A force $F$ acting on an object varies with distance $x$ as shown here. The force is in $N$ and $x$ in…
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:
$18.0 \mathrm{~J}$
$13.5 \mathrm{~J}$
$9.0 \mathrm{~J}$
$4.5 \mathrm{~J}$
Solution
Work done $=$ Area under $=f-x$ curve.
$\begin{aligned}
& =\text { area of trapezium }=\frac{1}{2} \times(6+3) \times 3 \\
& =\frac{27}{2}=13.5 \mathrm{~J}
\end{aligned}$
