New Notes Blog: Use Hooke's Law to determine the variable force in the spring problem. A force of 5 inches compresses a 15-inch spring a total of 3 inches. How...

Wednesday, 15 July 2015

Use Hooke's Law to determine the variable force in the spring problem. A force of 5 inches compresses a 15-inch spring a total of 3 inches. How...

Hooke's law states that a force is needed to stretch or compress a spring by a distance of x. The force is proportional to the distance x. is written as $\displaystyle{F}={k}{x}$

where:

F = force

k = proportionality constant or spring constant

x= length displacement from its natural length

Applying the given variable force: $\displaystyle{F}={5}$ to compress a $\displaystyle{15}$ -inch spring a total of $\displaystyle{3}$ inches, we get:

$\displaystyle{F}={k}{x}$

$\displaystyle{5}={k}\cdot{3}$

$\displaystyle{k}=\frac{{5}}{{3}}$

Plug-in $\displaystyle{k}\ldots$

Hooke's law states that a force is needed to stretch or compress a spring by a distance of x. The force is proportional to the distance x. is written as $\displaystyle{F}={k}{x}$

where:

F = force

k = proportionality constant or spring constant

x= length displacement from its natural length

Applying the given variable force: $\displaystyle{F}={5}$ to compress a $\displaystyle{15}$ -inch spring a total of $\displaystyle{3}$ inches, we get:

$\displaystyle{F}={k}{x}$

$\displaystyle{5}={k}\cdot{3}$

$\displaystyle{k}=\frac{{5}}{{3}}$

Plug-in $\displaystyle{k}=\frac{{5}}{{3}}$ on Hooke's law, we get:

$\displaystyle{F}={\left(\frac{{5}}{{3}}\right)}{x}$

Works is done when a force is applied to move an object to a new position, It can be defined with formula: $\displaystyle{W}={F}\cdot\delta{x}$ where:

$\displaystyle{F}$ = force or ability to do work.

$\displaystyle\delta{x}$ = displacement as

With variable force function: $\displaystyle{F}{\left({x}\right)}={\left(\frac{{5}}{{3}}\right)}{x}$ , we set-up the integral application for work as:

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

$\displaystyle{W}={\int_{{0}}^{{7}}}{\left(\frac{{5}}{{3}}\right)}{x}{\left.{d}{x}\right.}$

Apply basic integration property: $\displaystyle\int{c}\cdot{f{{\left({x}\right)}}}{\left.{d}{x}\right.}={c}\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}.$

$\displaystyle{W}={\left(\frac{{5}}{{3}}\right)}{\int_{{0}}^{{7}}}{x}{\left.{d}{x}\right.}$

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

$\displaystyle{W}={\left(\frac{{5}}{{3}}\right)}\cdot\frac{{{x}}^{{{1}+{1}}}}{{{1}+{1}}}{{\mid}_{{0}}^{{7}}}$

$\displaystyle{W}={\left(\frac{{5}}{{3}}\right)}\cdot\frac{{{x}}^{{2}}}{{2}}{{\mid}_{{0}}^{{7}}}$

$\displaystyle{W}=\frac{{{5}{{x}}^{{2}}}}{{6}}{{\mid}_{{0}}^{{7}}}$

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

$\displaystyle{W}=\frac{{{5}{{\left({7}\right)}}^{{2}}}}{{6}}-\frac{{{5}{{\left({0}\right)}}^{{2}}}}{{6}}$

$\displaystyle{W}=\frac{{245}}{{6}}-{0}$

$\displaystyle{W}=\frac{{245}}{{6}}$ or $\displaystyle{40.83}$ inch-lbs

New Notes Blog

Wednesday, 15 July 2015

Use Hooke's Law to determine the variable force in the spring problem. A force of 5 inches compresses a 15-inch spring a total of 3 inches. How...

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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, 15 July 2015

Use Hooke's Law to determine the variable force in the spring problem. A force of 5 inches compresses a 15-inch spring a total of 3 inches. How...

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?