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Grade: 11
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

Answers : (2)

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) + KE
on further solving,
KE = mg (mg/k)- k/2 (mg/k)^2 
KE = (mg)^2 /k - (mg)^2 /2k 
KE = 1/2 (mg)^2 /k 
KE = 1/2 mg a
would be the final kinetic energy.
2 years ago
Akriti bhatt
19 Points
Initially the block was left and had velocity zero
Using conservation of energy 
1/2m(0)^2+mgx =1/2mv^2+mg(0)+1/2kx^2
On putting x = mg/k we get
KE = 1/2mga
one year ago
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