A stationary body is lowered on to a massive platform moving horizontally with a speed of 4 ms^-1. The distance the body slides relative to the platform, if the coefficient of friction is 0.2, is(g=10ms^-2)a)16mb)8mc)4md)2m
Vivek Balakrishnan , 14 Years ago
Grade 11
1 Answers
vikas askiitian expert
Last Activity: 14 Years ago
applying v2 = u2 + 2as
finally particle comes to rest over the platform due to friction..
(retardation)a = (cofficient of friction)g =0.2*10=2m/s2
u2 = 2(2)s
s = u2/4 = 4m
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