As we know by work energy theorem
total work done = change in kinetic energy
so here we can say that wok done on the box will be equal to the change in kinetic energy of the system
[tex]W_p = KE_f - KE_i[/tex]
initial the box is at rest at position x = x1
so initial kinetic energy will be ZERO
at final position x = x2 final kinetic energy is given as
[tex]KE_f = \frac{1}{2}mv_1^2[/tex]
now work done is given as
[tex]W_p = \frac{1}{2}mv_1^2 - 0[/tex]
so we can say
[tex]W_p = \frac{1}{2}mv_1^2[/tex]
so above is the work done on the box to slide it from x1 to x2
