Question

A block of mass $2\text{kg}$ is attached to a spring of force constant $k=10\text{N/m}$ as shown in the figure. Find the range in which the block can be kept without slipping when the block is pulled or pushed towards the spring (spring is elongated or compressed). Take $g=10{\text{m/s}}^2$.

• A. $3.2\text{m}$
• B. $1.6\text{m}$
• C. $2.4\text{m}$
• D. $4.8\text{m}$

Answer 1

The correct option is A $3.2\text{m}$

When the spring is compressed:
By say $x_1$, from FBD we get (here, $R$ is the normal force)

Where $x_1$ is compression
$x_1=|x_0-x_0^{\prime }|$
To keep the block without slipping i.e., keeping it in equilibrium using
$\sum F_x=0$
$\Rightarrow f_s=k{x}_1$
and, $\sum F_y=0$
$\Rightarrow R=2g$
$\because f_s={\mu }_{s}R=0.8\times 2\times 10=16\text{N}$
$x_1=\frac{f_s}{k}=\frac{16}{10}=1.6\text{m}$

When the spring is elongated:
By say $x_2$, from FBD we get

Where $x_2$ is elongation
$x_2=|x_0^{\prime \prime }-x_0|$
At equilibrium,
$\sum F_x=0$
$\Rightarrow k{x}_2=f_s$
and, $\sum F_y=0$
$\Rightarrow R=mg$
$\because f_s={\mu }_{s}R=0.8\times 2g=16N$
$x_2=\frac{f_s}{k}=\frac{16}{10}=1.6\text{m}$
$\because$
Range in which block can be kept without slipping is
$x_1+x_2=1.6+1.6=3.2\text{m}$