A block of mass $5 \mathrm{~kg}$ moves along the $x$-direction subject to the force $F=(-20 x+10)…
A block of mass $5 \mathrm{~kg}$ moves along the $x$-direction subject to the force $F=(-20 x+10) \mathrm{N}$, with the value of $x$ in metre. At time $t=0 \mathrm{~s}$, it is at rest at position $x=1 \mathrm{~m}$. The position and momentum of the block at $t=(\pi / 4) \mathrm{s}$ are
$\mathrm{F}=-20\left(\mathrm{x}-\frac{1}{2}\right)=-20 \mathrm{X} \quad\left(\mathrm{X}=\mathrm{x}-\frac{1}{2}\right)$
$\therefore$ Particle will perform SHM about $\mathrm{x}=\frac{1}{2}$ with $\omega=2 \mathrm{rad} / \mathrm{sec} \Rightarrow \mathrm{T}=\pi \mathrm{sec}$.
$\therefore$ Phase covered in $t=\frac{\pi}{4}$ second $=90^{\circ}$.
Given particle is at rest at $\mathrm{x}=1 \mathrm{~m} \Rightarrow \mathrm{x}=1$ is extreme position.
$\therefore$ In $\frac{\pi}{4}$ sec, it will be at equilibrium
$\therefore \mathrm{x}=0.5 \mathrm{~m}$ and momentum $=\mathrm{m} \omega \mathrm{A}=5 \times 2 \times 0.5=5 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$ Direction will be towards -ve $\mathrm{x}$.
Hence option (3)