Question

A block of mass $M=4\text{kg}$ is kept on a smooth horizontal plane. A bar of mass $m=1\text{kg}$ is kept on it. They are connected to a spring as shown and the spring is compressed. Then, what is the maximum compression in the spring for which the bar will not slip on the block when released, if coefficient of friction between them is $0.2$ and spring constant $=1000\text{N/m}$ (Take $g=10{\text{m/s}}^2$)

• A. $1\text{cm}$
• B. $1\text{m}$
• C. $0.1\text{m}$
• D. $10\text{cm}$

Answer 1

The correct option is A $1\text{cm}$

For maximum elongation in spring, static friction will act at $f_{\text{max}}$, so that both blocks move together and there is no relative motion between the blocks.


Considering $M+m$ as a system:
Common acceleration
$a=\frac{F_{\text{spring}}}{M+m}=\frac{K{x}_{max}}{M+m}$
$\Rightarrow a=\frac{1000x_{max}}{5}=200x_{max}$ ... (i)

From FBD of $m:$


$\Rightarrow N=mg=10\text{N}$ ... (ii)
$f_{\text{max}}=\mu N=0.2\times 10=2\text{N}$

Applying Newton's 2nd law in the direction of acceleration:
$f_{\text{max}}=ma$ ... (iii)

From Eq. (i) and (iii)
$2=\left(m\right)\left(200x_{max}\right)$
$\Rightarrow 2=1\times 200x_{max}$
$\therefore x_{max}=\frac{2}{200}=\frac{1}{100}\text{m}$
$=1\text{cm}$