Last Activity: 14 Years ago
let the mass of block b is mb....
force on lower block is F=10N....
accleration of system is a...
(mb+5)a = 10
a= 10/(mb+5) m/s2
now a seudo force acts on the upper block in opposite direction and its magnitude is f=(mb)a...
due to this force block moves in opposite direction with accleration a1....
mba1=mba
a1=a=10/(5+mb)
when the block covers a distance of 20cm then it separates out...
0.2 = a1t2/2
t2 = (0.4/a1)=[0.04(5+mb)]
t= 0.2(5+mb)1/2
Last Activity: 7 Years ago
The block A will accelerate under the force 10 N applied on it. But the surfaces being frictionless, block B will remain stationary. Soblock B will separate from block A when block A has travelled a distance s equal to its length i.e. 20cm.For block A:u=0 m/s; a=Force/Mass=10/5=2NSo, v²=√(u²+2as)v=√2as=√(2*2*0.2)=0.894 m/sTherefore, v=u+at0.894=0+2t=>t=0.447 sec=0.45 sec
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