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

A block is fastened to a horizontal spring. The block is pulled to a distance \(x =10\,cm\) from its equilibrium position (at \(x =0\) ) on a frictionless surface from rest. The energy of the block at \(x =5\) \(cm\) is \(0.25\,J\). The spring constant of the spring is \(.........Nm ^{-1}\)

  1. A \(65\)
  2. B \(66\)
  3. C \(69\)
  4. D \(50\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(50\)

Step-by-step Solution

Detailed explanation

\(U _{ i } =\frac{1}{2} kx _0^2\) \(K _{ i } =0\) \(U _{ f }=\frac{1}{2} k \left(\frac{ x _0}{2}\right)^2\) \(K _{ f }=0.25\,J\) \(\frac{1}{2} kx _0^2+0=\frac{1}{2} k \frac{ x _0^2}{4}+0.25\) \(\frac{1}{2} k x _0^2 \frac{3}{4}=\frac{1}{4}\)…
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