A spring with spring constant k is ext | vedclass

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Question

(b) $F=-k x$

$d w=F \cdot d x$

$\int d x=\int_{0}^{x_1}-k x d x$

$w=-k \int_{0}^{x_1}xdx$

$=-k\left[\frac{x^{2}}{2}\right]_{0}^{x_{1}}$

Vedclass · Accepted Answer

$\frac{1}{2}kx_1^2$

Vedclass · Answer

(b) $F=-k x$

$d w=F \cdot d x$

$\int d x=\int_{0}^{x_1}-k x d x$

$w=-k \int_{0}^{x_1}xdx$

$=-k\left[\frac{x^{2}}{2}\right]_{0}^{x_{1}}$

$=-K\left[\frac{x_{1}^{2}}{2}\right]$

$w=-\frac{-k x_{1}^{2}}{2}$

$w=\frac{-1}{2} k x_{1}^{2}$

work  done $=-(w)$

$=-\left[\frac{-1}{2} k x_{1}^{2}\right]$

$=\frac{1}{2} k x_{1}^{2}$

A spring with spring constant $k $ is extended from $x = 0$to$x = {x_1}$. The work done will be

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