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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 $x=1 \mathrm{~m}$ to $x=5 \mathrm{~m}$ will be
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The correct answer is:
$15 \mathrm{~J}$
The work done in a $F-x$ graph is calculated by the area under the graph. The given graph can be shown as
$\therefore$ Work done in displacing the object from $x=1 \mathrm{~m}$ to $x=5 \mathrm{~m}$ will be
$\begin{aligned}
W &=\text { area under } F-x \text { graph } \\
=& \text { area } A B J K+\text { areaCLDJ } \\
& \quad+\text { area } D G F E+\text { area } G H I \\
=&(10 \times 1)+(5 \times 1)+(-5 \times 1)+\left(\frac{1}{2} \times 1 \times 10\right) \\
=& 10+5-5+5=15 \mathrm{~J}
\end{aligned}$
$\therefore$ Work done in displacing the object from $x=1 \mathrm{~m}$ to $x=5 \mathrm{~m}$ will be
$\begin{aligned}
W &=\text { area under } F-x \text { graph } \\
=& \text { area } A B J K+\text { areaCLDJ } \\
& \quad+\text { area } D G F E+\text { area } G H I \\
=&(10 \times 1)+(5 \times 1)+(-5 \times 1)+\left(\frac{1}{2} \times 1 \times 10\right) \\
=& 10+5-5+5=15 \mathrm{~J}
\end{aligned}$
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