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Question-26924




Question Number 26924 by Tinkutara last updated on 31/Dec/17
Answered by mrW1 last updated on 31/Dec/17
k=spring constant  u=deformation of spring under force  u_2 =max. deformation of spring after release  (1/2)ku^2 −m_A gu=(1/2)ku_2 ^2 +m_A gu_2    ...(i)  ku_2 =m_B g  ⇒u_2 =((m_B g)/k)=((mg)/k)  (i) becomes  ku^2 −2mgu=((3(mg)^2 )/k)  ⇒k^2 u^2 −2mgku−3(mg)^2 =0  ⇒(ku−3mg)(ku+mg)=0  ⇒ku=3mg    F+m_A g=ku=3mg  ⇒F=2mg    Answer (2)
$${k}={spring}\:{constant} \\ $$$${u}={deformation}\:{of}\:{spring}\:{under}\:{force} \\ $$$${u}_{\mathrm{2}} ={max}.\:{deformation}\:{of}\:{spring}\:{after}\:{release} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{ku}^{\mathrm{2}} −{m}_{{A}} {gu}=\frac{\mathrm{1}}{\mathrm{2}}{ku}_{\mathrm{2}} ^{\mathrm{2}} +{m}_{{A}} {gu}_{\mathrm{2}} \:\:\:…\left({i}\right) \\ $$$${ku}_{\mathrm{2}} ={m}_{{B}} {g} \\ $$$$\Rightarrow{u}_{\mathrm{2}} =\frac{{m}_{{B}} {g}}{{k}}=\frac{{mg}}{{k}} \\ $$$$\left({i}\right)\:{becomes} \\ $$$${ku}^{\mathrm{2}} −\mathrm{2}{mgu}=\frac{\mathrm{3}\left({mg}\right)^{\mathrm{2}} }{{k}} \\ $$$$\Rightarrow{k}^{\mathrm{2}} {u}^{\mathrm{2}} −\mathrm{2}{mgku}−\mathrm{3}\left({mg}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\left({ku}−\mathrm{3}{mg}\right)\left({ku}+{mg}\right)=\mathrm{0} \\ $$$$\Rightarrow{ku}=\mathrm{3}{mg} \\ $$$$ \\ $$$${F}+{m}_{{A}} {g}={ku}=\mathrm{3}{mg} \\ $$$$\Rightarrow{F}=\mathrm{2}{mg} \\ $$$$ \\ $$$${Answer}\:\left(\mathrm{2}\right) \\ $$
Commented by mrW1 last updated on 31/Dec/17
Commented by Tinkutara last updated on 31/Dec/17
Thank you very much Sir! I got the answer.

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