The system as shown in the figure is released from rest. The pulley, spring and string are ideal & friction is absent everywhere. If speed of 5 kg block when 2 kg block leaves the contact with ground is 
m/s, then value of x is: (spring constant k = 40 N/m & g = 10 m/s 2 )
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
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Sol. F.B.D of 2 kg block
When 2 kg block just leaves the contact
N = 0
Kx + 0 = 20 ⇒ x = 
= 
m
Applying WET for the whole system
w g + w sf = Δ k
⇒ 50 × 
– 
(40) 
= 
(5) [V 2 – 0]
⇒ V = 
m/s
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