Answer:
Here, \[U=A{{x}^{4}}\]
\[{{F}_{x}}=-\frac{dU}{dx}=\frac{d}{dx}\left(
A{{x}^{4}} \right)=-4A{{x}^{3}}\]
When \[x=-0.8m\]then
\[{{F}_{x}}=-4\times 1.2{{\left( -0.8 \right)}^{3}}=2.46N\]
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