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Question-27899

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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=μm_A g ...(i) (1/2)kd^2 −μm_B g(d+e)=(1/2)ke^2 ...(ii) kd^2 −2μm_B gd−(2μm_B g+ke)e=0 kd^2 −2μm_B gd−((m_A (2m_B +m_A )(μg)^2 )/k)=0 ⇒d=((2μm_B g+(√((2μgm_B )^2 +4m_A (2m_B +m_A )(μg)^2 )))/(2k)) ⇒d=((μg(m_A +2m_B ))/k) ⇒d=((0.5×10(2+2×4))/(50))=1 m
$${let}\:{e}={max}.\:{elongation}\:{of}\:{spring}\:{after}\:{release} \\ $$$${ke}=\mu{m}_{{A}} {g}\:\:\:\:\:…\left({i}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{kd}^{\mathrm{2}} −\mu{m}_{{B}} {g}\left({d}+{e}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ke}^{\mathrm{2}} \:\:\:\:…\left({ii}\right) \\ $$$${kd}^{\mathrm{2}} −\mathrm{2}\mu{m}_{{B}} {gd}−\left(\mathrm{2}\mu{m}_{{B}} {g}+{ke}\right){e}=\mathrm{0} \\ $$$${kd}^{\mathrm{2}} −\mathrm{2}\mu{m}_{{B}} {gd}−\frac{{m}_{{A}} \left(\mathrm{2}{m}_{{B}} +{m}_{{A}} \right)\left(\mu{g}\right)^{\mathrm{2}} }{{k}}=\mathrm{0} \\ $$$$\Rightarrow{d}=\frac{\mathrm{2}\mu{m}_{{B}} {g}+\sqrt{\left(\mathrm{2}\mu{gm}_{{B}} \right)^{\mathrm{2}} +\mathrm{4}{m}_{{A}} \left(\mathrm{2}{m}_{{B}} +{m}_{{A}} \right)\left(\mu{g}\right)^{\mathrm{2}} }}{\mathrm{2}{k}} \\ $$$$\Rightarrow{d}=\frac{\mu{g}\left({m}_{{A}} +\mathrm{2}{m}_{{B}} \right)}{{k}} \\ $$$$\Rightarrow{d}=\frac{\mathrm{0}.\mathrm{5}×\mathrm{10}\left(\mathrm{2}+\mathrm{2}×\mathrm{4}\right)}{\mathrm{50}}=\mathrm{1}\:{m} \\ $$
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 (say at instant t_1 ), then due to tension force in spring the friction force btw. A and floor increases from zero till μm_A g (say at instant t_2 ), 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_2 , when the tension force in spring just overcomes the static friction between block A and floor, i.e. ke=μm_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_1 . In this case the block B may move max. to a position where the spring has no elongation (e=0) so that the spring force ( and therefore the contact force to wall) 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. (1/2)kd^2 =μm_B gd This is the solution given in book. You can also get this solution if you put ke=0 instead of ke=μm_A g in my eqn. (i). 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”.
$${When}\:{block}\:{A}\:{is}\:{pressed}\:{against}\:{the} \\ $$$${wall},\:{the}\:{contact}\:{force}\:{between}\:{block} \\ $$$${A}\:{and}\:{the}\:{wall}\:{is}\:{the}\:{push}\:{force}\:{in} \\ $$$${spring},\:{which}\:{is}\:{kd}. \\ $$$${When}\:{block}\:{B}\:{is}\:{released},\:{the}\:{contact} \\ $$$${force}\:{btw}.\:{A}\:{and}\:{wall}\:{will}\:{be}\:{reduced} \\ $$$${till}\:{zero}\:\left({say}\:{at}\:{instant}\:{t}_{\mathrm{1}} \right),\:{then}\:{due}\:{to} \\ $$$${tension}\:{force}\:{in}\:{spring}\:{the}\:{friction}\:{force} \\ $$$${btw}.\:{A}\:{and}\:{floor}\:{increases}\:{from}\:{zero}\:{till}\:\mu{m}_{{A}} {g}\: \\ $$$$\left({say}\:{at}\:{instant}\:{t}_{\mathrm{2}} \right),\:{after}\:{that}\:{the}\:{block}\:{A} \\ $$$${begins}\:{to}\:{move}. \\ $$$$ \\ $$$${The}\:{answer}\:{depends}\:{very}\:{much}\:{on}\:{how} \\ $$$${you}\:{understand}\:“{block}\:{A}\:{loses}\:{contact} \\ $$$${to}\:{wall}''. \\ $$$$ \\ $$$${My}\:{answer}\:{is}\:{correct},\:{if}\:{man} \\ $$$${understands}\:{the}\:{lose}\:{of}\:{contact}\:{as}\:{the}\:{instant}\:{when} \\ $$$${block}\:{A}\:{is}\:{about}\:{to}\:{move}\:{away}\:{from}\:{wall}, \\ $$$${i}.{e}.\:{at}\:{instant}\:{t}_{\mathrm{2}} ,\:{when}\:{the}\:{tension}\:{force} \\ $$$${in}\:{spring}\:{just}\:{overcomes}\:{the}\:{static}\:{friction} \\ $$$${between}\:{block}\:{A}\:{and}\:{floor},\:{i}.{e}.\: \\ $$$${ke}=\mu{m}_{{A}} {g}. \\ $$$$ \\ $$$${The}\:{answer}\:{in}\:{book}\:{is}\:{correct},\:{if}\:{man} \\ $$$${understands}\:{the}\:{lose}\:{of}\:{contact}\:{as}\:{the}\:{instant}\:{when} \\ $$$${the}\:{contact}\:{force}\:{to}\:{the}\:{wall}\:{becomes}\:{zero}, \\ $$$${i}.{e}.\:{at}\:{instant}\:{t}_{\mathrm{1}} .\:{In}\:{this}\:{case}\:{the}\:{block} \\ $$$${B}\:{may}\:{move}\:{max}.\:\:{to}\:{a}\:{position}\:{where} \\ $$$${the}\:{spring}\:{has}\:{no}\:{elongation}\:\left({e}=\mathrm{0}\right)\:{so}\:{that} \\ $$$${the}\:{spring}\:{force}\:\left(\:{and}\:{therefore}\:{the}\:{contact}\:{force}\:{to}\:\right. \\ $$$$\left.{wall}\right)\:{is}\:{zero}.\:{The}\:{energy}\:{stored}\:{in}\:{the} \\ $$$${compressed}\:{spring}\:{is}\:{just}\:{enough}\:{to}\:{compensate} \\ $$$${the}\:{lose}\:{due}\:{to}\:{friction}\:{during}\:{the}\:{movement}\:{of}\:{block}\:{B}\:{in}\:{a}\:{distance}\:{d},\:{i}.{e}. \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{kd}^{\mathrm{2}} =\mu{m}_{{B}} {gd} \\ $$$${This}\:{is}\:{the}\:{solution}\:{given}\:{in}\:{book}. \\ $$$${You}\:{can}\:{also}\:{get}\:{this}\:{solution}\:{if}\:{you}\:{put} \\ $$$${ke}=\mathrm{0}\:{instead}\:{of}\:{ke}=\mu{m}_{{A}} {g}\:\:{in}\:{my}\:{eqn}.\:\left({i}\right). \\ $$$$ \\ $$$${I}\:{don}'{t}\:{agree}\:{that}\:{the}\:{book}'{s} \\ $$$${answer}\:{is}\:{the}\:{only}\:{correct}\:{answer}.\: \\ $$$${Because}\:{it}\:{is}\:{not}\:{clearly}\:{defined}\:{what} \\ $$$${is}\:{meant}\:{with}\:“{block}\:{A}\:{loses}\:{contact} \\ $$$${to}\:{wall}''. \\ $$
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=μm_A g ....(i) (1/2)kd^2 =(1/2)ke^2 +μm_B g(d+e) ...(ii) from (i): e=((0.5×2×10)/(50))=(1/5) replacing in (ii): e=(1/5), k=50 m_A =2 , m_B =4 25d^2 =1+20(d+(1/5)) ⇒ 5d^2 −4d−1=0 5(d−(2/5))^2 =(9/5) d=(2/5)+(3/5) = 1 .
$${ke}=\mu{m}_{{A}} {g}\:\:\:\:….\left({i}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{kd}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{ke}^{\mathrm{2}} +\mu{m}_{{B}} {g}\left({d}+{e}\right)\:\:\:…\left({ii}\right) \\ $$$${from}\:\left({i}\right):\:\:\:\:\:\:\:\:{e}=\frac{\mathrm{0}.\mathrm{5}×\mathrm{2}×\mathrm{10}}{\mathrm{50}}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$${replacing}\:{in}\:\left({ii}\right):\:\:{e}=\frac{\mathrm{1}}{\mathrm{5}},\:{k}=\mathrm{50} \\ $$$${m}_{{A}} =\mathrm{2}\:,\:{m}_{{B}} =\mathrm{4} \\ $$$$\mathrm{25}{d}^{\mathrm{2}} =\mathrm{1}+\mathrm{20}\left({d}+\frac{\mathrm{1}}{\mathrm{5}}\right) \\ $$$$\Rightarrow\:\:\mathrm{5}{d}^{\mathrm{2}} −\mathrm{4}{d}−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\mathrm{5}\left({d}−\frac{\mathrm{2}}{\mathrm{5}}\right)^{\mathrm{2}} =\frac{\mathrm{9}}{\mathrm{5}} \\ $$$${d}=\frac{\mathrm{2}}{\mathrm{5}}+\frac{\mathrm{3}}{\mathrm{5}}\:=\:\mathrm{1}\:. \\ $$
Commented by Tinkutara last updated on 17/Jan/18
Thank you very much Sir! I got the answer.

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