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In Fig. initialy the system is at rest .Find out minimum value of `F `for which sliding start between the two blocks Given`m_(A) = 10 kg` and `m_(B) = 20 kg`

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At just sliding condition limiting friction is acting

` F - 50 = 20a`….(i)
`f = 10a` ….(ii)
`50 = 10a`
`:. a = 5 ms^(-2)`
Hence, `F = 50 + 20 xx 5 = 150 N `
`:. F_(max) = 150 N`
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CENGAGE PHYSICS ENGLISH-NEWTON'S LAWS OF MOTION 2-Exercise 7.2
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