Question

A block of mass $200g$ is suspended by a vertical spring. The spring is stretched by $1.0cm$ when the block is in equilibrium. A particle of mass $120g$ is dropped on the block from a height of $45cm$. The particle sticks to the block after the impact. Find the maximum extension of the spring. Take $g=10m/s^2$.

Answer 1

Given that,

Block of mass $M=200g$

Particle of mass $m=120g$

Height $h=45cm$

We know that,

When the block is in equilibrium its all the forces are balanced

Let, $k$ is the spring constant and $x_0$ is the extension of the spring

We know that,

$k{x}_0=mg$

$k=\frac{mg}{x_0}$

$k=\frac{200\times {10}^{-3}\times 10}{{10}^{-2}}$

$k=200N/m$

Now, when the mass is dropped from some height now applying work energy theorem

Work done by all forces = change in kinetic energy of system

Initially system has K.E =0

And final K.E =0 because after maximum extension the system again comes to rest

Now,

$\Delta K.E=W_g+W_{springforce}$

$0=mgh+(M+m)gx-\frac{1}{2}k{x}^2$

$120\times {10}^{-3}\times 10\times 45\times {10}^{-2}+(200+120)\times {10}^{-3}\times 10x-\frac{1}{2}\times 200x^2=0$

$54\times {10}^{-2}+3.2x-100x^2=0$

$x=\frac{-(-3.2)+\sqrt{{\left(3.2\right)}^2-4\times 100\times -\left(0.54\right)}}{2\times 100}$

$x=\frac{3.2+\sqrt{10.24+216}}{200}$

$x=\frac{3.2+15.04}{200}$

$x=0.0912m$

$x=9.1cm$

Hence, the maximum extension of the spring is $9.1cm$