Gujarati
Hindi
4-1.Newton's Laws of Motion
medium

A block of mass $M_1$ is hanged by a light spring of force constant $k$ to the top bar of a reverse $U$ -frame of mass $M_2$ on the floor. The block is pulled down from its equilibrium position by a distance $x$ and then released. The minimum value of $x$ such that the reverse $U$ -frame will leave the floor momentarily, is :-

A

$x = (M_1 + M_2)g/k$

B

$x = (2M_1 + M_2)g/k$

C

$x = (M_1 + 2M_2)g/k$

D

$x = M_1g/k$

Solution

$2 \mathrm{N}+\mathrm{kx}_{1}=\mathrm{M}_{2} \mathrm{g}$

$\mathrm{N}=0 \Rightarrow \mathrm{kx}_{1}=\mathrm{M}_{2} \mathrm{g}$

$A=x_{0}+x_{1}=x$

$\frac{\mathrm{M}_{1} \mathrm{g}}{\mathrm{k}}+\frac{\mathrm{M}_{2} \mathrm{g}}{\mathrm{k}}=\mathrm{x}$

Standard 11
Physics

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