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After coming to a stop, the elevator then accelerates at 3.0m/s? as it begins to move downward.Calculate the tension in the cable.

1 Answer

6 votes

We know that

• The acceleration is 3.0 m/s^2.

Using Newton's Second Law, we have


mg-T=ma

The two forces involved are the weight of the elevator and the tension force. Let's solve got T.


\begin{gathered} mg-ma=T \\ T=m(g-a) \end{gathered}

The mass of the elevator is m = 5000kg, g = 9.8 m/s^2, and a = 3 m/s^2.


\begin{gathered} T=5000\operatorname{kg}(9.8((m)/(s^2))-3((m)/(s^2))) \\ T=5000\operatorname{kg}\cdot6.8((m)/(s^2)) \\ T=5000\operatorname{kg}\cdot7((m)/(s^2)) \\ T=35,000N \end{gathered}

Therefore, the tension in the cable is 35,000 N.

User Butterbrot
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