Question

A bungee cord exerts a nonlinear elastic force of magnitude F(z) distance the cord is stretched, and k be done on the cord to stretch it 16.7 m? = kiz+k2 where r is the = 204 N/m and ka =0.233 N/m, How much work must be done on the code to stretch it 16.7m?

Answer 1

Answer:

The work done on the cord is $2.845\times10^{4}\ J$.

Explanation:

Given that,

The function

$F(z)=k_{1}z+k_{2}$

Spring constant

$K_{1}=204 N/m$

$K_{2}=0.233\ N/m$

Stretch length b = 16.7 m

We need to calculate the work done

Using formula of work done

$W=\int_{a}^{b}{F(z)}dz$

Put the value into the formula

$W=\int_{a}^{b}{k_{1}z+k_{2}}dz$

$W=\int_{a}^{b}{k_{1}zdz+k_{2}dz}$

On integrating

$W=({\dfrac{204\times z^2}{2}})_{0}^{16.7}+({0.233\times z})_{0}^{16.7}$

$W={\dfrac{204\times16.7^2}{2}+0.233\times16.7}$

$W=2.845\times10^{4}\ J$

Hence, The work done on the cord is $2.845\times10^{4}\ J$.