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NEET 2026
Exam
NEET
Year
2026
Subject
Physics
Chapter
Chapter

NEET 2026 · Physics · Chapter

A block of mass m is attached to one end of a horizontal ideal...

A block of mass mm is attached to one end of a horizontal ideal spring of spring constant kk, while the other end is fixed. The block is pushed to compress the spring by a distance AA and then released from rest on a smooth horizontal surface. At a certain instant during its motion, the kinetic energy of the block is twice the elastic potential energy of the spring. The compression of the spring and the speed of the block at this instant are respectively

Options

  1. AA3,A2k3m\dfrac{A}{\sqrt{3}}, A\sqrt{\dfrac{2k}{3m}}
  2. B23A,Ak3m\sqrt{\dfrac{2}{3}}A, A\sqrt{\dfrac{k}{3m}}
  3. CA2,Ak2m\dfrac{A}{\sqrt{2}}, A\sqrt{\dfrac{k}{2m}}
  4. DA3,Ak3m\dfrac{A}{\sqrt{3}}, A\sqrt{\dfrac{k}{3m}}

Correct answer

A. A3,A2k3m\dfrac{A}{\sqrt{3}}, A\sqrt{\dfrac{2k}{3m}}

Step-by-step solution

Total energy of the system is E=12kA2E = \dfrac{1}{2}kA^2.

Let xx be the compression of the spring and vv be the speed of the block at the given instant.

The elastic potential energy is U=12kx2U = \dfrac{1}{2}kx^2 and the kinetic energy is K=12mv2K = \dfrac{1}{2}mv^2.

Given that K=2UK = 2U, by conservation of mechanical energy:

E=K+U=3UE = K + U = 3U

12kA2=3(12kx2)\dfrac{1}{2}kA^2 = 3 \left(\dfrac{1}{2}kx^2\right)

x2=A23⇒x=A3x^2 = \dfrac{A^2}{3} \Rightarrow x = \dfrac{A}{\sqrt{3}}

Now, substituting xx into K=2UK = 2U:

12mv2=2(12k(A3)2)\dfrac{1}{2}mv^2 = 2 \left(\dfrac{1}{2}k \left(\dfrac{A}{\sqrt{3}}\right)^2\right)

12mv2=kA23\dfrac{1}{2}mv^2 = \dfrac{kA^2}{3}

v=A2k3mv = A\sqrt{\dfrac{2k}{3m}}

Answer: A3,A2k3m\dfrac{A}{\sqrt{3}}, A\sqrt{\dfrac{2k}{3m}}
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