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An elevator cab of mass m=500 kg is desc...

An elevator cab of mass `m=500` kg is descending with speed `v_(i)=4.0m//s` when its supporting cable begins to slip, allowing it to fall with constant acceleration `vec(a)=vec(g)//5` (Fig. 8-11a).
(c) What is the net work W done on the cab during the fall?

Text Solution

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Calculation: The net work is the sum of the works done by the forces acting on the cab:
`W=W_(g)+W_(T)=5.88xx10^(4)J-4.70xx10^(4)J`
`=1.18xx10^(4)J~~12kJ`.
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