Question

The magnitude of acceleration of centre of mass of the system is 2√x m/s^2. The value of x is​

Answer 1

Answer:

Explanation:

pls find the attachment

Answer 2

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$\underline{\blacksquare\:\:\:\footnotesize{\red{\text{Figure:}}}}$

$\setlength{\unitlength}{1.5mm}\begin{picture}(30,20)\linethickness{0.08mm}\put(7,3.5){\line(1,0){22.5}}\put(10,6){\vector(1,0){4}}\put(7,3.5){\vector(1,0){15}}\put(0,3.5){\vector(-1,0){6}}\put(30.5,-16){\line(0,1){18.5}}\put(30.5,-16){\vector(0,1){10}}\put(25,0){\line(5,3){3.5}}\multiput(-10,0)(1,0){36}{\line(-2,-3){1.6}}\multiput(20,-33)(0,1){26}{\line(-1,0){2}}\multiput(20,-8)(1,1.3){3}{\line(-2,3){1.6}}\linethickness{0.5mm}\put(-10,0){\line(1,0){35}}\qbezier(25,0)(20,-8)(20,-8)\put(20,-8){\line(0,-1){25}}\put(0,0){\dashbox{0.1}(7,7)[l]}\put(27,-23){\dashbox{0.1}(7,7)[l]}\put(29.5,2.5){\circle{2}}\linethickness{0.2mm}\footnotesize{\put(-9,3.5){ F_k }}\footnotesize{\put(15,5.5){ \vec{a} }}\put(21,4.5){T}\footnotesize{\put(2.5,3){ m_1 }}\footnotesize{\put(29.5,-20){ m_2 }}\put(28,-6){T}\end{picture}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{5mm}\put(1,1){\line(1,0){6.9}}\end{picture}$

$\underline{\blacksquare\:\:\:\footnotesize{\red{\text{SolutioN:}}}}$

$\footnotesize{\bf{Let}\:,\text{ the acceleration of the arrangement is = a }}$

$\footnotesize{\text{Here , }m_1=m_2=m=5\:kg \:\:\:and\:\:\:\mu_k=0.2}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{1mm}\put(1,1){\line(1,0){6.9}}\end{picture}$

$\footnotesize{\bf{\underline{\red{\text{For the block }m_1\:\:}}}}$

$\setlength{\unitlength}{1.5mm}\begin{picture}(30,20)\linethickness{0.08mm}\put(0,0){\dashbox{0.1}(7,7)[l]}\put(7,3.5){\vector(1,0){15}}\footnotesize{\put(-9,3.5){ F_k }}\footnotesize{\put(15,5.5){ m\vec{a} }}\put(21,4.5){T}\put(0,3.5){\vector(-1,0){6}}\put(10,6){\vector(1,0){4}}\end{picture}$

$\footnotesize{\therefore\:\:ma=T-F_k}$

$\footnotesize{\implies\:ma=T-\mu_k R}$

$\footnotesize{\implies\:ma=T-\mu_k mg}$

$\footnotesize{\implies\:5a=T-0.2\times5g}$

$\footnotesize{\implies\:5a=T-g}$

$\footnotesize{\implies\:T=5a+g}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{1mm}\put(1,1){\line(1,0){6.9}}\end{picture}$

$\footnotesize{\bf{\underline{\red{\text{For the block }m_2\:\:}}}}$

$\setlength{\unitlength}{1.5mm}\begin{picture}(30,20)\linethickness{0.08mm}\put(27,-23){\dashbox{0.1}(7,7)[l]}\put(30.5,-16){\vector(0,1){5}}\put(30.5,-23){\vector(0,-1){5}}\put(37,-19){\vector(0,-1){3}}\put(28,-11){T}\put(29,-30){mg}\put(37,-24){ma}\end{picture}$

$\footnotesize{\therefore\:\:T=mg-ma}$

$\footnotesize{\implies\:\:T=m(g-a)}$

$\footnotesize{\implies\:\:T=5(g-a)}$

$\footnotesize{\implies\:\:5a+g=5g-5a}$

$\footnotesize{\implies\:\:5a+5a=5g-g}$

$\footnotesize{\implies\:\:10a=4g}$

$\footnotesize{\implies\:\:5a=2g}$

$\footnotesize{\implies\:\:5a=2\times10}$

$\footnotesize{\implies\:\:5a=20}$

$\footnotesize{\implies\:\:a=\dfrac{\cancel{20}}{\cancel5}}$

$\footnotesize{\implies\:\:a=4}$

$\footnotesize{\implies\:\:2\sqrt{x}=4}$

$\footnotesize{\implies\:\:\sqrt{x}=\dfrac{\cancel4}{\cancel2}}$

$\footnotesize{\implies\:\:\sqrt{x}=2}$

$\footnotesize{\implies\:\:x=2^2}$

$\footnotesize{\implies\:\:\bf{\red{\boxed{x=4}}}}$

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$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{3mm}\put(1,1){\line(1,0){6.9}}\end{picture}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\linethickness{5mm}\put(1,1){\line(1,0){6.9}}\end{picture}$