# 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. No, you left out the vertical acceleration
a up = 1/1.4 = .714 m/s^2

F = m a
F up - mg = m a
Fup = m (g+a)
Fup = 6.3 (9.8+.714)
Fup = 66.24 N

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posted by Damon

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