Question

A particle is moving along x axis such that net force acting on it is F=-10x+20 and mass of the particle is 2.5 kg. Find equation of it's motion if the particle is released from rest i.e x=3.​

Answer 1

Answer:

Given:

Force vs Position equation is as follows ;

F = -10x + 20

Mass of object = 2.5 kg

To find:

Equation of motion I velocity of particle is zero at x = 3

Concept:

Equation of motion can be expresed as a relationship between Velocity and Position .

Acceleration can be expressed on form of Velocity and Position on the following form :

$\boxed{ \red{ \huge{acc. = v \dfrac{dv}{dx} }}}$

Calculation:

$F = - 10x + 20$

$= > 2.5a = - 10x + 20$

$= > a = - 4x + 8$

$= > v \dfrac{dv}{dx} = - 4x + 8$

$\displaystyle = > \int v \: dv = \int \: ( - 4x + 8)dx$

$= > \dfrac{ {v}^{2} }{2} = - 2 {x}^{2} + 8x + c$

Now we know that v = 0 , for x = 3

$= > \dfrac{ {0}^{2} }{2} = - 2 ({3}^{2} ) + 8(3) + c$

$= > c = -24 + 18 = -6$

So equation of motion comes as :

$= > \dfrac{ {v}^{2} }{2} = - 2 {x}^{2} + 8x -6$

$= > {v}^{2} = - 4 {x}^{2} + 16x -12$

$= > v = \sqrt{16x- 4 {x}^{2} - 12}$

So final answer :

$\boxed{ \blue{ \sf{ \bold{ \huge{ v = \sqrt{16x- 4 {x}^{2} -12}}}}}}$