IIT JEE , AIEEE Forum - Physics , Mechanics - Newtons laws-tough question

acpassword

A small block is placecd on another block A of mass 5kg and length 20cm.Initially the block B is near the right end of block A const. horizontal force of 10N is applied to block A.Find the time elapsed before block separates from A.All surfaces r frictionless.

Ans:-0.45s

coolank2

The surfaces are frictionless. Hence, the above block will not move w.r.t ground, as no horizontal force is acting on it.

Thus, a = F/m = 2 ms^-2

S = 0.2 m

u = 0 ms^-1

S = ut + (0.5)at²

Thus, t = sqrt(0.2) = 0.447 s

sahilgupta_iit

since no force is acting on the small block, hence it will have no acceleration w.r.t. earth. now,

F=10 N

mass of block A=5kg

acc. = 2m/s^2

distance moved to separate from the small block=20cm=0.2m

initial velocity = 0m/s

s = ut + at^2/2

0.2 = 2.t^2/2

t^2 = 0.2

t = 0.45s (approx)

do rate

aneesh_laksh

very easy sir aagr aise question karo gay to nikal gaye iit
ans-0.44sec

Goalie

i think i answered a few days back same question

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