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Q.

Five forces F1→, F2→, F3→, F4→ and F5→ are acting on a particle of mass 2.0 kg so that it is moving with 4 m/s2 in east direction. If F1→ force is removed, then the acceleration becomes 7 m/s2 in north, then the acceleration of the block if only F1→ is acting will be:

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a

16 m/s2

b

65  m/s2

c

260 m/s2

d

33 m/s2

answer is .

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Detailed Solution

F1→+F2→ + F3→+F4→ + F5→  = 2(4i^)         -------(i)and F2→+F3→+F4→+F5→ = 2(7j^)      ------(ii)From (i) and (ii), F1→ = 8i^ - 14j^a1→ = F1→m = 4i^ -7 j^⇒  a1 = 16+49 = 65 m/s2
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