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An elevator starts from rest with a constant upward acceleration. It moves 2 m in the first 0.6 s. A passenger in the elevator is holding a 3 kg package by a vertical string. The tension in the string during acceleration is (Take g=9.8 m/s2)


  1. 60.7 N

  2. 63.0 N

  3. 62.7 N

  4. 61.7 N


Solution

The correct option is C

62.7 N


As we know, the equation of motion: 
      S=ut+12at2
Substituting the values, S=2, u=0 and t = 0.6, we get:
  2=0+12×a×0.6×0.6
a=4×1006×6=1009m/s2

Now, balancing the forces:
Tmg=ma
T=m(g+a)
=3×(9.8+1009)=62.7 N

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