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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? |
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Answer» Solution :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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