A block is fastened to a horizontal spring. block is pulled to a distance x =10,cm from its equilibr

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5.Work, Energy, Power and Collision

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A block is fastened to a horizontal spring. The block is pulled to a distance $x =10\,cm$ from its equilibrium position (at $x =0$ ) on a frictionless surface from rest. The energy of the block at $x =5$ $cm$ is $0.25\,J$. The spring constant of the spring is $.........Nm ^{-1}$

A

$65$

B

$66$

C

$69$

D

$50$

(JEE MAIN-2023)

Solution

$U _{ i } =\frac{1}{2} kx _0^2$

$K _{ i } =0$

$U _{ f }=\frac{1}{2} k \left(\frac{ x _0}{2}\right)^2$

$K _{ f }=0.25\,J$

$\frac{1}{2} kx _0^2+0=\frac{1}{2} k \frac{ x _0^2}{4}+0.25$

$\frac{1}{2} k x _0^2 \frac{3}{4}=\frac{1}{4}$

$\frac{1}{2} k \frac{3}{100}=1 \Rightarrow k =\frac{200}{3}\,N / m$

$=67\,N / m$

Standard 11

Physics

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