# 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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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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