Q128 / 160 · AP EAMCET 2022 (05 Jul Shift 2) · Physics
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 $\left(\mathrm{K}=2 \mathrm{Nm}^{-3}\right)$
- A$10 \mathrm{~J}$
- B$4 \mathrm{~J}$
- C$100 \mathrm{~J}$
- Correct answerD$8 \mathrm{~J}$
Solution
Work done $=\int \overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{dx}}$
$=\int_0^2 \mathrm{Kx}^3 \mathrm{dx}=2 \times\left[\frac{\mathrm{x}^4}{4}\right]_0^2=\frac{2}{4}\left[2^4-0^4\right]=8 \mathrm{~J}$
Hence, the correct answer is option: 'D'.