Answer
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Hint: As seen, force in itself is dependent on the position of the particle hence the acceleration is not a constant. For such a changing force, the equation of motion can be used to solve the problem, with an integral sign before the acceleration.
Formula used: In this solution we will be using the following formulae;
\[{v^2} = {u^2} + 2\int {adx} \] where \[v\] is the final velocity (or instantaneous velocity at a particular position) of a body, \[u\] is the initial velocity at the start point, \[a\] is the acceleration of the object, \[x\] signifies the distance, and \[\int {adx} \] signifies an integral of a changing acceleration with respect to the position of the object.
Acceleration \[a = \dfrac{F}{m}\] where \[F\] is the force acting on a body of mass \[m\].
Complete Step-by-Step solution:
The force acting on a body is in itself proportional to the position of the body. Hence, the acceleration of the body is changing. For such a situation, we have the equation of motion to be given as
\[{v^2} = {u^2} + 2\int {adx} \] where \[v\] is the final velocity of a body, \[u\] is the initial velocity at the start point, \[a\] is the acceleration of the object, \[x\] signifies the distance, and \[\int {adx} \] signifies an integral of a changing acceleration with respect to the position of the object.
But, generally, acceleration is given by
\[a = \dfrac{F}{m}\] where \[F\] is the force acting on a body of mass \[m\].
Hence,
\[{v^2} = {u^2} + 2\int {\dfrac{F}{m}dx} \]
Assuming the body begins from rest, and inserting known values and limit of integration, we have
\[{v^2} = 2\int_2^{10} {\dfrac{{3x}}{8}dx} = \dfrac{3}{4}\int_2^{10} {xdx} \]
\[ \Rightarrow {v^2} = \dfrac{3}{4}\left( {\dfrac{{{{10}^2}}}{2} - \dfrac{{{2^2}}}{2}} \right) = \dfrac{3}{8}\left( {100 - 4} \right)\]
By computation,
\[{v^2} = 36\]
\[v = 6m/s\]
Hence, the correct option is A.
Note: For understanding, the equation, \[{v^2} = {u^2} + 2\int {adx} \] can be derived from the definition of instantaneous acceleration, which is
\[a = \dfrac{{dv}}{{dt}}\]
\[ \Rightarrow dv = adt\]
But also, we know that
\[v = \dfrac{{dx}}{{dt}}\], hence,
\[dt = \dfrac{{dx}}{v}\]
Hence, inserting into \[dv = adt\], we get
\[dv = a\dfrac{{dx}}{v}\]
\[ \Rightarrow vdv = adx\]
Integrating both sides from initial to final value, we have
\[\dfrac{{{v^2} - {u^2}}}{2} = \int_{{x_o}}^{{x_1}} {adx} \]
Rearranging we have,
\[{v^2} = {u^2} + 2\int_{{x_o}}^{{x_1}} {adx} \]
Formula used: In this solution we will be using the following formulae;
\[{v^2} = {u^2} + 2\int {adx} \] where \[v\] is the final velocity (or instantaneous velocity at a particular position) of a body, \[u\] is the initial velocity at the start point, \[a\] is the acceleration of the object, \[x\] signifies the distance, and \[\int {adx} \] signifies an integral of a changing acceleration with respect to the position of the object.
Acceleration \[a = \dfrac{F}{m}\] where \[F\] is the force acting on a body of mass \[m\].
Complete Step-by-Step solution:
The force acting on a body is in itself proportional to the position of the body. Hence, the acceleration of the body is changing. For such a situation, we have the equation of motion to be given as
\[{v^2} = {u^2} + 2\int {adx} \] where \[v\] is the final velocity of a body, \[u\] is the initial velocity at the start point, \[a\] is the acceleration of the object, \[x\] signifies the distance, and \[\int {adx} \] signifies an integral of a changing acceleration with respect to the position of the object.
But, generally, acceleration is given by
\[a = \dfrac{F}{m}\] where \[F\] is the force acting on a body of mass \[m\].
Hence,
\[{v^2} = {u^2} + 2\int {\dfrac{F}{m}dx} \]
Assuming the body begins from rest, and inserting known values and limit of integration, we have
\[{v^2} = 2\int_2^{10} {\dfrac{{3x}}{8}dx} = \dfrac{3}{4}\int_2^{10} {xdx} \]
\[ \Rightarrow {v^2} = \dfrac{3}{4}\left( {\dfrac{{{{10}^2}}}{2} - \dfrac{{{2^2}}}{2}} \right) = \dfrac{3}{8}\left( {100 - 4} \right)\]
By computation,
\[{v^2} = 36\]
\[v = 6m/s\]
Hence, the correct option is A.
Note: For understanding, the equation, \[{v^2} = {u^2} + 2\int {adx} \] can be derived from the definition of instantaneous acceleration, which is
\[a = \dfrac{{dv}}{{dt}}\]
\[ \Rightarrow dv = adt\]
But also, we know that
\[v = \dfrac{{dx}}{{dt}}\], hence,
\[dt = \dfrac{{dx}}{v}\]
Hence, inserting into \[dv = adt\], we get
\[dv = a\dfrac{{dx}}{v}\]
\[ \Rightarrow vdv = adx\]
Integrating both sides from initial to final value, we have
\[\dfrac{{{v^2} - {u^2}}}{2} = \int_{{x_o}}^{{x_1}} {adx} \]
Rearranging we have,
\[{v^2} = {u^2} + 2\int_{{x_o}}^{{x_1}} {adx} \]
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