In the figure shown the spring is compressed by `'x_(0)'` and released . Two block `'A'` and `'B'` of masses `'m'` and `'2m'` respectively are attached at the ends of the spring. Blocks are kept on a smooth horizontal surface and released. Find the work done by the spring on `'A'` by the time compression of the spring reduced to `(x_(0))/(2)`.

Answer

`(1)/(4)kx_(0)^(2)`

Explanation

Let the block `A ` shift to left by `x_(1)` and block `B` shift to right by `x_(2)`. The centre of mass of the two block system is at rest.
Hence `mx_(1)=2ms_(2)`.
or `x_(2)=(x_(1))/(2) …..(1)`
and the spring force on either block is `k(x_(0)-[x_(1)+x_(2)]),` where `x_(0)` is the initial compression in the spring.
Let the block `A` shift further left by `dx_(1)`
`:. `work donw on block by spring is
`dW=(k(x_(0)-x_(1)-x_(2))dx_(1) .....(2)`
`=k(x_(0)-x_(1)-(x_(1))/(2))dx_(1)`
`:. `Net work done
`int dW=underset(x_(1)=0)overset(x_(0)//3)intk(x_(0)-(3)/(2)x_(1))dx_(1)=(kx_(0)^(2))/(4)`
`Ans. (1)/(2) kx_(0)^(2)`
Alternative Solution
Let the speeds of blocks `A` and `B` at the instant compression is `(x_(0))/(2)` be `v_(A)` and`v_(B)` as shown in figure
`[l_(0)]` natural length of spring `]`

No external forces act on the system in the horizontal direction
Applying conservation of momentum in horizontal direction
Initial momentum `=` final momentum
`0=m(-v_(A))+2mv_(B)`
or `v_(A)=2v_(B) ......(1)`
from conservation of energy
`(1)/(2)kx_(0)^(2)=(1)/(2)k((x_(0))/(2))^(2)+(1)/(2) mv_(A)^(2)+(1)/(2) 2mv_(B)^(2) ....(2)`
from `(1)` and `(2)` we get
`(1)/(2)mv_(A)^(2)=(1)/(4)kx_(0)^(2)`
work done on block `A` by spring `=` change in kinetic energy of block `A`
`=(1)/(2)mv_(A)^(2)=(1)/(4)kx_(0)^(2)`