Correct Answer - B
Acceleration of the platform
`overset(vec)(a_(p) ) - (doverset(vec)v)/(dt) - 2hati + hat(j) + 3hat(k)`
Horizontal force on the block
Normal force on the block
`a_(H) = sqrt(4 + 1) = sqrt(5) m//s^(2)`
`a_(v)= 3 m//s^(2)`
Normal force on the block
`N = m (g + a_(v)) = 1 xx 13 = 13N`
Maximum acceleration that friction can positive
`a_(mass) = mu(g + a_(y)) = 0.2 xx 13 = 206 m//s^(2)`
`:. a_(max) gt a_(H)`
`:.` Value of frication force on the block
`f = ma_(H) = sqrt(5)N`
`:.` Force by the platform on the block is
`F = sqrt(N^(2) t^(2)) = sqrt(169 + 5) = sqrt(174)N`