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A block of mass of 2kg placed on the floor of on an elevator. Acceleration of elevator is $6 \hat{i} + 7 \hat{j} {m/s}^{2}$ . If $μ = 0 .5, g = 10 {ms}^{- 2}$ and horizontal, vertically upward directions are taken as +ve x, +ve y axes. Then

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Friction is 12N

N = 34 N

Friction is 17 N

Maximum friction is 17N

answer is D, A, B.

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Detailed Solution

When the block is at rest

F - f = 0

$F = f$

$f = ma$

$= 2 (6) = 12 N$

When elevator moving vertically upward $N = m (g + a)$

$= 2 (10 + 7)$

$= 34 N$

$f_{\max} = μ_{s} N = 0.5 \times 34 = 17 N$

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