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Q.
The relationship between the force $F$ and position $x$ of a body is as shown in the figure. The work done in displacing the body from $x = 1\,m$ to $x = 5\,m$ will be
Step1: Calculation of work done
As work done = area enclosed by F-x graph
Work done in displacing the body from $x =1 m$ to $x =5 m$ will be:
$W =$ Area of ABNM + Area of CDEN-Area of EFGH + Area of HIJ
(Area of EF GH is taken negative as the force is in negative direction)
$$
\Rightarrow W =(1 \times 10)+(1 \times 5)-(1 \times 5)+\left(\frac{1}{2} \times 10 \times 1\right)=15 J
$$
So, the work done in displacing the body from $x=1 m$ to $x=5 m$ will be $15 J$.
