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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?
one year ago

Answers : (2)

Vikas TU
6874 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 
or
KE = 1/2 mg a
would be the final kinetic energy.
 
 
one year 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 month ago
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