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Physics · Work, Power & Energy

JEE Main 2025 — 8 April, Evening Shift — Question 59

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 x m(x<2)x \mathrm{~m}(x<2) from the wall, the speed of the block will be

  1. Option A:

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

  2. Option B:

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

    Correct
  3. Option C:

    10[1−(2−x)]3/2 m/s10[1-(2-x)]^{3 / 2} \mathrm{~m} / \mathrm{s}

  4. Option D:

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

Answer: B

Step-by-step solution

figure

Energy conservation 12k(1)2=12mv2+12k(2−x)2\frac{1}{2} k(1)^{2}=\frac{1}{2} m v^{2}+\frac{1}{2} k(2-x)^{2}

Compression in the spring =(2−x)=(2-x)

⇒12k(1−(2−x)2)=12mv2⇒v=[100(1−(2−x)2)]1/2v=10[1−(2−x)]1/2\begin{aligned} \Rightarrow & \frac{1}{2} k\left(1-(2-x)^{2}\right)=\frac{1}{2} m v^{2} \Rightarrow \\ & v=\left[100\left(1-(2-x)^{2}\right)\right]^{1 / 2} \\ & v=10[1-(2-x)]^{1 / 2} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
8 April, Evening Shift
Subject
Physics
Chapter
Work, Power & Energy
Topic
Applications of Conservation of Mechanical Energy