Question 74 of 183

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

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Answer: 67

At first the block is pulled to a distance ${x}_{0}=10cm$(extreme position).
Potential energy of block is ${U}_{i}=\dfrac{1}{2}k{x}_{0}^{2}$ and kinetic energy os block, ${K}_{i}=0$.
Now, when the block is at distance $5cm$.

Potential energy is ${U}_{f}=\dfrac{1}{2}k{\left(\dfrac{{x}_{0}}{2}\right)}^{2}$ and kinetic energy is ${K}_{f}=0.25J$
Total energy of block will be conserved, so
$\dfrac{1}{2}k{x}_{0}^{2}+0=\dfrac{1}{2}k\dfrac{{x}_{0}^{2}}{4}+0.25$
$\Rightarrow \dfrac{1}{2}k{x}_{0}^{2}\dfrac{3}{4}=\dfrac{1}{4}$
$\Rightarrow \dfrac{1}{2}k\dfrac{3}{100}=1 \Rightarrow k=\dfrac{200}{3}N{m}^{-1}$
$=67N{m}^{-1}$