Question

Two blocks of mass $1\text{kg and 2 kg}$ are connected by a vertical spring of spring constant $k=2000\text{N/m}$ such that the $2\text{kg}$ block is resting on the floor (assume frictionless). The upper block of mass $2\text{kg}$ is pushed down by a distance $x$ and released, and it executes $SHM$. What is the maximum value of $x$ so that the $2\text{kg}$ block does not leave the floor?

• A. $5\text{mm}$
• B. $10\text{mm}$
• C. $15\text{mm}$
• D. none of the above

Answer 1

The correct option is B $10\text{mm}$


If $x_0$ is the compreesion in the spring at equilibrium position, then

$k{x}_0=mg$
$x_0=\frac{1\times 10}{2000}=\frac{1}{2}$

Now extension at the top most position in spring will be
$x^{\prime }=(x-\frac{1}{2})cm$

So as we can understand from the diagram, block of mass $2kg$ will leave the floor when $N=0$.
So,
$k{x}^{\prime }=m^{\prime }g$
$2000\times \frac{x-\frac{1}{2}}{100}=2\times 10$
$x=1.5cm=15mm$