Question

For the situation shown in the given figure, the horizontal surface is rough. The coefficient of friction between the block of mass $m_{1}=4 \,kg$ 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

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{2 m_{2} g}{k}$
Or $ \frac{\mu m_{1} g}{k}=\frac{2 m_{2} g}{k} $
$ m_{2}=\frac{\mu m_{1}}{2}=\frac{0.5 \times 4}{2}=1 \,kg $