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

4

Explanation

Let the speed of the blocks A and B at the instant compression `(x_(0))/4` be `v_(A)` and `v_(B)` respectively as shown. (`l_(0)` is natural length)

Since, No external force acts on the system in Hz direction, therefore applying linear momentum conservation in Hz direction.
`0=m(-v_(A))+2m(v_(B))` or
`v_(A)=2v_(B)......(1)`
Now, By conservation of energy,
`1/2 kx_(0)^(2)=1/2 k((x_(0))/2)^(2)+1/2 (m) v_(A)^(2)+1/2 (2m) v_(B)^(2)`
From (1) and (2) we get
`1/2 mv_(A)^(2)=1/2 kx_(0)^(2)`
Using work energy theorem,
W.d by spring on A =change in K.E. of A
or `W_(S)=1/2 mv_(A)^(2)=5/16 kx_(0)^(2)`