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A 0.5kg block slides from the point A on...

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 confficient of static and kinetic friction as `0.20` respectively. If the distancences 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) ).

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The correct Answer is:
B, D
As the track AB us frictionless, the block moves the distance without loss in its initial `KE =1/2 mv^(2) =1/2 xx 0.5 xx 3^(2) =2.25J`. In the path BD as friction is present, so work done against friction.
`=mu ._(k)mgd =0.2xx0.5xx10xx2.14 =2.14J`
So ,at D the KE of the block is `=2.25-2.14=0.11J`
.
Now, if the spring is compressed by `x`
`0.11 =1/2 xx kx xx x^(2) + mu_(r)._(k)mgx`
i.e. `0.11 =1/2 xx 2 xx x^(2)+ 0.2 xx 0.5 xx 10x`
or `x^(2)+x-0.11 =0`
which on solving gives positive value of `x=0.1`
After moving the distance `x=0.m` the block comes to rest.
Now the compressed spring compressed spring exerts a force:
`F = kx = rx =2 xx 0.1 =0.2N`
on the block while limiting frictional force between block and track is `f_(L) =mu_(s)mg`
`=0.22xx0.5xx10 =1.1N`. Since, `Fltf_(L)`. The block will not move back. So, the total distance moved by the block
`=AB+BD+0.1 =2+2.14+0.1`
`=4.24m`.
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