Question

Block $A$ of mass $M$ in the figure is released from rest when the extension in the spring is $x_0$. The maximum downward displacement of the block is (assume $x_0<Mg/K$).

• A. $\frac{Mg}{2K}+x_0$
• B. $\frac{2Mg}{K}+x_0$
• C. $\frac{2Mg}{K}-x_0$
• D. $2(\frac{MG}{k}-x_0)$

Answer 1

The correct option is D $2(\frac{MG}{k}-x_0)$

As we know, according to work energy theorem, as per the given diagram,
Loss in gravitational potential energy of block $=$ gain in elastic potential energy of spring
$Mgx=\frac{1}{2}k(x+x_0{)}^2-\frac{1}{2}k{x}_0^2$
$Mgx=\frac{1}{2}k(x^2+x_0^2+2x{x}_0)-\frac{1}{2}k{x}_0^2$
$Mgx=\frac{1}{2}k{x}_0(x_0+2x)$
Hence, value of $x$
$x=2(\frac{Mg}{k}-x_0)$

Final Answer: (a)