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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 0.25 J. The spring constant of the spring is ________Nm^{-1}

Option: 1


Option: 2


Option: 3


Option: 4


Answers (1)




At any instant total energy for free oscillation remains constant=\frac{1}{2} \mathrm{kA}^2

\begin{aligned} & \Rightarrow \frac{1}{2} \mathrm{kA}^2=0.25 \mathrm{~J} \\ & \Rightarrow \frac{1}{2} \mathrm{kA}^2=0.25 \mathrm{~J} \quad \Rightarrow \mathrm{K}=\frac{0.25 \times 2}{\mathrm{~A}^2} \\ & \Rightarrow \mathrm{k}=\frac{0.50}{(10 \mathrm{~cm})^2}=\frac{0.50}{\left(10 \times 10^{-2}\right)}=\frac{0.50 \times 10^4}{100} \\ & \mathrm{k}=0.50 \times 100=50 \mathrm{~N} / \mathrm{m} \end{aligned}

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Divya Prakash Singh

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