A block is fastened to a horizontal spring. The block is pulled to a distance from its equilibrium position (at ) on a frictionless surface from rest. The energy of the block at is The spring constant of the spring is ________ . [2023]

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Solution

Q.
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 ________ ${Nm}^{- 1}$ . [2023]

Ans.
(67)

$U_{i} = \frac{1}{2} k x_{0}^{2} and K_{i} = 0$

$U_{f} = \frac{1}{2} k {(\frac{x_{0}}{2})}^{2} and K_{f} = 0.25 J$

$\frac{1}{2} k x_{0}^{2} + 0 = \frac{1}{2} k \frac{x_{0}^{2}}{4} + 0.25$

$\frac{1}{2} k x_{0}^{2} \cdot \frac{3}{4} = \frac{1}{4}$

$\frac{1}{2} k \cdot \frac{3}{100} = 1 \Rightarrow k = \frac{200}{3} N/m \approx 67 N/m$

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