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. 3.23 m
B. 4.14 m
C. 5.24 m
D. 4.67 m

Answer

C

Explanation

As the track AB is frictionless, the block moves this distance = 2m without loss in its initial KE
`=1/2 mv^(2) = 1/2 xx 0.5 xx 3^(2) = 2.25 J`
In the path BD (= 2.14 m) as friction is present, so work done against friction = `mu_(K)` mgs
`= 0.2 xx 0.5 xx 10 xx 2.14 = 2.14 J`.
So at D the KE of the block is = `2.25 - 2.14 = 0.11 J`.
Now if the spring is compressed by x,
`0.11 = 1/2 xx k xx x^(2) + mu_(K)mgx`
i.e. `0.11 = 1/2 xx 2 xx x^(2) xx 0.5 xx 10 x`
or `x^(2) + x - 0.11 =0`
which on solving gives x = 0.1 m (as x = - 1.1 is inadmissible).After moving the distance x = 0.1 m the block comes to rest. Now the compressed spring exerts a force
`F = kx = 2 xx 0.1 = 0.2 N`
on the block while limiting frictional force between block and track `f_(L) = mu_(s) mg = 0.22 xx 0.5 xx 10 = 1.1 N`
Since, `F lt f_(L)`, the block will not move back. So, the total distance moved by the block
`= 2 + 2.14 + 0.1 = 5.24 m`