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.

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

B)How much work must be done to compress this spring by 0.050 m from its unstretched length?

C)Is it easier to stretch or compress this spring?

Answer 1

Hi,

Hope you are doing well.

PART A:

Here,the spring doesn't obey Hooke's law and the force is variable. So to find the work done for stretching or compressing the spring, we have to integrate all the way from zero to x.

Work done for stretching the spring is given by,

W-İ F.da

Here, Given that,
r = +0.050 m

$F=kx-bx^{2}+cx^{3}$

$\therefore$ Work done on the spring for stretching is given by,

0.050 r2r3) dr

0.050 0.050 0.050 kr.dr ca3d

$W=k\left [\frac{x^{2}}{2} \right ]_{0}^{0.050}-b\left [ \frac{x^{3}}{3} \right ]_{0}^{0.050}+c\left [ \frac{x^{4}}{4} \right ]_{0}^{0.050}$

W-k (1.25x 10-3 m?)-b (6.25x 10-5 rn3) + c (3. 125 10-6 ที่

Given that,

$k=100\;N/m$

b = 700 N/ n ?

$c=12000\;N/m^{3}$

W = 100 Λ7m × (1.25 × 10-3 rn 2)-700 yrn 2 × (6.25 × 10-5 ㎡) + 12000 N/m3 x (3.125 x 10-6 m)

W 0.125 J-0.04375 J 0.0375 J

$\mathbf{\therefore W=0.11875\:J}$

Similarly,

PART B:

Work done to compress the spring is given by,

W2-F.dr

Here, Given that, $x=-0.050\;m$

$F=kx-bx^{2}+cx^{3}$

$\therefore$ Work done for compressing is given by,

$W_{2}=\int_{0}^{-0.050}\left ( kx-bx^{2}+cx^{3} \right )dx$

$W_{2}=\int_{0}^{-0.050}kx.dx-\int_{0}^{-0.050}bx^{2}dx+\int_{0}^{-0.050}cx^{3}dx$

3 ך0.050-

$\therefore W_{2}=k\left (1.25\times10^{-3}\;m^{2} \right )+b\left (6.25\times10^{-5}\;m^{3} \right )+c\left (3.125\times10^{-6}\;m^{4} \right )$

Given that,

$k=100\;N/m$

b = 700 N/ n ?

$c=12000\;N/m^{3}$

$\therefore W_{2}=100\;N/m\times\left (1.25\times10^{-3}\;m^{2} \right )+700\;N/m^{2}\times\left (6.25\times10^{-5}\;m^{3} \right )+12000\;N/m^{3}\times\left (3.125\times10^{-6}\;m^{4} \right )$

$\therefore W_{2}=0.125\;J+0.04375\;J+0.0375\:J$

W20.1625 J

PART C:

We have got that work done for compressing the spring (W₂) is greater than work done for stretching the spring (W).

$\mathbf{ie;}\;W_{2}>W$

$\therefore$ It is easier to stretch the spring.

Hope this helped for your studies. Keep learning. Have a good day.

Feel free to clear any doubts at the comment section.

Please don't forget to give a thumbs up.

Thank you. :)

Answer 2

Hi,

Hope you are doing well.

PART A:

Here,the spring doesn't obey Hooke's law and the force is variable. So to find the work done for stretching or compressing the spring, we have to integrate all the way from zero to x.

Work done for stretching the spring is given by,

W-İ F.da

Here, Given that,
r = +0.050 m

$F=kx-bx^{2}+cx^{3}$

$\therefore$ Work done on the spring for stretching is given by,

0.050 r2r3) dr

0.050 0.050 0.050 kr.dr ca3d

$W=k\left [\frac{x^{2}}{2} \right ]_{0}^{0.050}-b\left [ \frac{x^{3}}{3} \right ]_{0}^{0.050}+c\left [ \frac{x^{4}}{4} \right ]_{0}^{0.050}$

W-k (1.25x 10-3 m?)-b (6.25x 10-5 rn3) + c (3. 125 10-6 ที่

Given that,

$k=100\;N/m$

b = 700 N/ n ?

$c=12000\;N/m^{3}$

W = 100 Λ7m × (1.25 × 10-3 rn 2)-700 yrn 2 × (6.25 × 10-5 ㎡) + 12000 N/m3 x (3.125 x 10-6 m)

W 0.125 J-0.04375 J 0.0375 J

$\mathbf{\therefore W=0.11875\:J}$

Similarly,

PART B:

Work done to compress the spring is given by,

W2-F.dr

Here, Given that, $x=-0.050\;m$

$F=kx-bx^{2}+cx^{3}$

$\therefore$ Work done for compressing is given by,

$W_{2}=\int_{0}^{-0.050}\left ( kx-bx^{2}+cx^{3} \right )dx$

$W_{2}=\int_{0}^{-0.050}kx.dx-\int_{0}^{-0.050}bx^{2}dx+\int_{0}^{-0.050}cx^{3}dx$

3 ך0.050-

$\therefore W_{2}=k\left (1.25\times10^{-3}\;m^{2} \right )+b\left (6.25\times10^{-5}\;m^{3} \right )+c\left (3.125\times10^{-6}\;m^{4} \right )$

Given that,

$k=100\;N/m$

b = 700 N/ n ?

$c=12000\;N/m^{3}$

$\therefore W_{2}=100\;N/m\times\left (1.25\times10^{-3}\;m^{2} \right )+700\;N/m^{2}\times\left (6.25\times10^{-5}\;m^{3} \right )+12000\;N/m^{3}\times\left (3.125\times10^{-6}\;m^{4} \right )$

$\therefore W_{2}=0.125\;J+0.04375\;J+0.0375\:J$

W20.1625 J

PART C:

We have got that work done for compressing the spring (W₂) is greater than work done for stretching the spring (W).

$\mathbf{ie;}\;W_{2}>W$

$\therefore$ It is easier to stretch the spring.

Hope this helped for your studies. Keep learning. Have a good day.

Feel free to clear any doubts at the comment section.

Please don't forget to give a thumbs up.

Thank you. :)