A `2kg` block slides on a horizontal floor with the a speed of `4m//s` it strikes a uncompressed spring , and compresses it till the block is motionless . The kinetic friction force is compresses is `15 N` and spring constant is `10000 N//m` . The spring by

Options

A. `8.5 cm`
B. `5.5 cm`
C. `2.5 cm`
D. 11.0 cm`

Answer

B

Explanation

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`