Question

Consider a spring that does not obey Hooke's law very faithfully. One end of the spring is fixed. To keep the spring stretched or compressed an amount x, a force along the x-axis with x-component Fx=kx−bx2+cx3 must be applied to the free end. Here k=100N/m, b=700N/m2, and c=12000N/m3. Note that x>0 when the spring is stretched and x<0 when it is compressed.

Answer 1

Answer:

The work done will be 0.115 J.

Explanation:

Given that,

$k=100\ N/m$

$b=700\ N/m^2$

$c=12000\ N/m^3$

A force along the x-axis with x-component,

$F(x)=kx-bx^2+cx^3$

Suppose, How much work must be done to stretch this spring by 0.050 m from its unstretched length?

We need to calculate the work done

Using formula of work done

$W(x)=\int{F(x)dx}$

Put the value into the formula

$W(x)=\int{(kx-bx^2+cx^3)dx}$

$W(x)=\dfrac{kx^2}{2}-\dfrac{bx^3}{3}+\dfrac{cx^4}{4}$

Put the value of k,b,c and x

$W(0.050)=\dfrac{100\times(0.050)^2}{2}-\dfrac{700\times(0.050)^3}{3}+\dfrac{12000\times(0.050)^4}{4}$

$W(0.050)=0.115\ J$

Hence, The work done will be 0.115 J.