A 0.5 kg block slides from the point A on a horizontal track with an initial speed 3 m/s towards a weightless horizontal spring of length 1 m and force constant 2 N/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 distance AB and BD are 2 m and 2.14 m respectively ,then the total distance through which the block moves before it comes to rest completely is –[g = 10 m/s2].
As the track AB is frictionless, the block moves this distance (= 2m) without loss in its initial K.E. = 12mv2 = 12∗0.532 = 2.25J. In the path BD (= 2.14)m as friction is present, so work done against friction = μkmg = 0.2×0.5×10×2.14 = 2.14J.
So at D the K.E. of the block is = 2.25 - 2.14 = 0.11 J
Now if the spring is compressed by x,
0.11 = 12×k×x2 + μk mgx
i.e. 0.11 = 12×2×x2 + 0.2×0.5×10x,
or x2 + x − 0.11 = 0.
Which on solution gives x = 0.1 m (as x = -1.1 is inadmissible).