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# A body of 8 kg is moving under the force $F = 3xN$ when $x$ is the distance covered, if the initial position of the particle is $x = 2m$ and final $x = 10m$. What is the speed of the particle?(A) 6 m/s(B) 12 m/s(C) 36 m/s(D) 144 m/s

Last updated date: 13th Jun 2024
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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,
${v^2} = {u^2} + 2\int_{{x_o}}^{{x_1}} {adx}$