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Block B of mass 2kg rests on block A of ...

Block `B` of mass `2kg` rests on block `A` of mass `10kg`. All surfaces are rought with the value of coefficient of friction as shown in the figure. Find the minimum force `F` that should be applied on block `A` to cause relative motion between `A` and `B`. `(g=10m//s^(2))`

Text Solution

Verified by Experts

The correct Answer is:
`48N`
`FBD` of `B` ,
`(a_(B))_(max)=(f_(max))/(m_(B))=mu_(S)g=2.5m//s^(2)`
FBD of combined system

`f_(k)=0.15(2+10)g=18N`
`F_(max)-f_(k)=(m_(A)+m_(B))(a_(B))_(max)`
`rArr F_(max)=t_(k)+12xx2.5=48N. Ans. 48N`.
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