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

The work done to stretch the rubber band at a distance of x can be calculated by integrating the force function F(x) for x. However, the integral for this function is complex and cannot be solved analytically. To calculate the work numerically, divide the range of x into small intervals and approximate the integral using numerical integration methods.

Step-by-step explanation:

The work done to stretch the rubber band at a distance of x can be calculated by integrating the force function F(x) for x. In this case, the force function is given by F(x) = F0[L0−x/L0−L0²/(L0+x)²]. To find the work, you integrate this function concerning x over the desired range. However, the integral for this function is complex and cannot be solved analytically.

To calculate the work numerically, you can divide the range of x into small intervals and approximate the integral using numerical integration methods such as the trapezoidal rule or Simpson's rule. This will give you an estimate of the work needed to stretch the rubber band.

Answer 2

`F_0(x+(1)/(2L_0)x^2+(L^2_0)/(L_0+x)-L_0)`

Step-by-step explanation:

We are given that

Force exerted by a rubber band is given approximately by

`F=F_0((L_0+x)/(L_0)-(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((L_0+x)/(L_0)-(L^2_0)/((L_0+x)^2)dx`

`W=\int_(0)^(x)((F_0)/(L_0)(L_0+x)-(L^2_0F_0)/((L_0+x)^2))dx`

`W=(F_0)/(L_0)[L_0x+(x^2)/(2)]^(x)_(0)+F_0L^2_0[(1)/(L_0+x)]^(x)_(0)`

`W=(F_0)/(L_0)(L_0x+(x^2)/(2))+L^2_0F_0((1)/(L_0+x)-(1)/(L_0))`

`W=F_0(x+(1)/(2L_0)x^2+(L^2_0)/(L_0+x)-L_0)`