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Q.
The work done by a force acting on a body is as shown in the graph. The total work done in covering an initial distance of $20\, m$ is
Work, Energy and Power
Solution:
Work done $W=$ area under $F-S$ graph
$=$ area of trapezium $A B C D+$ area of trapezium $C E F D$
$=\frac{1}{2} \times(10+15) \times 10+\frac{1}{2} \times(10+20) \times 5$
$=125+75=200\, J$
