Correct Answer - B
Here, `m=2kg, upsilon=4m//s`
Force of kinetic friction, `F=15N`
spring constant, `K=10000N//m`
Let the spring be compressed by `x` metre.
`K.E.` supplied `=` Work done against friction
`+` P.E. of spring
`(1)/(2)m upsilon^(2)=Fx+(1)/(2)Kx^(2)`
`(1)/(2)xx2xx4^(2)=15x+(10000)/(2)x^(2)`
or `5000x^(2)+15x-16=0`
On solving, we get `x=0.055m=5.5cm`