AP EAMCET · PHYSICS · Laws of Motion
To the free end of spring hanging from a rigid support, a block of mass \(m\) is hung and slowly allowed to come to its equilibrium position. Then stretching in the spring is \(d\). If the same block is attached to the same spring and allowed to fall suddenly, the amount of stretching is : (force constant, \(k\) )
- A \(\frac{m g}{k}\)
- B \(2 d\)
- C \(\frac{m g}{3 k}\)
- D \(4 d\)
Answer & Solution
Correct Answer
(B) \(2 d\)
Step-by-step Solution
Detailed explanation
To leave the block, it oscillates in vertical plane. If maximum extension in spring in extreme position of block is \(x_1\), then Work done by weight of the block \(=\) potential energy stored in spring \(m g x=\frac{1}{2} k x^2\)…
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