The engine of a small lift performs work on the cabin in the following manner: • to :…

PHYSICS · Multi-part · CBSE · Physics · Work, Energy and Power · Work, Energy and Power

Problem

The engine of a small lift performs work on the cabin in the following manner: • t=0 st = 0\,\text{s} to t=2 st = 2\,\text{s} : work rises uniformly from 0 J0\,\text{J} to 80 J80\,\text{J}. • t=2 st = 2\,\text{s} to t=4 st = 4\,\text{s} : work rises uniformly from 80 J80\,\text{J} to 160 J160\,\text{J}. • t=4 st = 4\,\text{s} to t=5 st = 5\,\text{s} : work remains constant at 160 J160\,\text{J}. • t=5 st = 5\,\text{s} to t=7 st = 7\,\text{s} : work decreases uniformly to 120 J120\,\text{J}. Based on this data, answer the following. (i) Sketch a neat work–time graph for the entire 00–7 s7\,\text{s} period. (ii) Calculate the power developed in each of the four time intervals. (iii) Determine the average power of the engine for the whole 00–7 s7\,\text{s} duration. (iv) Explain, using the graph, why the power is zero during the third interval.

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