Question

6. Two masses m, and m, are joined by a spring as
shown. The system is dropped to the ground from
a certain height. The spring will be :-​

Answer 1

Let the system move downward with an acceleration $\sf{a.}$

Let the spring extends by $\sf{x}$ during the motion.

Let the spring constant be $\sf{k}$ so the restoring force developed in it is $\sf{kx.}$

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\put(0,15){\framebox(7.5,7.5){ \sf{m_1} }}\put(0,-15){\framebox(7.5,7.5){ \sf{m_2} }}\multiput(3.75,15)(0,-17.5){2}{\line(0,-1){5}}\multiput(3.75,10)(0,-1.5){7}{\qbezier(0,0)(4,-1.5)(0,-2.5)\qbezier(0,-2.5)(-4,-1.5)(0,-1.5)}\put(3.75,-0.4){\qbezier(0,0)(4,-1.5)(0,-2.5)}\put(30,0){\framebox(7.5,7.5){ \sf{m_1} }}\put(60,0){\framebox(7.5,7.5){ \sf{m_2} }}\multiput(33.75,0)(30,0){2}{\vector(0,-1){10}}\put(25.3,-14){ \sf{m_1g+kx} }\put(60.5,-14){ \sf{m_2g} }\put(61.9,19.6){ \sf{kx} }\put(63.75,7.5){\vector(0,1){10}}\multiput(18.2,8.75)(30,0){2}{\vector(0,-1){10}\put(1.5,-5){\sf{a}}}\end{picture}$

From free body diagram of $\sf{m_1,}$ the net force acting on it is given by,

$\longrightarrow\sf{m_1g+kx=m_1a\quad\quad\dots(1)}$

From free body diagram of $\sf{m_2,}$ the net force acting on it is given by,

$\longrightarrow\sf{m_2g-kx=m_2a\quad\quad\dots(2)}$

Adding (1) and (2),

$\longrightarrow\sf{(m_1+m_2)g=(m_1+m_2)a}$

$\longrightarrow\sf{a=g}$

So from (1),

$\longrightarrow\sf{kx=m_1a-m_1g}$

$\longrightarrow\sf{kx=m_1g-m_1g}$

$\longrightarrow\sf{kx=0}$

Since $\sf{k}$ is non - zero,

$\longrightarrow\sf{\underline{\underline{x=0}}}$

Hence the spring neither gets extended nor gets compressed.