Question

Find at what values of $\Delta l$, the initial compression of the spring, the lower cube will bounce up after the thread has been burned through.

• A. $\Delta l<\frac{3mg}{\kappa }$
• B. $\Delta l>\frac{3mg}{\kappa }$
• C. $\Delta l>=\frac{3mg}{\kappa }$
• D. $\Delta l<=\frac{3mg}{\kappa }$

Answer 1

The correct option is C $\Delta l>=\frac{3mg}{\kappa }$

Let x be the length above mean it extends so that just lower block can be lifted.

$\Rightarrow kx=mg$

$\Rightarrow x=\frac{mg}{k}$--------->this is the least value of $x$.

Now apply energy equation to the syatem,

$\frac{1}{2}k(\Delta l{)}^2>=mg(\Delta l+x)+\frac{1}{2}k(x^2)$

$\frac{1}{2}k(\Delta l{)}^2>=mg(\Delta l)+(mg{)}^2[\frac{1}{2k}+\frac{1}{k}]$

$k(\Delta l{)}^2-mg(\Delta l)-\frac{3(mg{)}^2}{k}$>=0

$\Delta l>=\frac{3mg}{k}$