A small block b is placed on another block a of mass 5 kg and 20 cm and initially the block b is near it and f block a constant force 10 newton is applied on block a all surfaces are frictionless time find time elapsing for the be separated from a
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given,
mass of block A , m = 5kg
from Newton's 2nd law,
F = ma
10 = 5a
a = 2m/s²
because there is no friction force between A and B. when the block A moves , block B remains at rest in its position.
take initial velocity of A, u = 0
distance to cover so that block B separate out, s = 0.2m
so, S = ut + 1/2at²
0.2 = 0 + 1/2 × 2 × t²
t² = 0.2
t = 0.45 sec
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