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Question Number 21670 by Tinkutara last updated on 30/Sep/17

The block of mass 2 kg and 3 kg are

placed one over the other . The contact

surfaces are rough with coefficient of

friction μ_{1} = 0.2, μ_{2} = 0.06 . A force F =

\frac{1}{2} t N (where t is in second) is applied

on upper block in the direction . (Given

that g = 10 m / s^{2})

1 . The relative slipping between the

blocks occurs at t =

2 . Friction force acting between the two

blocks at t = 8 s

3 . The acceleration time graph for 3 kg

block is

Commented by Tinkutara last updated on 30/Sep/17

Answered by ajfour last updated on 30/Sep/17

A t t = 6 s, F = 3 N

f_{1} = 3 N b a c k w a r d o n 2 k g a n d

f o r w a r d o n 3 k g .

f_{2} = 3 N (l i m i t i n g v a l u e) b a c k w a r d

o n 3 k g .

\Rightarrow o n s e t o f a c c e l e r a t i o n a t t = 6 s

A t t = 8 s, F = 4 N

f_{1} - f_{2} = 3 a \Rightarrow f_{1} - 3 = 3 a \dots (i)

F - f_{1} = 2 a \Rightarrow 4 - f_{1} = 2 a \dots . (i i)

s o 2 f_{1} - 6 = 12 - 3 f_{1}

\Rightarrow 5 f_{1} = 18; f_{1} = 3.6 N .

Commented by Tinkutara last updated on 30/Sep/17

Thank you very much Sir!

Can you answer other 2 ? Because they

do not match answer .