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JEE Main 20258 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
8 April
Shift
Evening Shift
Subject
Physics
Chapter
Work, Power and Energy

JEE Main 2025 · 8 April · Evening Shift · Physics · Work, Power and Energy

A block of mass 2 kg is attached to one end of a massless spring...

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 N/m200 \mathrm{~N} / \mathrm{m}. The block is pushed such that the length of the spring becomes 1 m and then released. At distance xm(x<2)\mathrm{x} \mathrm{m}(\mathrm{x} \lt 2) from the wall. the speed of the block will be :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

Options

  1. A10[1−(2−x)]3/2 m/s10[1-(2-x)]^{3 / 2} \mathrm{~m} / \mathrm{s}
  2. B10[1−(2−x)2]1/2 m/s10\left[1-(2-x)^2\right]^{1 / 2} \mathrm{~m} / \mathrm{s}
  3. C10[1−(2−x)2]m/s10\left[1-(2-\mathrm{x})^2\right] \mathrm{m} / \mathrm{s}
  4. D10[1−(2−x)2]2 m/s10\left[1-(2-\mathrm{x})^2\right]^2 \mathrm{~m} / \mathrm{s}

Correct answer

B. 10[1−(2−x)2]1/2 m/s10\left[1-(2-x)^2\right]^{1 / 2} \mathrm{~m} / \mathrm{s}

Step-by-step solution

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
Given, Natural length of spring =2 m=2 \mathrm{~m}
Initial compression in spring \left(\mathrm{x}_{\mathrm{i}}\right)=1 \mathrm{~m}<br/>Finalcompressioninspring(xf)=(2−x)m<br/>Final compression in spring \left(\mathrm{x}_{\mathrm{f}}\right)=(2-\mathrm{x}) \mathrm{m}
Using energy conservation
<br/>Ki+Ui=Kf+Uf0+12Kxi2=12mv2+12Kxf212mv2=12 K(xi2−xf2)12×2×v2=12×200×(12−(2−x)2)v2=100[1−(2−x)2]v=10[1−(2−x)2]1/2<br/>\begin{aligned}<br/>& \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}<br/>\end{aligned}
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