A mass on a vertical spring begins its motion at rest at $y = 0\ cm$. It reaches a maximum height of $y = 10\ cm$. The two forces acting on the mass are gravity and the spring force. The graph of its kinetic energy ($KE$) versus position is given below. Net force on the mass varies with $y$ as
Diffcult
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Amplitude of $\mathrm{SHM}=5 \mathrm{cm}$
$\frac{1}{2} \mathrm{m} \omega^{2} \mathrm{A}^{2}=4$
$\frac{1}{2} m \omega^{2} A^{2} \times 25=4$
$\mathrm{m} \omega^{2}=\frac{8}{25}$
$F=-m \omega^{2}(y-5)$
$=-\frac{8}{25}(y-5)$
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