Question 44 of 183

A block of mass 2 kg is attached to one end of a massless spring whose other end is fixed at a wall. The spring-mass system moves on a frictionless horizontal table. The spring's natural length is 2 m and spring constant is $200 \mathrm{~N} / \mathrm{m}$. The block is pushed such that the length of the spring becomes 1 m and then released. At distance $\mathrm{x} \mathrm{m}(\mathrm{x} \lt 2)$ from the wall. the speed of the block will be :

A$10[1-(2-x)]^{3 / 2} \mathrm{~m} / \mathrm{s}$B$10\left[1-(2-x)^2\right]^{1 / 2} \mathrm{~m} / \mathrm{s}$C$10\left[1-(2-\mathrm{x})^2\right] \mathrm{m} / \mathrm{s}$D$10\left[1-(2-\mathrm{x})^2\right]^2 \mathrm{~m} / \mathrm{s}$

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Answer: (B) $10\left[1-(2-x)^2\right]^{1 / 2} \mathrm{~m} / \mathrm{s}$

Given, Natural length of spring $=2 \mathrm{~m}$
Initial compression in spring $\left(\mathrm{x}_{\mathrm{i}}\right)=1 \mathrm{~m}$
Final compression in spring $\left(\mathrm{x}_{\mathrm{f}}\right)=(2-\mathrm{x}) \mathrm{m}$
Using energy conservation
$\begin{aligned}
& \mathrm{K}_{\mathrm{i}}+\mathrm{U}_{\mathrm{i}}=\mathrm{K}_{\mathrm{f}}+\mathrm{U}_{\mathrm{f}} \\
& 0+\frac{1}{2} \mathrm{Kx}_{\mathrm{i}}^2=\frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{Kx}_{\mathrm{f}}^2 \\
& \frac{1}{2} \mathrm{mv}^2=\frac{1}{2} \mathrm{~K}\left(\mathrm{x}_{\mathrm{i}}^2-\mathrm{x}_{\mathrm{f}}^2\right) \\
& \frac{1}{2} \times 2 \times \mathrm{v}^2=\frac{1}{2} \times 200 \times\left(1^2-(2-\mathrm{x})^2\right) \\
& \mathrm{v}^2=100\left[1-(2-\mathrm{x})^2\right] \\
& \mathrm{v}=10\left[1-(2-\mathrm{x})^2\right]^{1 / 2}
\end{aligned}$