A `0.5kg` block slides from the point A on a horizontal track with an initial speed `3 m//s` towards a weightless horizontal spring of length `1m` and force constant `2N//m.` The part AB of the track is frictionless and the part BC has the coefficient of static and kinetic friction as '0.22' and `0.20` respectively. If the distances AB and BD are `2m` and `2.14m` respectively, find total distance through which the block moves before it comes to rest completely. `(g=10 m//s^(2) ).

Options

A. `4.20 m`
B. `4.14 m`
C. `4.24 m`
D. `4.26 m`

Answer

C

Explanation

From `A` to `B`, there will be no loss energy. Now let block compresses the spring by an amount `x` and comes momentarily to rest. Then, loss of energy will be equal to the work done against friction. Therefore,

`mu_(k)mg(BD+x) =(1)/(2) mv^(2) -(1)/(2) kx^(2)`
Substituting the values `(0.2)(0.5)(10)(2.14 + x)`
=`(1)/(2) (0.5)(3)^(2) -(1)/(2)(2)(x)^(2)`
Solving this equation, we get `x = 0.1 m`.