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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 $20 \,m$ isPhysics Question Image

ManipalManipal 2009Work, Energy and Power

Solution:

Word done $W=$ Area $A B C E F D A$
Area $A B C D+$ Area $C E F D$
image
$=\frac{1}{2} \times(15+10) \times 10+\frac{1}{2} \times(10+20) \times 5$
$=125+75=200\, J$