Work done by Diff type of forces

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Question

The work done by external agent in stretching a spring of force constant k = 100 N/m from deformation $x_{1} = 10 to deformation x_{2} = 20 cm .$

Easy

Solution

$U_{2} = \frac{1}{2} {kx}_{2}^{2}$
$U_{1} = \frac{1}{2} {kx}^{2}_{1}$
$W_{ext} = ∆ U = U_{2} - U_{1}$
= $\frac{1}{2} {kx}_{2}^{2} - \frac{1}{2} {kx}_{1}^{2} = \frac{1}{2} k ({x_{2_{}}}^{2} - {x_{1}}^{2})$
= $\frac{1}{2} \times 100 \times ({(\frac{20}{100})}^{2} - {(\frac{10}{100})}^{2}) = 150 J$

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