Question

For the situation shown in the given figure, the horizontal surface is rough. The coefficient of friction between the block of mass $m_1=4kg$ and the horizontal surface is $\mu =0.5$. The blocks are connected with combination of light spring of spring constant $k$ and a light string. The block of mass $m_2$ is released from rest, when spring is in its natural length. Find minimum value of $m_2$ in $kg$ such that the block of mass $m_1$ will start moving.

Answer 1

Answer: 1

Let for just shifting of the block of mass $m_1$, the elongation in spring is $x_1$
$\therefore k{x}_1>\mu m_{1}g$
$\because x_{1\min }=\frac{\mu m_{1}g}{k}$
For minimum value of $m_2$, the speed of block of mass $m_2$ is zero when the elongation in spring is $x_{1\min }$.
According to the conservation principle of mechanical energy, Loss in gravitational potential energy of the block mass $m_2$ is equal to gain in elastic potential energy in the spring.
$\therefore m_{2}g{x}_{1\min }=\frac{1}{2}k{x}_{\min }^2$
$\therefore x_{1\min }=\frac{2m_{2}g}{k}$
Or $\frac{\mu m_{1}g}{k}=\frac{2m_{2}g}{k}$
$m_2=\frac{\mu m_1}{2}=\frac{0.5\times 4}{2}=1kg$