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A body of $0.5 \mathrm{~kg}$ moves along the positive $\mathrm{x}$-axis under the influence of a varying force $\mathrm{F}$ (in Newtons) as shown below.
If the speed of the object at $x=4 \mathrm{~m}$ in $3.16 \mathrm{~ms}^{-1}$ then its speed at $x=8 \mathrm{~m}$ is
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If the speed of the object at $x=4 \mathrm{~m}$ in $3.16 \mathrm{~ms}^{-1}$ then its speed at $x=8 \mathrm{~m}$ is
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Verified Answer
The correct answer is:
$6.8 \mathrm{~ms}^{-1}$
According to work-energy principle
$\begin{array}{l}
\mathrm{W}_{\mathrm{C}}+\mathrm{W}_{\mathrm{nc}}+\mathrm{W}_{\mathrm{ext}}=\Delta \mathrm{KE} \\
\int_{4}^{8} \mathrm{Fdx}=\frac{1}{2} \mathrm{mv}_{\mathrm{f}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{i}}^{2}
\end{array}$
$\frac{1}{2} \times 3 \times 8-\frac{1}{2} \times 1.5 \times 4=\frac{1}{2} \times \frac{1}{2}\left[v_{f}^{2}-(3.16)^{2}\right]$
$v_{f}=6.8 \mathrm{~m} / \mathrm{s}$
$\begin{array}{l}
\mathrm{W}_{\mathrm{C}}+\mathrm{W}_{\mathrm{nc}}+\mathrm{W}_{\mathrm{ext}}=\Delta \mathrm{KE} \\
\int_{4}^{8} \mathrm{Fdx}=\frac{1}{2} \mathrm{mv}_{\mathrm{f}}^{2}-\frac{1}{2} \mathrm{mv}_{\mathrm{i}}^{2}
\end{array}$
$\frac{1}{2} \times 3 \times 8-\frac{1}{2} \times 1.5 \times 4=\frac{1}{2} \times \frac{1}{2}\left[v_{f}^{2}-(3.16)^{2}\right]$
$v_{f}=6.8 \mathrm{~m} / \mathrm{s}$
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