mathrm{0} \\ $$$$\Rightarrow{gx}\mathrm{sin}\:\alpha=\varphi{g}\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\mathrm{cos}\:\alpha \\ $$$${x}=\frac{\mathrm{2tan}\:\alpha}{\varphi} \\ $$$${we}\:{got} \\ $$$${v}=\sqrt{\mathrm{2}{gx}\mathrm{sin}\:\alpha−{g}\varphi{x}^{\mathrm{2}} \mathrm{cos}\:\alpha}=\sqrt{{u}}\left[{let}\right] \\ $$$$\frac{{dv}}{{dx}}=\frac{{dv}}{{du}}×\frac{{du}}{{dx}} \\ $$$$\frac{{du}}{{dx}}=\frac{{d}}{{dx}}\left(\mathrm{2}{gx}\mathrm{sin}\:\alpha−{g}\varphi{x}^{\mathrm{2}} \mathrm{cos}\:\alpha\right) \\ $$$$=\mathrm{2}{g}\mathrm{sin}\:\alpha−\mathrm{2}{g}\varphi{x}\mathrm{cos}\:\alpha \\ $$$$\frac{{dv}}{{du}}=\frac{{d}}{{du}}\sqrt{{u}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{u}}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}{gx}\mathrm{sin}\:\alpha−{g}\varphi{x}^{\mathrm{2}} \mathrm{cos}\:\alpha}} \\ $$$$\therefore\frac{{dv}}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}{gx}\mathrm{sin}\:\alpha−{g}\varphi{x}^{\mathrm{2}} \mathrm{cos}\:\alpha}}×\mathrm{2}{g}\mathrm{sin}\:\alpha−\mathrm{2}{g}\varphi{x}\mathrm{cos}\:\alpha \\ $$$$=\frac{{g}\left(\mathrm{sin}\:\alpha−\varphi{x}\mathrm{cos}\:\alpha\right)}{\:\sqrt{\mathrm{2}{gx}\mathrm{sin}\:\alpha−{g}\varphi{x}^{\mathrm{2}} \mathrm{cos}\:\alpha}} \\ $$$${at}\:{v}_{{max}} \frac{{dv}}{{dx}}=\mathrm{0} \\ $$$$\Rightarrow{g}\left(\mathrm{sin}\:\alpha−\varphi{x}\mathrm{cos}\:\alpha\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}}{\varphi}\mathrm{tan}\:\alpha \\ $$