A body of mass 8kg is moved by a force F=3x N,where is the distance covered. Initial position is x=2 m and the final position is x=10 m.The initial speed is 0.0m/s.The final speed is
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a
6 m/s
b
12 m/s
c
18 m/s
d
14 m/s
answer is A.
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Detailed Solution
Increment in kinetic energy = work done⇒12m(v2−u2)=∫x1x2F.dx=∫210(3x) dx ⇒12mv2=32[x2]210=32[100−4] ⇒12×8×v2=32×96⇒v=6m/s