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Two blocks a and B of mass m(A)=1kg and ...

Two blocks a and B of mass `m_(A)=1kg` and `m_(B)=3kg` are kept on the table as shown in figure the coefficient of friction between A and B is 0.2 and between B and the surface of the table is also 0.2 the maximum force F that can be applied on B horizontally, so that the block A does not slide over the block B is [Take `g=10m//s^(2)`]

Text Solution

Verified by Experts

The correct Answer is:
16
max. acceleration of 1 kg block
`= (f_("max"))/(1) = (mu xx 1 xx g)/(1) = 2m//s^(2)`
When maximum F is applied on 3 kg
`f_(k) = 0.2 xx 4 xx 10 = 8`
`F - 8 = (3 + 1) xx 2`
`F = 16N`
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