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A 2kg block is connected with two spring...

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))`

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
`therefore mg 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)=10 m//s^(2)` upwards
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