Q.
The relationship between the force F and position a: of a body is as shown in figure. The work done in displacing the body from x = 1 m to x = 5 m will be
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Solution:
Work done = area enclosed by F-x graph.
= area of OAM + area of ABNM + area of CDEN area of EFGH + area of HIJ $ =\frac{1}{2}\times 1\times 10+1\times 10+1\times 5-1\times 5+\frac{1}{2}\times 1\times 10 $ $ =5+10+5-5+5=20J $