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Q.

Five forces F→1,F→2,F→3,F→4 and F→5, are acting on a particle of mass 2.0 kg so that it is moving with 4 m/s2 in east direction. If F→1 force is removed, then the acceleration becomes 7 m/s2 in north, then the acceleration of the block if only F→1 is acting will be

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a

16m/s2

b

65m/s2

c

260m/s2

d

33m/s2

answer is B.

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Detailed Solution

F→1+F→2+F→3+F→4+F→5=2(4i^)     .......(i)and    F→2+F→3+F→4+F→5=2(7j^)   ........(ii) From (i) and (ii), F→1=8i^−14j^a→1=F→1m=4i^−7j^⇒ a1=16+49=65m/s2
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