Question

The figure shows two blocks $\text{A}$ and $\text{B}$, each having a mass of $320\text{g}$ connected by a light string passing over a smooth light pulley. The horizontal surface on which the block $\text{A}$ can slide is smooth. The block $\text{A}$ is attached to a spring of spring constant $40\text{N/m}$ whose other end is fixed to a support $40\text{cm}$ above the horizontal surface. Initially, the spring is vertical and unstretched when the system is released to move. Find the horizontal distance covered by the block $\text{A}$ at the instant it breaks off the surface below it.
Take $g=10{\text{m/s}}^2$.

• A. $1.5\text{m/s}$
• B. $2.5\text{m/s}$
• C. $3.0\text{m/s}$
• D. $0.5\text{m/s}$

Answer 1

The correct option is A $1.5\text{m/s}$

Schematic diagram of the system at when block $\text{A}$ is breaking off the surface.

From above diagram, in $△PQR$,

$\cos \theta =\frac{0.4}{0.4+x}$

FBD of block $\text{A}$ at the instant of breaking off the surface:


Applying equation of equilibrium along vertical direction,

$Kx\cos \theta =mg(\text{no surface contact,}N=0)$

$\cos \theta =\frac{mg}{Kx}$

So, we get,

$\cos \theta =\frac{0.4}{0.4+x}=\frac{mg}{Kx}=\frac{0.32\times 10}{40x}$

$\therefore x=0.1\text{m}$

So, horizontal distance covered by block $\text{A}$ is

$s=\sqrt{(0.4+x{)}^2-(0.4)}=\sqrt{(0.4+0.5{)}^2-(0.4)}$

$\therefore s=0.3\text{m}$

Therefore, option $\left(D\right)$ is correct.