Answer:
option B
Explanation:
given,
mass of the box = 1 Kg
spring constant = k = 1000 N/m
spring is compressed upto = 15 cm = 0.15 m
coefficient of friction = μk =0.4
using conservation of energy
   [tex]\dfrac{1}{2}kx^2 = \dfrac{1}{2}mv^2[/tex]
   [tex]kx^2 =mv^2[/tex]
   [tex]v = \sqrt{\dfra{k}{m}}x[/tex]
   [tex]v = \sqrt{\dfra{1000}{1}}\times 0.15[/tex]
     v = 4.74 m/s
so, the velocity of the box = v = 4.74 m/s
hence, the correct answer is option B