Figure shows two block `A` and `B`, each having a mass of 320 g connected by a light string passing over a smooth light pulley. The horizontal surface on which the block `A` can slide is smooth. The block `A` is attached to spring constant `40 N//m` whose other end is fixed to a support 40 cm above the horizontal surface. Initially, the spring is vertical and unstretched when the system is released to move. Find the velocity of the block A at the instant it breaks off the surface below it. Take `g = 10 m//s^(2)` .

Explanation

At the instant of break-off ,`N = 0`
`implies kx cos theta = mg`
Where extension in the spring
`x=l-0.4=(0.4)/(costheta)-0.4`
`therefore k((0.4)/(costheta)-0.4)costheta=mg`
`k(0.4)(1-costheta)=mg`
`impliescostheta=1-(mg)/(0.4k)`

`=1-(0.320xx10)/(0.4xx40)`
`costheta=(4)/(5)impliesx=(0.4xx5)/(4)-0.4=0.1m`
Also, `tantheta=(3)/(4)impliesd=0.4tantheta=0.3m`
`therefore` By work energy theorem for motion of both the blocks.
`W_(g)+W_(s)=DeltaKE`
`impliesmgd-(1)/(2)kx^(2)=((1)/(2)mv^(2))xx2`
`implies0.32xx10xx0.3-(1)/(2)(40)(0.1)^(2)=(0.32)v^(2)`
`impliesv=1.54m//s`