Question

The work done on a particle of mass "m" by a force k [ xî / ( x² + y² )³/² + yî/ ( x² + y² ) ³/² ] (where k is a constant of approximate dimensions ) when the particle is taken from the point ( a , 0 ) along a circular path of radius about the origin in the x - y plane

Answer 1

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${\maltese \; \; {\underline{\underline{\textsf{\textbf{Given that :}}}}}}$

• The force acting upon a particle of mass " m " is

${\red{\bigstar \; {\underline{\boxed{\rm{ K \bigg[ \dfrac{x \hat{i} }{(x^2 + y^2)^{3/2}} + \dfrac{y \hat{j}}{(x^2 + y^2)^{3/2}} \bigg] }}}}}}$    

• The particle is taken from the point ( a , 0 ) along a circular path of radius about the origin in the x - y plane

${\maltese \; \; {\underline{\underline{\textsf{\textbf{To Find :}}}}}}$

• The work done on the particle by the force given

${\maltese \; \; {\underline{\underline{\textsf{\textbf{Understanding Concept :}}}}}}$

➤ Now we observe that the force given over here is considered as a variable force which isn't a fixed amount , so let's adapt the method of integration to find out the amount of work done on the particle

${\maltese \; \; {\underline{\underline{\textsf{\textbf{Formula Used :}}}}}}$

• The work done by a variable force is given by :

${ \bigstar { \underline{ \boxed{\bf \: Work \; done = {\displaystyle{\int \limits^{r_2}_{r_1}}\vec{\bf F} . \vec{\bf dr}}}}}}$      

★ Now we know that the particle moves in the circular path along the co - ordinates { a , 0 } that means that the particle starts from the distance a at x - axis and ends at the point { 0 , a } at the y - axis so, let us assume that  a , a  are the limits

${\maltese \; \; {\underline{\underline{\textsf{\textbf{Required Solution :}}}}}}$  

➛ Here we have the force but we don't have the components of the unit displacement $\vec{\sf dr}$ so , let us assume the components of

• Now the work done will be :

$\: {\red{\bullet \; {\bf {Work \; done = {\displaystyle{\int \limits^{r_2}_{r_1}}}}}} \red{\vec{\bf F} . \vec{\bf dr}}}$  

• Integrating it we get the result :

${\longrightarrow} \red { \displaystyle{\rm{ \int \limits^ {a}_{a} K \bigg[ \dfrac{x \hat{i} }{(x^2 + y^2)^{3/2}} + \dfrac{y \hat{j}}{(x^2 + y^2)^{3/2}} \bigg] . \bigg( dx \hat{i} + dy\hat{j} \bigg)}}}$

${\longrightarrow} {\displaystyle{\rm{ \int \limits^ {a}_{a} \bigg[ \dfrac{x \hat{i} }{(x^2 + y^2)^{3/2}} + \dfrac{y \hat{j}}{(x^2 + y^2)^{3/2}} \bigg] . \bigg( dx \hat{i} + dy\hat{j} \bigg)}}}$  

$\:{\longrightarrow} { \displaystyle{ \rm{k \int \limits^ {a}_{a} \bigg[ \dfrac{ 1 }{(x^2 + y^2)^{3/2}}} \bigg] . \bigg( dx \hat{i} + dy\hat{j} \bigg)  }}$    

• let's integrate the second term using  formula :

$\bullet \; {\underline{\boxed{ {\bf { \int \limits^ {}_{}} x^n dx = \dfrac{x^{n+1}}{n+1} }}}}$  

${ \longrightarrow}{ \displaystyle{ \rm{ K\int \limits^ {a}_{a} \bigg[ \dfrac{ 1 }{(x^2 + y^2)^{3/2}}} \bigg] \int \limits^ {a}_{a}\bigg( dx \hat{i} + dy\hat{j} \bigg)  }}$  

${\longrightarrow } { \displaystyle {\rm{ K \int \limits^ {a}_{a} \bigg[ \dfrac{ 1 }{(x^2 + y^2)^{3/2}}} \bigg] \bigg\{ \bigg( \int \limits^ {a}_{a} dx \bigg) + \bigg(\int \limits^ {a}_{a} dy\bigg)\bigg\}}}$

${\longrightarrow} { \displaystyle {\rm{K \int \limits^ {a}_{a} \bigg[ \dfrac{ 1 }{(x^2 + y^2)^{3/2}}} \bigg] \bigg\{d \bigg( \int \limits^ {a}_{a} x \bigg) +d \bigg(\int \limits^ {a}_{a} y\bigg)\bigg\}}}$

${\longrightarrow} {\displaystyle{\rm{ k \int \limits^ {a}_{a} \bigg[ \dfrac{ 1 }{(x^2 + y^2)^{3/2}}} \bigg] \bigg\{d \bigg( \frac{x^2}{2} \bigg) +d \bigg( \dfrac{y^2}{2} \bigg)\bigg\}}}$  

Simplifying the rest we get ;

${\longrightarrow } {\rm{ \bigg[ \dfrac{K }{(x^2 + y^2)^{3/2}}} \bigg]^a_a \; \bigg[\dfrac{x^2}{2} \bigg] ^{a}_{a}+ \bigg[ \dfrac{y^2}{2} \bigg] ^{a}_{a}}$  

$\longrightarrow {\rm {\bigg[ \dfrac{K }{(a^4)^{3/2}}} \bigg] \bigg[\dfrac{a^2 - a^2}{2} \bigg] + \bigg[ \dfrac{a^2- a^2}{2} \bigg]}$

$\longrightarrow {\rm {\bigg[ \dfrac{K }{(a)^{3/2}}} \bigg] \bigg[\dfrac{0}{2} \bigg] + \bigg[ \dfrac{0}{2} \bigg]}$

$\longrightarrow {\rm {\bigg[ \dfrac{K }{(a^2 + y^2)^{3/2}}} \bigg] \times 0 + 0}$  

$\longrightarrow { \rm { \bigg[ \dfrac{K }{(a)^{3/2}}} \bigg] \times 0}$  

$\longrightarrow {\red{\rm{\underline{\underline{ Work \; done _{(by \; the \; variable \; force) } = 0 }}}}}$

${\maltese \; \; {\underline{\underline{\textsf{\textbf{Therefore :}}}}}}$

• The work done on the particle by the force is 0

${\maltese \; \; {\underline{\underline{\textsf{\textbf{More to know :}}}}}}$

• Integration formulae of Trigonometric Functions :

$\longrightarrow {\displaystyle {\rm{ \int Sin(x) \; dx = - cos ( x) + c}}}$

$\longrightarrow {\displaystyle {\rm{ \int Cos(x) \; dx = Sin ( x) + c}}}$

$\longrightarrow {\displaystyle{\rm{ \int Sec^2(x) \; dx = tan ( x) + c}}}$

$\longrightarrow {\displaystyle{\rm{ \int Sec(x) \;tan(x) \; dx = Sec ( x) + c}}}$

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