Question #727

An elevator starts from rest with a constant
upward acceleration and moves 1m in the first
1.5 s. A passenger in the elevator is holding a
9.6 kg bundle at the end of a vertical cord.
The acceleration of gravity is 9.8 m/s2 .
What is the tension in the cord as the ele-
vator accelerates?
Answer in units of N.

Expert's answer

From the equation of the motion we can get the acceleration of the elevator:h = x_{0}+ v_{0}t + at^{2}/2; v_{0}=0, x_{0}=0;thusa = 2h/t^{2} = 2*1/1.5^{2} = 0.9 m/s^{2}. Then we can find the tension in the cord:

T = mg + ma = 9.6 (9.8 + 0.9) = 102.7 N

Answer: 102.7 N.

T = mg + ma = 9.6 (9.8 + 0.9) = 102.7 N

Answer: 102.7 N.

## Comments

Assignment Expert16.02.12, 16:06Because we have gravity force and elevator acting on the object and gravity also contributes into value of tension

Lila16.02.12, 02:41why does T=mg+ma instead of just T=ma?

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