Elastic force in string is conservative in nature .

`:.`       `W = Dela  u`

where  W = work  done by elastic force of string

     `Delta u` = Change in elastic potential  energy

`because `  ` W = - (u_f - u_i) = u_i - U_f`

or        `w = 1/2 kx^2 - 1/2 k(x + y)^2`

or        `w = 1/2kx^2 - 1/2 k(x^2 + y^2  + 2xy)`

           =     `1/2kx^2 - 1/2 kx^2 - 1/2ky^2 - 1/2 K( 2xy)`

           = ` - kxy - 1/2 ky^2`

           = `1/2 ky( - 2x - y)`

The work done against elastic force is

`w_(external) =  - w = (ky)/(2)  (3x + y)`            

Here         u  = 100 m/s

            v = 0,       s = 1 cm = `10^-2` m

`:.`        ` v^2 = u^2 - 2as`

`:.`         `a =( u^2)/(2s)  = (100 xx 100)/(2 xx 1 xx 10^-2`

     = `5 xx 10^5` m / `s^2`

`:.`    F = ma = `10 xx 10^-3 xx 5 xx 10^5` = 5000  N

Alternative method

              `W = Delta `K

`or      - Fs = 1/2 mv^2 - 1/2 mu^2`

                   = `0 - 1/2 xx 10 xx 10^-3 xx 10000`

or      `- F xx 10^-2 = - 1/2 xx 10  xx 10^-3 xx 10 ^4`

`:.`              F = 5000 N.

According to work energy theorem, net work done is given by change in kinetic energy ./ Since no change takes place in kinetic energy . Hence net work  done should be zero

(A) This statement is correct only when forces are of conservative nature.

(B) In the absence of non -conservative force mechanical energy and non mechanical energy and non machanical energy are separately constant .

(C) The work done by internal  force may not be zero.