Question

derive an expression to find the work done by the spring when the spring is to The spring is pulled through a distance of xm.​

Answer 1

Explanation:

if the spring is pulled from x1 to x2, you get the above expression

Answer 2

To derive:-

Formula for work done by a spring when spring is pulled through a distance of x m. $(\sf W.)$

Answer:-

Let the initial position of the spring be $\sf x = x_i,$ and the final position be $\sf x = x_f.$

From Hooke's law,

$\sf Spring\;force = F_s = - kx\quad\dots (1)$

$\sf dW = F\:dx$

$\longrightarrow \displaystyle \sf \int_{0}^{W} dW = \int_{x_i}^{x_f} F\:dx$

$\longrightarrow \displaystyle \sf W = \int_{x_i}^{x_f} -kx\:dx\quad [From\:(1)]$

$\longrightarrow \displaystyle \sf W = \bigg[-k\; \; \dfrac{x^{(1+1)}}{1+1}\bigg]_{x_i}^{x_f}$

$\longrightarrow \displaystyle \sf W = \bigg[-\dfrac{1}{2}kx^2\bigg]_{x_i}^{x_f}$

$\longrightarrow \sf W = -\dfrac{1}{2}k\big(x_f { }^2 - x_i { }^2 \big)$

Now, as we have to find work done from pulling a spring through a distance of x metres considering from its natural position,

$\quad \quad\bullet \sf x_i = 0$

$\quad \quad \bullet \sf x_f = x$

$\longrightarrow \sf W = - \dfrac{1}{2}k\big(x^2 - 0^2\big)$

$\longrightarrow \underline{\underline{ \sf{\green{W = - \dfrac{1}{2}kx^2}}}}$