MCQ
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
    $69$
  • $50$

Answer

Correct option: D.
$50$
d
$U _{ i } =\frac{1}{2} kx _0^2$

$K _{ i } =0$

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

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

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

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

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

$=67\,N / m$

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