A block of mass 5 kg moves along the x-direct | vedclass

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Question

$F =-20\left( x -\frac{1}{2}ight)=-20 X \quad\left( X = x -\frac{1}{2}ight)$

$	herefore$ Particle will perform $SHM$ about $x =\frac{1}{2}$ wi

Vedclass · Accepted Answer

$5 \ m ,-5 \ kg m / s$

Vedclass · Answer

$F =-20\left( x -\frac{1}{2}ight)=-20 X \quad\left( X = x -\frac{1}{2}ight)$

$	herefore$ Particle will perform $SHM$ about $x =\frac{1}{2}$ with $\omega=2 rad / sec \Rightarrow T =\pi sec$.

$	herefore$ Phase covered in $t =\frac{\pi}{4}$ second $=90^{\circ}$.

Given particle is at rest at $x=1 m \Rightarrow x =1$ is extreme position.

$	herefore$ In $\frac{\pi}{4} sec$, it will be at equilibrium

$	herefore x =0.5 m$ and momentum $= m \omega A =5 	imes 2 	imes 0.5=5 kg m / s$

Direction will be towards -ve $x$.

Hence option $(C)$

A block of mass $5 kg$ moves along the $x$-direction subject to the force $F=(-20 x+10) N$, with the value of $x$ in metre. At time $t=0 s$, it is at rest at position $x=1 m$. The position and momentum of the block at $t=(\pi / 4) s$ are

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