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Question Number 53330 by Kunal12588 last updated on 20/Jan/19

Commented by Kunal12588 last updated on 20/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 20/Jan/19

F×S=(1/2)(2m)×V_(2m) ^2 +μ(2mg)s  F×S=(1/2)(m)×V_m ^2 +μ(mg)s  μ=fricion coefficient between ice and boat  (K.E)_(2m) +μ(2mg)s=(K.E)_m +μmgs  μ×2mgs−μmgs=(K.E)_m −(K.E)_(2m)   so (K.E)_m >(K.E)_(2m)   pls check is it correct...if not pls comment with  answer...

$${F}×{S}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{m}\right)×{V}_{\mathrm{2}{m}} ^{\mathrm{2}} +\mu\left(\mathrm{2}{mg}\right){s} \\ $$$${F}×{S}=\frac{\mathrm{1}}{\mathrm{2}}\left({m}\right)×{V}_{{m}} ^{\mathrm{2}} +\mu\left({mg}\right){s} \\ $$$$\mu={fricion}\:{coefficient}\:{between}\:{ice}\:{and}\:{boat} \\ $$$$\left({K}.{E}\right)_{\mathrm{2}{m}} +\mu\left(\mathrm{2}{mg}\right){s}=\left({K}.{E}\right)_{{m}} +\mu{mgs} \\ $$$$\mu×\mathrm{2}{mgs}−\mu{mgs}=\left({K}.{E}\right)_{{m}} −\left({K}.{E}\right)_{\mathrm{2}{m}} \\ $$$${so}\:\left({K}.{E}\right)_{{m}} >\left({K}.{E}\right)_{\mathrm{2}{m}} \\ $$$${pls}\:{check}\:{is}\:{it}\:{correct}...{if}\:{not}\:{pls}\:{comment}\:{with} \\ $$$${answer}... \\ $$

Commented by Kunal12588 last updated on 21/Jan/19

I think what you are doing is correct but the question ask something else.(great conceptual question)

Commented by Kunal12588 last updated on 21/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 21/Jan/19

thank you for answer...but  i thought the frictuon  between ice and boat..which cause the ice to melt  localised thus help to move forward.  if friction ignored then   △K.E=work done  (1/2)×2m×v_(2m ) ^2 −0=F×s for boat 2m mass  (1/2)×m×v_m ^2 −0=F×S  for boat m mass  thus  (K.E)_(2m) ^(final) =(K.E)_(m ) ^(final)

$${thank}\:{you}\:{for}\:{answer}...{but}\:\:{i}\:{thought}\:{the}\:{frictuon} \\ $$$${between}\:{ice}\:{and}\:{boat}..{which}\:{cause}\:{the}\:{ice}\:{to}\:{melt} \\ $$$${localised}\:{thus}\:{help}\:{to}\:{move}\:{forward}. \\ $$$${if}\:{friction}\:{ignored}\:{then}\: \\ $$$$\bigtriangleup{K}.{E}={work}\:{done} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}{m}×{v}_{\mathrm{2}{m}\:} ^{\mathrm{2}} −\mathrm{0}={F}×{s}\:{for}\:{boat}\:\mathrm{2}{m}\:{mass} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}×{m}×{v}_{{m}} ^{\mathrm{2}} −\mathrm{0}={F}×{S}\:\:{for}\:{boat}\:{m}\:{mass} \\ $$$${thus} \\ $$$$\left({K}.{E}\right)_{\mathrm{2}{m}} ^{{final}} =\left({K}.{E}\right)_{{m}\:} ^{{final}} \\ $$$$ \\ $$

Commented by Kunal12588 last updated on 21/Jan/19

yeah I was thinking why are you adding the frictional force

Commented by tanmay.chaudhury50@gmail.com last updated on 21/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 21/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 21/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 21/Jan/19

Commented by Kunal12588 last updated on 21/Jan/19

Thank you very much sir

Commented by tanmay.chaudhury50@gmail.com last updated on 21/Jan/19

which book you read...i think at present days  physics galaxy books by Ashsish Arora is excellent

$${which}\:{book}\:{you}\:{read}...{i}\:{think}\:{at}\:{present}\:{days} \\ $$$${physics}\:{galaxy}\:{books}\:{by}\:{Ashsish}\:{Arora}\:{is}\:{excellent} \\ $$

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