Question

The force exerted by a rubber band is given approximately by F=F0[L0−xL0−L20(L0+x)2]F=F0[L0−xL0−L02(L0+x)2] where L0L0 is the unstretched length, xx is the stretch, and F0F0 is a constant.Find the work needed tostretch the rubber band the distance, x.

Answer 1

Answer:

$F_0(x+\frac{1}{2L_0}x^2+\frac{L^2_0}{L_0+x}-L_0)$

Explanation:

We are given that

Force exerted by a rubber band is given approximately by

$F=F_0(\frac{L_0+x}{L_0}-\frac{L^2_0}{(L_0+x)^2})$

Where $L_0$=Unstretched  length

x=Stretch length

$F_0$=Constant

We have to find the work needed to stretch the rubber band the distance x.

Work done=$\int_{0}^{x}Fdx$

$W=\int_{0}^{x}F_0(\frac{L_0+x}{L_0}-\frac{L^2_0}{(L_0+x)^2}dx$

$W=\int_{0}^{x}(\frac{F_0}{L_0}(L_0+x)-\frac{L^2_0F_0}{(L_0+x)^2})dx$

$W=\frac{F_0}{L_0}[L_0x+\frac{x^2}{2}]^{x}_{0}+F_0L^2_0[\frac{1}{L_0+x}]^{x}_{0}$

$W=\frac{F_0}{L_0}(L_0x+\frac{x^2}{2})+L^2_0F_0(\frac{1}{L_0+x}-\frac{1}{L_0})$

$W=F_0(x+\frac{1}{2L_0}x^2+\frac{L^2_0}{L_0+x}-L_0)$