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Potential energy of a body of mass $1 \mathrm{~kg}$ free to move along $X$-axis is given by $U(x)=\left(\frac{x^2}{2}-x\right) J$. If the total mechanical energy of the body is $2 \mathrm{~J}$, then the maximum speed of the body is (Assume only conservative force acts on the body)
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Verified Answer
The correct answer is:
$\sqrt{5} \mathrm{~ms}^{-1}$
Key Idea Total mechanical energy of a system is the addition of potential and kinetic energy
$$
E=U+\mathrm{KE}
$$
Here, mass of the body, $m=1 \mathrm{~kg}$,
$$
E_{\text {mech }}=2 \mathrm{~J}
$$
So, for $U_{\min }=\frac{d U(x)}{d x}=\frac{d}{d x}\left[\frac{x^2}{2}-x\right]=0$
$$
\begin{aligned}
\Rightarrow & x-1 & =0 \\
\Rightarrow & x & =1
\end{aligned}
$$
Hence, $U_{\min }=\frac{(1)^2}{2}-1=-\frac{1}{2}$
$$
\text { Kinetic energy, } \mathrm{KE}=\frac{1}{2} m v^2=\frac{v^2}{2} \quad(\because m=1 \mathrm{~kg})
$$
Now, putting the values in above expression,
$$
\begin{aligned}
& \Rightarrow & 2 & =-\frac{1}{2}+\frac{v^2}{2} \\
& \Rightarrow & v^2 & =5 \\
& \Rightarrow & v & =\sqrt{5} \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
Hence, the correct option is (a).
$$
E=U+\mathrm{KE}
$$
Here, mass of the body, $m=1 \mathrm{~kg}$,
$$
E_{\text {mech }}=2 \mathrm{~J}
$$
So, for $U_{\min }=\frac{d U(x)}{d x}=\frac{d}{d x}\left[\frac{x^2}{2}-x\right]=0$
$$
\begin{aligned}
\Rightarrow & x-1 & =0 \\
\Rightarrow & x & =1
\end{aligned}
$$
Hence, $U_{\min }=\frac{(1)^2}{2}-1=-\frac{1}{2}$
$$
\text { Kinetic energy, } \mathrm{KE}=\frac{1}{2} m v^2=\frac{v^2}{2} \quad(\because m=1 \mathrm{~kg})
$$
Now, putting the values in above expression,
$$
\begin{aligned}
& \Rightarrow & 2 & =-\frac{1}{2}+\frac{v^2}{2} \\
& \Rightarrow & v^2 & =5 \\
& \Rightarrow & v & =\sqrt{5} \mathrm{~m} / \mathrm{s}
\end{aligned}
$$
Hence, the correct option is (a).
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