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A force $F$ acting on an object varies with distance $x$ as shown here. The force is in $N$ and $x$ in $\mathrm{m}$. The work done by the force in moving the object $x=0$ to $x=6 \mathrm{~m}$ is:
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Verified Answer
The correct answer is:
$13.5 \mathrm{~J}$
Work done $=$ Area under $=f-x$ curve.
$$
\begin{aligned}
& =\text { area of trapezium }=\frac{1}{2} \times(6+3) \times 3 \\
& =\frac{27}{2}=13.5 \mathrm{~J}
\end{aligned}
$$
$$
\begin{aligned}
& =\text { area of trapezium }=\frac{1}{2} \times(6+3) \times 3 \\
& =\frac{27}{2}=13.5 \mathrm{~J}
\end{aligned}
$$
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