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JEE Mains · Physics · STD 11 - 5. work,energy,power and collision

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 :

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

Answer & Solution

Correct Answer

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

Step-by-step Solution

Detailed explanation

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…
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