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# A 0.5kg block slides from the pts A on a horizontal track with an initial speed of 3m/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 coefficient of static and kinetic friction as 0.22 and 0.2 respec .Find the total distance through which the block moves before it comes to rest completely. Asked by m.nilu 21st August 2018, 2:57 PM
Initially the block moves with constant speed  3 m/s on a smooth surface.
Then for a distance 2.14 m, it is subjected to friction provided by rough surface.

friction force = μN = μmg = 0.2×0.5×9.8 = 0.98 N
Retardation experienced by the block = force/mass = 0.98/0.5 = 1.96 m/s2

speed after travelling 2.14 m rough path:-  v2 = 9 - 2×1.96×2.14  , we get v = 0.78 m/s

after travelling 2.14 m rough path, the block is stopped by spring. Let the spring has compressed by a distance x.
Then the stopping force provided by the spring is kx, where k is force constant of the spring.

since the block is stopped, we get the distance x moved from,  x = u2 / (2×ar )..............(1)
where u is initialvelocity and ar is the retardation provided by the spring.  Retardation is given by,  ar = kx/m  = 2x/0.5  = 4x
hence we rewrite eqn.(1) to get x :   x = u2 / (2×4x)  or x2 = u2 / 8 = (0.78×0.78)/8 = 0.07605
hence x = 0.28 m   = 28 cm
Answered by Expert 26th August 2018, 12:24 PM
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Tags: distance