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The potential energy for a force field $\overrightarrow{\mathrm{F}}$ is given by $U(x, y)=\cos (x+y)$. The force acting on a particle at position given by coordinates $(0, \pi / 4)$ is $-$
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Verified Answer
The correct answer is:
$\frac{1}{\sqrt{2}}(\hat{i}+\hat{j})$
$\quad \mathrm{F}_{\mathrm{x}}=-\frac{\partial \mathrm{U}}{\partial \mathrm{x}}=\sin (\mathrm{x}+\mathrm{y})$
$$
\begin{array}{l}
\mathrm{F}_{\mathrm{y}}=-\frac{\partial \mathrm{U}}{\partial \mathrm{x}}=\sin (\mathrm{x}+\mathrm{y}) \\
\left.\mathrm{F}_{\mathrm{x}}=\sin (\mathrm{x}+\mathrm{y})\right]_{(0, \pi / 4)}=\frac{1}{\sqrt{2}} \\
\left.\mathrm{~F}_{\mathrm{y}}=\sin (\mathrm{x}+\mathrm{y})\right]_{(0, \pi / 4)}=\frac{1}{\sqrt{2}} \\
\therefore \mathrm{F}=\frac{1}{\sqrt{2}}[\hat{\mathrm{i}}+\hat{\mathrm{j}}]
\end{array}
$$
$$
\begin{array}{l}
\mathrm{F}_{\mathrm{y}}=-\frac{\partial \mathrm{U}}{\partial \mathrm{x}}=\sin (\mathrm{x}+\mathrm{y}) \\
\left.\mathrm{F}_{\mathrm{x}}=\sin (\mathrm{x}+\mathrm{y})\right]_{(0, \pi / 4)}=\frac{1}{\sqrt{2}} \\
\left.\mathrm{~F}_{\mathrm{y}}=\sin (\mathrm{x}+\mathrm{y})\right]_{(0, \pi / 4)}=\frac{1}{\sqrt{2}} \\
\therefore \mathrm{F}=\frac{1}{\sqrt{2}}[\hat{\mathrm{i}}+\hat{\mathrm{j}}]
\end{array}
$$
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