A 2kg block is connected with two springs of force constants K_(1) = 100 N/m and K_(2) = 300 N/m as show in figure. The block is released from rest with the springs unstreched. Find the acceleration of the block in its lowest position (g = 10 m//s^(2))

A 2kg block is connected with two springs of force constants K_(1) = 1

1.	A 2kg block is connected with two springs of force constants K_(1) = 100 N/m and K_(2) = 300 N/m as show in figure. The block is released from rest with the springs unstreched. Find the acceleration of the block in its lowest position (g = 10 m//s^(2))
Answer» Solution :LET .X. be the maximum displacement of block downwards. Then from conservation of MECHANICAL enery : decrease in potential energy of 2 kg block = increase in ELASTIC potential energy of both the springs `thereforemg x =(1)/(2)(k_(1)+k_(2))x^(2)` `orx = (2mg)/(k_(1)+k_(2))=((2)(2)(10))/(100+300)=0.1m` ACCELERATION of block in this position is `a=((k_(1)+k_(2))x-mg)/(m)("upwards")` `=((400)(0.1)-(2)(10))/(2)=10m//s^(2)"upwards"`