A 2 kg block is connected with two springs of force constants k 1 = 100 N/m and k 2 = 300 N/m as shown in figure. The block is released from rest with the springs unstretched. The acceleration of the block in its lowest position is: (g = 10 m/s 2 ) –
Text Solution
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Let x be the maximum displacement of block downwards. Then from conservation of mechanical energy:
decrease in potential energy of 2 kg block = increase in elastic potential energy of both the springs
mg x = 
(k 1 + k 2 ) x 2
or x = 
= 
= 0.1 m
Acceleration of block in this position is –
a = 
(upwards)
= 
= 10 m/s 2 (upwards)
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