JEE Mains · Physics · STD 11 - Work,energy,power and collision
- JEE Mains 2023
- Medium
- Numerical
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 \(.........Nm ^{-1}\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(50\)
Step-by-step Solution
Detailed explanation
\(U _{ i } =\frac{1}{2} kx _0^2\)
\(K _{ i } =0\)
\(U _{ f }=\frac{1}{2} k \left(\frac{ x _0}{2}\right)^2\)
\(K _{ f }=0.25\,J\)
\(\frac{1}{2} kx _0^2+0=\frac{1}{2} k \frac{ x _0^2}{4}+0.25\)
\(\frac{1}{2} k x _0^2 \frac{3}{4}=\frac{1}{4}\)
\(\frac{1}{2} k \frac{3}{100}=1 \Rightarrow k =\frac{200}{3}\,N / m\)
\(=67\,N / m\)
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