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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
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The correct answer is:
$52.5 \times 10^{-4}$
Work done $=\int \mathrm{Fdx}=\int_{0.02}^{0.05} \mathrm{~m}(\beta \mathrm{x}) \mathrm{dx}$
$\begin{aligned} & =\frac{1 \times 5}{2} \times\left[\mathrm{x}^2\right]_{0.02}^{0.05} \\ & =\frac{5}{2}\left[25 \times 10^{-4}-4 \times 10^{-4}\right] \\ & =\frac{5}{2} \times 21 \times 10^{-4} \\ & =52.5 \times 10^{-4} \text { Joule. }\end{aligned}$
$\begin{aligned} & =\frac{1 \times 5}{2} \times\left[\mathrm{x}^2\right]_{0.02}^{0.05} \\ & =\frac{5}{2}\left[25 \times 10^{-4}-4 \times 10^{-4}\right] \\ & =\frac{5}{2} \times 21 \times 10^{-4} \\ & =52.5 \times 10^{-4} \text { Joule. }\end{aligned}$
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