AP EAMCET · PHYSICS · Work Power Energy
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 \(100 \mathrm{~J}\)
- D \(8 \mathrm{~J}\)
Answer & Solution
Correct Answer
(D) \(8 \mathrm{~J}\)
Step-by-step Solution
Detailed explanation
Work done \(=\int \overrightarrow{\mathrm{F}} \cdot \overrightarrow{\mathrm{dx}}\) \(=\int_0^2 \mathrm{Kx}^3 \mathrm{dx}=2 \times\left[\frac{\mathrm{x}^4}{4}\right]_0^2=\frac{2}{4}\left[2^4-0^4\right]=8 \mathrm{~J}\)
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