A 2kg 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. Find the acceleration of the block in its lowest position `(g=10 m//s^(2))`

Explanation

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)` `or x = (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"`