Consider a force $\mathrm{F}=\mathrm{Kx}^3$, which acts on a particle at rest. The work done by the force for displacement of $2 \mathrm{~m}$ is $\left(\mathrm{K}=2 \mathrm{Nm}^{-3}\right)$

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Aicharya · Accepted Answer

Work done $=\int \overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{dx}}$ $=\int_0^2 \mathrm{Kx}^3 \mathrm{dx}=2 	imes\left[\frac{\mathrm{x}^4}{4}ight]_0^2=\frac{2}{4}\left[2^4-0^4ight]=8 \mathrm{~J}$

Consider a force $\mathrm{F}=\mathrm{Kx}^3$, which acts on a particle at rest. The work done by the force…

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