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) ).

Answer

4.24

Explanation

The block strikes the spring and compresses it by the distance x.
Using work energy theorem
`W_("total") = Delta K implies W_("friction") + W_("spring") = K_("final") - K_("initial")`
`(0 - 1/2 m v^2) = - mu_(k) mg(2.14 + x) - 1/2 kx^2`
`1/2 xx 0.5 xx 3^2 = 0.2 xx 0.5 xx 9.8 (2.14 + x) + 1/2 2 xx x^2`
`2.14 + x + x^2 = 2.25 " " :. " " x^2 + x - 0.11 = 0`
On solving we get `x = (-11)/10 "or" x = 1/10 = 0.1 m`
Here the body stops momentarily.
Restoring force at this position
`F_s = kx = 2 xx 0.1 = 0.2 N`
Frictional force at this position

`F_(lim) = mu_s mg = 0.22 xx 0.5 xx 9.8 = 1.078 N`
Since frictional force > Restoring force the body will stop here. The total distance travelled
= AB + BD + DY = 2 + 2.14 + 0.1 = 4.24 m.