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`        A block of mass m is hung vertically from an elastic thread of force constant mg/a. Initially the thread was at it natural length and the block is allowed to fall freely. The kinetic energy of the block when it passes through the equilibrium position will be?`
2 years ago

Vikas TU
9484 Points
```							From energy conservation,1/2 k (x)^2 - mg (x) + 1/2 m (v)^2 = 1/2 k (x)^2 - mg (x) + KEon further solving,KE = mg (mg/k)- k/2 (mg/k)^2 KE = (mg)^2 /k - (mg)^2 /2k KE = 1/2 (mg)^2 /k orKE = 1/2 mg awould be the final kinetic energy.
```
2 years ago
Akriti bhatt
19 Points
```							Initially the block was left and had velocity zeroUsing conservation of energy 1/2m(0)^2+mgx =1/2mv^2+mg(0)+1/2kx^2On putting x = mg/k we getKE = 1/2mga
```
one year ago
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• 18 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions