Question

A particle of mass m= 1 kg lying on x-axis experiences a force given by law F=x(3x-2) Newton, where x is the x-coordinate of the particle in meters. The points on x-axis where the particle is in equilibrium are

Answer 1

Explanation:

A particle of mass

lying on x-axis experiences a force given by law <br>

Newton. <br> Where x is the x-coordinate of the particle in meters. <br>

<br> (a) Locate the point on x-axis where the particle is in equilibrium. <br> (b) Draw the graph of variation of force F (y-axis) with x-coordinate of the particle (x-axis). Hence or otherwise indicate at which positions the particle is in stable or unstable equilibrium. <br> (c ) What is the minimum speed to be imparted to the particle placed at

meters such that it reaches tha origin.

Answer 2

At a distance of $0\ m$ and $\dfrac{2}{3}\ m$ , the particle is in equilibrium .

Given :

Mass of particle , m = 1 kg .

Distance dependent force equation , $F=x(3x-2)$

We need to find the points on x-axis where the particle is in equilibrium.

We know a particle will be in equilibrium when net force acting on it is zero.

So , putting zero in force equation and equating we get :

$x(3x-2)=0$

So , solution of above equation are :

x = 0 m and $x=\dfrac{2}{3}\ m$ .

Therefore , at a distance of $0\ m$ and $\dfrac{2}{3}\ m$ , the particle is in equilibrium .

Hence , this is the require solution .

Learn More :

Coordinate Axis

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