Question

If force $F=(x^2-3x+2)N$ acting on particle, then stable equilibrium of particle is at (Where $x$ is in meter and $F$ is in newton).

• A. $x=-1m$
• B. $x=1m$
• C. $x=2m$
• D. $x=-2m$

Answer 1

The correct option is B $x=1m$

$\frac{dF}{dt}=2x-3<0$ for equilibrium stable.

$x<\frac{3}{2}$

At equilibrium ,$F=0$

$x^2-3x+2=0$

$x^2-2x-x+2=0$

$x(x-2)-1(x-2)=0$

$(x-1)(x-2)=0$

$x=1$ and $x=2$

For stable equilibrium $x=1$