New Notes Blog: Eighteen foot-pounds of work is required to stretch a spring 4 inches from its natural length. Find the work required to stretch the spring an...

Wednesday, 13 November 2013

Eighteen foot-pounds of work is required to stretch a spring 4 inches from its natural length. Find the work required to stretch the spring an...

Hooke's law is written as $\displaystyle{F}={k}{x}$

where:

$\displaystyle{F}$ = force

$\displaystyle{k}$ = proportionality constant or spring constant

$\displaystyle{x}$ = length displacement from its natural length

Apply Hooke's Law to the integral application for work: $\displaystyle{W}={\int_{{a}}^{{b}}}{F}{\left.{d}{x}\right.}$ , we get:

$\displaystyle{W}={\int_{{a}}^{{b}}}{k}{x}{\left.{d}{x}\right.}$

$\displaystyle{W}={k}\cdot{\int_{{a}}^{{b}}}{x}{\left.{d}{x}\right.}$

Apply Power rule for integration: $\displaystyle\int{{x}}^{{n}}{\left({\left.{d}{x}\right.}\right)}=\frac{{{x}}^{{{n}+{1}}}}{{{n}+{1}}}.$

$\displaystyle{W}={k}\cdot\frac{{{x}}^{{{1}+{1}}}}{{{1}+{1}}}{{\mid}_{{a}}^{{b}}}$

$\displaystyle{W}={k}\cdot\ldots$

Hooke's law is written as $\displaystyle{F}={k}{x}$

where:

$\displaystyle{F}$ = force

$\displaystyle{k}$ = proportionality constant or spring constant

$\displaystyle{x}$ = length displacement from its natural length

Apply Hooke's Law to the integral application for work: $\displaystyle{W}={\int_{{a}}^{{b}}}{F}{\left.{d}{x}\right.}$ , we get:

$\displaystyle{W}={\int_{{a}}^{{b}}}{k}{x}{\left.{d}{x}\right.}$

$\displaystyle{W}={k}\cdot{\int_{{a}}^{{b}}}{x}{\left.{d}{x}\right.}$

Apply Power rule for integration: $\displaystyle\int{{x}}^{{n}}{\left({\left.{d}{x}\right.}\right)}=\frac{{{x}}^{{{n}+{1}}}}{{{n}+{1}}}.$

$\displaystyle{W}={k}\cdot\frac{{{x}}^{{{1}+{1}}}}{{{1}+{1}}}{{\mid}_{{a}}^{{b}}}$

$\displaystyle{W}={k}\cdot\frac{{{x}}^{{2}}}{{2}}{{\mid}_{{a}}^{{b}}}$

From the required work 18 ft-lbs, note that the units has "ft" instead of inches. To be consistent, apply the conversion factor: 12 inches = 1 foot then:

$\displaystyle{4}$ inches = $\displaystyle\frac{{1}}{{3}}$ ft

$\displaystyle{7}$ inches = $\displaystyle\frac{{7}}{{12}}$ ft

To solve for k, we consider the initial condition: W =18 ft-lbs to stretch a spring $\displaystyle{4}$ inches or $\displaystyle\frac{{1}}{{3}}$ ft from its natural length. Stretching $\displaystyle\frac{{1}}{{3}}$ ft of it natural length implies the boundary values: $\displaystyle{a}={0}$ to $\displaystyle{b}=\frac{{1}}{{3}}$ ft.

Applying $\displaystyle{W}={k}\cdot\frac{{{x}}^{{2}}}{{2}}{{\mid}_{{a}}^{{b}}}$ , we get:

$\displaystyle{18}={k}\cdot\frac{{{x}}^{{2}}}{{2}}{{\mid}_{{0}}^{{\frac{{1}}{{3}}}}}$

Apply definite integral formula: $\displaystyle{F}{\left({x}\right)}{{\mid}_{{a}}^{{b}}}={F}{\left({b}\right)}-{F}{\left({a}\right)}$ .

$\displaystyle{18}={k}{\left[\frac{{{\left(\frac{{1}}{{3}}\right)}}^{{2}}}{{2}}-\frac{{{\left({0}\right)}}^{{2}}}{{2}}\right]}$

$\displaystyle{18}={k}\cdot{\left[\frac{{\frac{{1}}{{9}}}}{{2}}-{0}\right]}$

$\displaystyle{18}={k}\cdot{\left[\frac{{1}}{{18}}\right]}$

$\displaystyle{18}=\frac{{k}}{{18}}$

$\displaystyle{k}={18}\cdot{18}$

$\displaystyle{k}={324}$

To solve for the work need to stretch the spring with additional 3 inches, we plug-in: $\displaystyle{k}={324}$ , $\displaystyle{a}=\frac{{1}}{{3}},$ and $\displaystyle{b}=\frac{{7}}{{12}}$ on $\displaystyle{W}={k}\cdot\frac{{{x}}^{{2}}}{{2}}{{\mid}_{{a}}^{{b}}}$ .

Note that stretching "additional 3 inches" from its initial stretch of 4 inches is the same as stretching 7 inches from its natural length.

$\displaystyle{W}={324}\cdot\frac{{{x}}^{{2}}}{{2}}{{\mid}_{{{\left(\frac{{1}}{{3}}\right)}}}^{{{\left(\frac{{7}}{{12}}\right)}}}}$

$\displaystyle{W}={324}{\left[\frac{{{\left(\frac{{7}}{{12}}\right)}}^{{2}}}{{2}}-\frac{{{\left(\frac{{1}}{{3}}\right)}}^{{2}}}{{2}}\right]}$

$\displaystyle{W}={324}{\left[\frac{{49}}{{288}}-\frac{{1}}{{18}}\right]}$

$\displaystyle{W}={324}{\left[\frac{{11}}{{96}}\right]}$

$\displaystyle{W}=\frac{{297}}{{8}}$ or $\displaystyle{37.125}$ ft-lbs

New Notes Blog

Wednesday, 13 November 2013

Eighteen foot-pounds of work is required to stretch a spring 4 inches from its natural length. Find the work required to stretch the spring an...

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In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?

Wednesday, 13 November 2013

Eighteen foot-pounds of work is required to stretch a spring 4 inches from its natural length. Find the work required to stretch the spring an...

No comments:

Post a Comment

In &quot;By the Waters of Babylon,&quot; under the leadership of John, what do you think the Hill People will do with their society?

In "By the Waters of Babylon," under the leadership of John, what do you think the Hill People will do with their society?