Question

A massless spring $S$ can be compressed $1\text{m}$ by a force of $100\text{N}$ in equilibrum. The same spring is placed at the bottom of a frictionless plane inclined at ${30}^{\circ }$ to the horizontal. A block of $10\text{kg}$ is released from rest at the top of incline and is brought to rest momentarily after compressing the spring by $2\text{m}$. If the speed of block just before it touches the spring is $\sqrt{10x}\text{m/s}$. Value of $x$ is .
$[g=10{\text{ms}}^{-2}]$

Answer 1

$F=Kx\Rightarrow k=\frac{F}{x}=\frac{100}{1}=100{\text{N/m}}^2$

Applying energy conservation principle at initial and final positions of block:
$\frac{1}{2}m{v}^2+mgh=0+\frac{1}{2}K{x}^2$

$\Rightarrow \frac{1}{2}\times 10\times v^2+10\times 10\times 1=\frac{1}{2}\times 100\times (2{)}^2$

$\Rightarrow 5v^2+100=200$

$\Rightarrow 5v^2=100$

$\Rightarrow v^2=20$

$\Rightarrow v=\sqrt{20}\text{m/s}=\sqrt{10x}\text{m/s}$

$\Rightarrow x=2$