Physics, asked by Anonymous, 3 months ago

Hi! Please explain how to solve this step-by-step​

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Answered by Ekaro
17

Question :

A force F = -k/x² (x ≠ 0) acts on a particle in x-direction. Find the work done by this force in displacing the particle from x = + a to x = + 2a. Here, k is a positive constant.

Solution :

❖ Work done is measured as the product of force and displacement.

But we can't use formula directly to solve this question as force is variable. We can directly use formula only when force is said to be constant.

Here we have to integrate the given force equation from a to 2a.

\displaystyle\dag\:\underline{\boxed{\bf{\orange{W=\int\:F\cdot dx}}}}

\displaystyle\sf:\implies\:W=\int\:\left(-\dfrac{k}{x^2}\right)\cdot dx

We know that, 1 / x² = x‾²

\displaystyle\sf:\implies\:W=\int\:-kx^{-2}\cdot dx

Rule : \displaystyle\sf\int\:x^n\:dx=\dfrac{x^{n+1}}{n+1}

\sf:\implies\:W=\left[\dfrac{-kx^{-2+1}}{(-2+1)}\right]

\sf:\implies\:W=\left[\dfrac{-kx^{-1}}{(-1)}\right]

\sf:\implies\:W=[kx^{-1}]

\sf:\implies\:W=\left[\dfrac{k}{x}\right]_a^{2a}

\sf:\implies\:W=\dfrac{k}{2a}-\dfrac{k}{a}

\sf:\implies\:W=\dfrac{k-2k}{2a}

:\implies\:\underline{\boxed{\bf{\gray{W=-\dfrac{k}{2a}}}}}


Anonymous: Exemplary answer!
Ekaro: Thank you! :)
Anonymous: nice answer
QueenSaanvi: awesome
BrainlyEmpire: Good!
Anonymous: Awesome
llMrDevilll: well answered
Answered by Anonymous
10

\large{\bold{\underline{ Question -\: }}}

A force F = - k/x² ( x ≠ 0) acts on a particle in x-direction . find the work done by his force in displacing the particle from x = + a to x = + 2a . here , k is a positive constant .

\large{\bold{\underline{Answer - \: }}}

\sf \red{ :\longrightarrow\sf{\displaystyle \sf w=  \frac{  - k}{ 2a}}}

\large{\bold{\underline{Step \:  by \:  step  \: explanation  - \: }}}

\sf{ dw = Fdx}

\sf :\longrightarrow{ \displaystyle\sf\int dw =  \int  Fdx}

\sf :\longrightarrow{  w= \displaystyle\sf\int_{x = a}^{x = 2a} Fdx}

\sf :\longrightarrow{  w= \displaystyle\sf\int_{ a}^{ 2a}  \dfrac{ - k}{ {x}^{2} } dx}

\sf :\longrightarrow{  w= - k \displaystyle\sf\int_{ a}^{ 2a}  {x}^{ - 2} dx}

\sf :\longrightarrow{\displaystyle w= - k\left[ \frac{ {x}^{ - 2}  + 1}{ - 2 + 1} \right]_{a}^{a}}

\sf :\longrightarrow{\displaystyle\sf w= \frac{ - k}{ - 1} \left[ \frac{  1}{ x} \right]_{2a}^{a}}

\sf :\longrightarrow{\displaystyle\sf w= k \: \frac{  1}{ 2a} -  \frac{1}{a}}

\sf \blue{ :\Longrightarrow{\displaystyle\sf w=  \frac{  - k}{ 2a}}}


Anonymous: Thank you! Amazing explanation. :)
Anonymous: :)
QueenSaanvi: great
llMrDevilll: good explanation
Anonymous: :)
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