Question-218221

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Question Number 218221 by Tawa11 last updated on 01/Apr/25

Commented by Tawa11 last updated on 01/Apr/25

$$ \\ $$Ground is rough But resistance not given
how come sir.

Commented by mr W last updated on 02/Apr/25

similar to the question i asked you in Q218030

$${similar}\:{to}\:{the}\:{question}\:{i}\:{asked}\:{you}\: \\ $$$${in}\:{Q}\mathrm{218030} \\ $$

Commented by mr W last updated on 02/Apr/25

$${acc}.\:{to}\:{that}\:{solved}\:{question}\:{what}'{s} \\ $$$${your}\:{answer}\:{to}\:{this}\:{question}? \\ $$

Commented by Tawa11 last updated on 02/Apr/25

Yes sir. That is why I posted this sir. For help.

$$\mathrm{Yes}\:\mathrm{sir}. \\ $$$$\mathrm{That}\:\mathrm{is}\:\mathrm{why}\:\mathrm{I}\:\mathrm{posted}\:\mathrm{this}\:\mathrm{sir}. \\ $$$$\mathrm{For}\:\mathrm{help}. \\ $$

Commented by Tawa11 last updated on 02/Apr/25

Tension = ((2mg)/(0.5)) × 0.1 = 0.4mg Friction = 0.8mg Permanent.

$$\mathrm{Tension}\:\:=\:\:\frac{\mathrm{2mg}}{\mathrm{0}.\mathrm{5}}\:\:×\:\mathrm{0}.\mathrm{1}\:\:=\:\:\mathrm{0}.\mathrm{4mg} \\ $$$$\mathrm{Friction}\:\:=\:\:\mathrm{0}.\mathrm{8mg} \\ $$$$\mathrm{Permanent}. \\ $$

Commented by mr W last updated on 02/Apr/25

why not Tension = ((2mg)/(0.5))×0.2=0.8mg Friction = 0.09mg then instantaneous?

$${why}\:{not}\: \\ $$$${Tension}\:=\:\frac{\mathrm{2}{mg}}{\mathrm{0}.\mathrm{5}}×\mathrm{0}.\mathrm{2}=\mathrm{0}.\mathrm{8}{mg} \\ $$$${Friction}\:=\:\mathrm{0}.\mathrm{09}{mg} \\ $$$${then}\:{instantaneous}? \\ $$

Commented by mr W last updated on 03/Apr/25

$${i}\:{think}\:{the}\:{question}\:{doesn}'{t}\:{have}\:{an} \\ $$$${unique}\:{answer}. \\ $$

Answered by mr W last updated on 02/Apr/25

Commented by mr W last updated on 04/Apr/25

say ∣BC∣=x f=friction force (=μmg) k=(λ/l_0 )=((2mg)/l_0 ) v=1.4 m/s KE_B −PE_C =work done by friction ((mv^2 )/2)−((kx^2 )/2)=fx=μmgx ((mv^2 )/2)−((2mgx^2 )/(2l_0 ))=μmgx (x^2 /l_0 )+μx−(v^2 /(2g))=0 ⇒x=(−μ+(√(μ^2 +((2v^2 )/(gl_0 )))))(l_0 /2) tension in string T=kx=((2mg)/l_0 )×(−μ+(√(μ^2 +((2v^2 )/(gl_0 )))))(l_0 /2) =(−μ+(√(μ^2 +((2v^2 )/(gl_0 )))))mg if T>f, particle will move further. if T≤f, particle rests permanently. T−f=(−2μ+(√(μ^2 +((2v^2 )/(gl_0 )))))mg −2μ+(√(μ^2 +((2v^2 )/(gl_0 ))))=0 ⇒μ=(√((2v^2 )/(3gl_0 )))=(√((2×1.4^2 )/(3×9.81×0.5)))≈0.516 if μ<0.516, position of rest is instantaneous. if μ≥0.516, position of rest is permanent. since μ is unknown in question, we can not uniquely determine if the position of rest is permanent or instantaneous.

$${say}\:\mid{BC}\mid={x} \\ $$$${f}={friction}\:{force}\:\left(=\mu{mg}\right) \\ $$$${k}=\frac{\lambda}{{l}_{\mathrm{0}} }=\frac{\mathrm{2}{mg}}{{l}_{\mathrm{0}} } \\ $$$${v}=\mathrm{1}.\mathrm{4}\:{m}/{s} \\ $$$${KE}_{{B}} −{PE}_{{C}} ={work}\:{done}\:{by}\:{friction} \\ $$$$\frac{{mv}^{\mathrm{2}} }{\mathrm{2}}−\frac{{kx}^{\mathrm{2}} }{\mathrm{2}}={fx}=\mu{mgx} \\ $$$$\frac{{mv}^{\mathrm{2}} }{\mathrm{2}}−\frac{\mathrm{2}{mgx}^{\mathrm{2}} }{\mathrm{2}{l}_{\mathrm{0}} }=\mu{mgx} \\ $$$$\frac{{x}^{\mathrm{2}} }{{l}_{\mathrm{0}} }+\mu{x}−\frac{{v}^{\mathrm{2}} }{\mathrm{2}{g}}=\mathrm{0} \\ $$$$\Rightarrow{x}=\left(−\mu+\sqrt{\mu^{\mathrm{2}} +\frac{\mathrm{2}{v}^{\mathrm{2}} }{{gl}_{\mathrm{0}} }}\right)\frac{{l}_{\mathrm{0}} }{\mathrm{2}} \\ $$$${tension}\:{in}\:{string} \\ $$$${T}={kx}=\frac{\mathrm{2}{mg}}{{l}_{\mathrm{0}} }×\left(−\mu+\sqrt{\mu^{\mathrm{2}} +\frac{\mathrm{2}{v}^{\mathrm{2}} }{{gl}_{\mathrm{0}} }}\right)\frac{{l}_{\mathrm{0}} }{\mathrm{2}} \\ $$$$\:\:\:=\left(−\mu+\sqrt{\mu^{\mathrm{2}} +\frac{\mathrm{2}{v}^{\mathrm{2}} }{{gl}_{\mathrm{0}} }}\right){mg} \\ $$$${if}\:{T}>{f},\:{particle}\:{will}\:{move}\:{further}. \\ $$$${if}\:{T}\leqslant{f},\:{particle}\:{rests}\:{permanently}. \\ $$$${T}−{f}=\left(−\mathrm{2}\mu+\sqrt{\mu^{\mathrm{2}} +\frac{\mathrm{2}{v}^{\mathrm{2}} }{{gl}_{\mathrm{0}} }}\right){mg} \\ $$$$−\mathrm{2}\mu+\sqrt{\mu^{\mathrm{2}} +\frac{\mathrm{2}{v}^{\mathrm{2}} }{{gl}_{\mathrm{0}} }}=\mathrm{0} \\ $$$$\Rightarrow\mu=\sqrt{\frac{\mathrm{2}{v}^{\mathrm{2}} }{\mathrm{3}{gl}_{\mathrm{0}} }}=\sqrt{\frac{\mathrm{2}×\mathrm{1}.\mathrm{4}^{\mathrm{2}} }{\mathrm{3}×\mathrm{9}.\mathrm{81}×\mathrm{0}.\mathrm{5}}}\approx\mathrm{0}.\mathrm{516} \\ $$$${if}\:\mu<\mathrm{0}.\mathrm{516},\:{position}\:{of}\:{rest}\:{is}\: \\ $$$${instantaneous}.\:{if}\:\mu\geqslant\mathrm{0}.\mathrm{516},\:{position} \\ $$$${of}\:{rest}\:{is}\:{permanent}. \\ $$$$ \\ $$$${since}\:\mu\:{is}\:{unknown}\:{in}\:{question}, \\ $$$${we}\:{can}\:{not}\:{uniquely}\:{determine}\:{if} \\ $$$${the}\:{position}\:{of}\:{rest}\:{is}\:{permanent} \\ $$$${or}\:{instantaneous}. \\ $$

Commented by Tawa11 last updated on 02/Apr/25

Commented by Tawa11 last updated on 02/Apr/25

Sir, see ∣BC∣ = 0.1. Is this approach correct?. This will tally with your working sir. If ∣BC∣ ≤ 0.13m it is permanent.

$$\mathrm{Sir},\:\mathrm{see}\:\mid\mathrm{BC}\mid\:=\:\mathrm{0}.\mathrm{1}. \\ $$$$\mathrm{Is}\:\mathrm{this}\:\mathrm{approach}\:\mathrm{correct}?. \\ $$$$\mathrm{This}\:\mathrm{will}\:\mathrm{tally}\:\mathrm{with}\:\mathrm{your}\:\mathrm{working}\:\mathrm{sir}. \\ $$$$\mathrm{If}\:\mid\mathrm{BC}\mid\:\leqslant\:\mathrm{0}.\mathrm{13m}\:\:\:\mathrm{it}\:\mathrm{is}\:\mathrm{permanent}. \\ $$

Commented by mr W last updated on 03/Apr/25

$${i}\:{asked}\:{you}\:{what}\:{you}\:{got},\:{but}\:{you} \\ $$$${answered}\:{me}\:{what}\:{an}\:{other}\:{person} \\ $$$${did}\:{and}\:{you}\:{are}\:{not}\:{sure}\:{if}\:{he} \\ $$$${or}\:{she}\:{is}\:{right}.\:{i}\:{have}\:{showed}\:{my} \\ $$$${complete}\:{working}\:{with}\:{all}\:{steps} \\ $$$${which}\:{you}\:{can}\:{follow}\:{and}\:{check}\:{to} \\ $$$${make}\:{your}\:{own}\:{judge}. \\ $$$${as}\:{i}\:{have}\:{said}\:{with}\:{given}\:{conditions} \\ $$$${the}\:{question}\:{doesn}'{t}\:{have}\:{an}\:{unique} \\ $$$${answer}.\:{i}\:{have}\:{showed}\:{why}. \\ $$$${in}\:{the}\:{solution}\:{you}\:{presented}\:{above} \\ $$$${a}\:{friction}\:{coefficient}\:\mu=\mathrm{0}.\mathrm{8}\:{is} \\ $$$${taken},\:{but}\:\underline{{this}\:{is}\:{not}\:{given}\:{in}\:{the}} \\ $$$$\underline{{original}\:{question}}.\:{in}\:{this}\:{case}\:{how} \\ $$$${can}\:{i}\:{answer}\:{you}\:{if}\:{the}\:{approach}\: \\ $$$${is}\:{correct}?\:{why}\:{didn}'{t}\:{he}/{she}\:{take}\: \\ $$$$\mu=\mathrm{0}.\mathrm{1}?\:{then}\:{he}/{she}\:{would}\:{have}\: \\ $$$${got}\:{a}\:{totally}\:{different}\:{result}\:{and}\: \\ $$$${say}\:{that}\:{the}\:{stop}\:{is}\:{instantaneous}. \\ $$

Commented by Tawa11 last updated on 03/Apr/25

Sir, I am only showing you sir. After I see your solution. I see the mistake I also made. I followed their concept before. But I understand your point sir. And I asked the question because I doubt their workings. I just need to show you that you are right since I have their duplicate workings. I only doubt their workings. Thanks for your time sir

$$\mathrm{Sir}, \\ $$$$\mathrm{I}\:\mathrm{am}\:\mathrm{only}\:\mathrm{showing}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{After}\:\mathrm{I}\:\mathrm{see}\:\mathrm{your}\:\mathrm{solution}. \\ $$$$\mathrm{I}\:\mathrm{see}\:\mathrm{the}\:\mathrm{mistake}\:\mathrm{I}\:\mathrm{also}\:\mathrm{made}. \\ $$$$\mathrm{I}\:\mathrm{followed}\:\mathrm{their}\:\mathrm{concept}\:\mathrm{before}. \\ $$$$\mathrm{But}\:\mathrm{I}\:\mathrm{understand}\:\mathrm{your}\:\mathrm{point}\:\mathrm{sir}. \\ $$$$\mathrm{And}\:\mathrm{I}\:\mathrm{asked}\:\mathrm{the}\:\mathrm{question}\:\mathrm{because} \\ $$$$\mathrm{I}\:\mathrm{doubt}\:\mathrm{their}\:\mathrm{workings}. \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{just}\:\mathrm{need}\:\mathrm{to}\:\mathrm{show}\:\mathrm{you}\:\mathrm{that}\:\mathrm{you}\:\mathrm{are} \\ $$$$\mathrm{right}\:\mathrm{since}\:\mathrm{I}\:\mathrm{have}\:\mathrm{their}\:\mathrm{duplicate}\:\mathrm{workings}. \\ $$$$\mathrm{I}\:\mathrm{only}\:\mathrm{doubt}\:\mathrm{their}\:\mathrm{workings}. \\ $$$$ \\ $$$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir} \\ $$

Commented by mr W last updated on 03/Apr/25