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Question Number 40058 by rahul 19 last updated on 15/Jul/18

Answered by rahul 19 last updated on 16/Jul/18

By Energy conservation 10.10.3 + (1/2) .100.(1)^2 = (1/2). 10.v^2 ⇒ v= (√(70)) m/s What′s wrong in this ?

$$\mathrm{By}\:\mathrm{Energy}\:\mathrm{conservation} \\ $$$$\mathrm{10}.\mathrm{10}.\mathrm{3}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:.\mathrm{100}.\left(\mathrm{1}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{2}}.\:\mathrm{10}.\mathrm{v}^{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{v}=\:\sqrt{\mathrm{70}}\:\mathrm{m}/\mathrm{s} \\ $$$$\:\mathrm{What}'\mathrm{s}\:\mathrm{wrong}\:\mathrm{in}\:\mathrm{this}\:? \\ $$

Commented by rahul 19 last updated on 16/Jul/18

ok sir.

$$\mathrm{ok}\:\mathrm{sir}. \\ $$

Commented by MrW3 last updated on 15/Jul/18

your answer is correct, if there is no friction between ring and rod. but in the question there is friction, so the velocity must be less than (√(70)) m/s, since a part of energy is lost due to friction.

$${your}\:{answer}\:{is}\:{correct},\:{if}\:{there}\:{is} \\ $$$${no}\:{friction}\:{between}\:{ring}\:{and}\:{rod}. \\ $$$${but}\:{in}\:{the}\:{question}\:{there}\:{is}\:{friction}, \\ $$$${so}\:{the}\:{velocity}\:{must}\:{be}\:{less}\:{than}\:\sqrt{\mathrm{70}}\:{m}/{s}, \\ $$$${since}\:{a}\:{part}\:{of}\:{energy}\:{is}\:{lost}\:{due}\:{to} \\ $$$${friction}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jul/18

mgh+(1/2)k(x_2 ^2 −x_1 ^2 )=(1/2)mv^2 10.10.3+(1/2).100.(5^2 −4^2 )=(1/2).10.v^2 300+450=5v^2 v=(√(150)) =12.25

$${mgh}+\frac{\mathrm{1}}{\mathrm{2}}{k}\left({x}_{\mathrm{2}} ^{\mathrm{2}} −{x}_{\mathrm{1}} ^{\mathrm{2}} \right)=\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} \\ $$$$\mathrm{10}.\mathrm{10}.\mathrm{3}+\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{100}.\left(\mathrm{5}^{\mathrm{2}} −\mathrm{4}^{\mathrm{2}} \right)=\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{10}.{v}^{\mathrm{2}} \\ $$$$\mathrm{300}+\mathrm{450}=\mathrm{5}{v}^{\mathrm{2}} \\ $$$${v}=\sqrt{\mathrm{150}}\:=\mathrm{12}.\mathrm{25} \\ $$

Commented by rahul 19 last updated on 15/Jul/18

Ans. given is 10m/s.

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jul/18

if 4m is the natural length of spring then eqn is 10.10.3+(1/2).100.(5−4)^2 =(1/2).10.v^2 v=(√(70))

$${if}\:\mathrm{4}{m}\:{is}\:{the}\:{natural}\:{length}\:{of}\:{spring}\:{then} \\ $$$${eqn}\:{is} \\ $$$$\mathrm{10}.\mathrm{10}.\mathrm{3}+\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{100}.\left(\mathrm{5}−\mathrm{4}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{10}.{v}^{\mathrm{2}} \\ $$$${v}=\sqrt{\mathrm{70}}\: \\ $$

Commented by rahul 19 last updated on 15/Jul/18

yes sir, i am also getting same ans. does it mean ans. given in book is wrong ?

$$\mathrm{yes}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{am}\:\mathrm{also}\:\mathrm{getting}\:\mathrm{same}\:\mathrm{ans}. \\ $$$$\mathrm{does}\:\mathrm{it}\:\mathrm{mean}\:\mathrm{ans}.\:\mathrm{given}\:\mathrm{in}\:\mathrm{book}\:\mathrm{is} \\ $$$$\mathrm{wrong}\:? \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jul/18

yes i think so

$${yes}\:{i}\:{think}\:{so} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jul/18

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jul/18

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