Question Number 27899 by Tinkutara last updated on 16/Jan/18
Answered by mrW2 last updated on 17/Jan/18
$${let}\:{e}={max}.\:{elongation}\:{of}\:{spring}\:{after}\:{release} \\ $$$${ke}=\mu{m}_{{A}} {g}\:\:\:\:\:...\left({i}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{kd}^{\mathrm{2}} −\mu{m}_{{B}} {g}\left({d}+{e}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ke}^{\mathrm{2}} \:\:\:\:...\left({ii}\right) \\ $$$${kd}^{\mathrm{2}} −\mathrm{2}\mu{m}_{{B}} {gd}−\left(\mathrm{2}\mu{m}_{{B}} {g}+{ke}\right){e}=\mathrm{0} \\ $$$${kd}^{\mathrm{2}} −\mathrm{2}\mu{m}_{{B}} {gd}−\frac{{m}_{{A}} \left(\mathrm{2}{m}_{{B}} +{m}_{{A}} \right)\left(\mu{g}\right)^{\mathrm{2}} }{{k}}=\mathrm{0} \\ $$$$\Rightarrow{d}=\frac{\mathrm{2}\mu{m}_{{B}} {g}+\sqrt{\left(\mathrm{2}\mu{gm}_{{B}} \right)^{\mathrm{2}} +\mathrm{4}{m}_{{A}} \left(\mathrm{2}{m}_{{B}} +{m}_{{A}} \right)\left(\mu{g}\right)^{\mathrm{2}} }}{\mathrm{2}{k}} \\ $$$$\Rightarrow{d}=\frac{\mu{g}\left({m}_{{A}} +\mathrm{2}{m}_{{B}} \right)}{{k}} \\ $$$$\Rightarrow{d}=\frac{\mathrm{0}.\mathrm{5}×\mathrm{10}\left(\mathrm{2}+\mathrm{2}×\mathrm{4}\right)}{\mathrm{50}}=\mathrm{1}\:{m} \\ $$
Commented by Tinkutara last updated on 16/Jan/18
Commented by Tinkutara last updated on 16/Jan/18
This is the solution in book. Can you please explain this? And why your method is wrong? This question was integer type so they approximated it to 1 m.
Commented by mrW2 last updated on 17/Jan/18
$${When}\:{block}\:{A}\:{is}\:{pressed}\:{against}\:{the} \\ $$$${wall},\:{the}\:{contact}\:{force}\:{between}\:{block} \\ $$$${A}\:{and}\:{the}\:{wall}\:{is}\:{the}\:{push}\:{force}\:{in} \\ $$$${spring},\:{which}\:{is}\:{kd}. \\ $$$${When}\:{block}\:{B}\:{is}\:{released},\:{the}\:{contact} \\ $$$${force}\:{btw}.\:{A}\:{and}\:{wall}\:{will}\:{be}\:{reduced} \\ $$$${till}\:{zero}\:\left({say}\:{at}\:{instant}\:{t}_{\mathrm{1}} \right),\:{then}\:{due}\:{to} \\ $$$${tension}\:{force}\:{in}\:{spring}\:{the}\:{friction}\:{force} \\ $$$${btw}.\:{A}\:{and}\:{floor}\:{increases}\:{from}\:{zero}\:{till}\:\mu{m}_{{A}} {g}\: \\ $$$$\left({say}\:{at}\:{instant}\:{t}_{\mathrm{2}} \right),\:{after}\:{that}\:{the}\:{block}\:{A} \\ $$$${begins}\:{to}\:{move}. \\ $$$$ \\ $$$${The}\:{answer}\:{depends}\:{very}\:{much}\:{on}\:{how} \\ $$$${you}\:{understand}\:``{block}\:{A}\:{loses}\:{contact} \\ $$$${to}\:{wall}''. \\ $$$$ \\ $$$${My}\:{answer}\:{is}\:{correct},\:{if}\:{man} \\ $$$${understands}\:{the}\:{lose}\:{of}\:{contact}\:{as}\:{the}\:{instant}\:{when} \\ $$$${block}\:{A}\:{is}\:{about}\:{to}\:{move}\:{away}\:{from}\:{wall}, \\ $$$${i}.{e}.\:{at}\:{instant}\:{t}_{\mathrm{2}} ,\:{when}\:{the}\:{tension}\:{force} \\ $$$${in}\:{spring}\:{just}\:{overcomes}\:{the}\:{static}\:{friction} \\ $$$${between}\:{block}\:{A}\:{and}\:{floor},\:{i}.{e}.\: \\ $$$${ke}=\mu{m}_{{A}} {g}. \\ $$$$ \\ $$$${The}\:{answer}\:{in}\:{book}\:{is}\:{correct},\:{if}\:{man} \\ $$$${understands}\:{the}\:{lose}\:{of}\:{contact}\:{as}\:{the}\:{instant}\:{when} \\ $$$${the}\:{contact}\:{force}\:{to}\:{the}\:{wall}\:{becomes}\:{zero}, \\ $$$${i}.{e}.\:{at}\:{instant}\:{t}_{\mathrm{1}} .\:{In}\:{this}\:{case}\:{the}\:{block} \\ $$$${B}\:{may}\:{move}\:{max}.\:\:{to}\:{a}\:{position}\:{where} \\ $$$${the}\:{spring}\:{has}\:{no}\:{elongation}\:\left({e}=\mathrm{0}\right)\:{so}\:{that} \\ $$$${the}\:{spring}\:{force}\:\left(\:{and}\:{therefore}\:{the}\:{contact}\:{force}\:{to}\:\right. \\ $$$$\left.{wall}\right)\:{is}\:{zero}.\:{The}\:{energy}\:{stored}\:{in}\:{the} \\ $$$${compressed}\:{spring}\:{is}\:{just}\:{enough}\:{to}\:{compensate} \\ $$$${the}\:{lose}\:{due}\:{to}\:{friction}\:{during}\:{the}\:{movement}\:{of}\:{block}\:{B}\:{in}\:{a}\:{distance}\:{d},\:{i}.{e}. \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{kd}^{\mathrm{2}} =\mu{m}_{{B}} {gd} \\ $$$${This}\:{is}\:{the}\:{solution}\:{given}\:{in}\:{book}. \\ $$$${You}\:{can}\:{also}\:{get}\:{this}\:{solution}\:{if}\:{you}\:{put} \\ $$$${ke}=\mathrm{0}\:{instead}\:{of}\:{ke}=\mu{m}_{{A}} {g}\:\:{in}\:{my}\:{eqn}.\:\left({i}\right). \\ $$$$ \\ $$$${I}\:{don}'{t}\:{agree}\:{that}\:{the}\:{book}'{s} \\ $$$${answer}\:{is}\:{the}\:{only}\:{correct}\:{answer}.\: \\ $$$${Because}\:{it}\:{is}\:{not}\:{clearly}\:{defined}\:{what} \\ $$$${is}\:{meant}\:{with}\:``{block}\:{A}\:{loses}\:{contact} \\ $$$${to}\:{wall}''. \\ $$
Commented by Tinkutara last updated on 17/Jan/18
Thank you very much Sir! I got the answer.
Answered by ajfour last updated on 17/Jan/18
$${ke}=\mu{m}_{{A}} {g}\:\:\:\:....\left({i}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{kd}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{ke}^{\mathrm{2}} +\mu{m}_{{B}} {g}\left({d}+{e}\right)\:\:\:...\left({ii}\right) \\ $$$${from}\:\left({i}\right):\:\:\:\:\:\:\:\:{e}=\frac{\mathrm{0}.\mathrm{5}×\mathrm{2}×\mathrm{10}}{\mathrm{50}}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$${replacing}\:{in}\:\left({ii}\right):\:\:{e}=\frac{\mathrm{1}}{\mathrm{5}},\:{k}=\mathrm{50} \\ $$$${m}_{{A}} =\mathrm{2}\:,\:{m}_{{B}} =\mathrm{4} \\ $$$$\mathrm{25}{d}^{\mathrm{2}} =\mathrm{1}+\mathrm{20}\left({d}+\frac{\mathrm{1}}{\mathrm{5}}\right) \\ $$$$\Rightarrow\:\:\mathrm{5}{d}^{\mathrm{2}} −\mathrm{4}{d}−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\mathrm{5}\left({d}−\frac{\mathrm{2}}{\mathrm{5}}\right)^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{5}} \\ $$$${d}=\frac{\mathrm{2}}{\mathrm{5}}+\frac{\mathrm{3}}{\mathrm{5}}\:=\:\mathrm{1}\:. \\ $$
Commented by Tinkutara last updated on 17/Jan/18
Thank you very much Sir! I got the answer.