Question

One end of a light spring of spring constant $200\text{N/m}$ is fixed to a block $\text{A}$ of mass $2\text{kg}$ placed on horizontal frictionless table. The other end of the spring is fixed to a wall. A smaller block $\text{B}$ of mass $3\text{kg}$ is placed on block $\text{A}$. The system is displaced by a small amount and released. What is the magnitude of friction coefficient $\mu$ at the surface of block $\text{A}$ and $\text{B}$ so that the upper block does not slip over lower block $?$
(Amplitude $\left(\text{A}\right)=0.1\text{m})$

• A. $\mu =0.7$
• B. $\mu =0.5$
• C. $\mu =0.66$
• D. $\mu =0.25$

Answer 1

The correct option is C $\mu =0.66$

Given:
$K=200\text{N/m};m_1=2\text{kg};m_2=3\text{kg}$

The angular frequency of the system is,

$\omega =\sqrt{\frac{K}{m_1+m_2}}$


The upper block of mass $3\text{kg}$ will not slip over the lower block if the maximum force on the upper block does not exceed the limiting frictional force $\mu mg$.

$\text{FBD}$ of upper block at extreme right


$\text{FBD}$ of lower block at extreme right,


As at extreme position, the velocity of blocks is zero, the net force on lower block must be zero

$\therefore f=KA=\left(200\right)\left(0.1\right)=20\text{N}$

For upper block

$f=\text{Limiting friction force}=\mu m_{2}g$

$\therefore \mu =\frac{20}{m_{2}g}=\frac{20}{3\times 10}=\frac{2}{3}=0.666$

Hence, option $\left(\text{C}\right)$ is correct.