AP EAMCET · PHYSICS · Work Power Energy
A \(1 \mathrm{~kg}\) box placed at the origin starts sliding along \(\mathrm{x}\)-axis under the action of a force \(\vec{F}=F \hat{i}\). Its acceleration as a function of \(x\) is given by \(a(x)=\beta . x\) where \(\beta=5 \mathrm{~s}^{-2}\). The work done by \(\overrightarrow{\mathrm{F}}\) in moving the box from \(\mathrm{x}=2 \mathrm{~cm}\) to \(\mathrm{x}=5 \mathrm{~cm}\) in joule is
- A \(52.5 \times 10^{-4}\)
- B \(105.5 \times 10^{-4}\)
- C \(17.0 \times 10^{-4}\)
- D \(34.0 \times 10^{-4}\)
Answer & Solution
Correct Answer
(A) \(52.5 \times 10^{-4}\)
Step-by-step Solution
Detailed explanation
Work done \(=\int \mathrm{Fdx}=\int_{0.02}^{0.05} \mathrm{~m}(\beta \mathrm{x}) \mathrm{dx}\)…
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