physics

An elevator starts from rest with a constant upward acceleration and moves 1m in the first 1.4 s. A passenger in the elevator is holding a 6.3 kg bundle at the end of a vertical chord. what is the tension in the chord as the elevator accelerates? The acceleration of gravity is 9.8 m/s^2. Answer in units of N

Im not sure if i solved this right, but here is my work:

6.3(9.8)=61.74 N

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asked by alexa
  1. a = 1/1.4 = .714 m/s^2 upward acceleration

    Total force on bundle = m(.714) = tension on string - m g

    so tension = m (.714 + 9.8) = 66.2 N

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    posted by Damon
  2. the acceleration is not right.
    you need to use this equation:
    Xf=Xi+Vi*t+(1/2)a*t^2
    Xi=0,Vi=0,t=1.4,Xf=1,a=?
    1=0+0(1.4)+(1/2)a(1.4^2)
    1(2)= a(1.96)
    2/1.96= a
    1.0204 = a

    Then you follow the rest of the equation
    T = m(g+a)

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    posted by john

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