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Determine position of a particle when it turns back if total energy is $2 \mathrm{~J}$ , and the particle is moving under the potential energy function $\mathrm{V}(\mathrm{x})=\left(\mathrm{k} x^2 / 2\right)$ , where $\mathrm{K}=0.5(\mathrm{~N} / \mathrm{m})$
Option: 1
$\sqrt{8}$

Option: 2
$\sqrt{13}$

Option: 3
$4$

Option: 4
$3$

Answers (1)

best_answer

Potential energy function:

$\begin{aligned} & V(x)=\frac{\mathrm{k} x^2}{2} \\ & =\frac{(0.5)\left(x^2\right)}{2} \end{aligned}$

Total energy of the particle: $\mathrm{E}$ $=2 \mathrm{~J}$

The total energy of the particle is the sum of KE and PE.

$\mathrm{V}(\mathrm{x})=\frac{\mathrm{k} x^2}{2}$

$=\frac{(0.5)\left(x^2\right)}{2}$

$\mathrm{E}=\mathrm{K} . \mathrm{E}+\mathrm{P} . \mathrm{E}$

$2=0+\frac{(0.5)\left(x^2\right)}{2} \text { (At Turning } \mathrm{v}=0 \text { ) }$

$4=0.5 x^2$

$x=\sqrt{8} m$

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rishi.raj

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