The potential energy of a particle under a conservative force is given by U(x)=(x2 - 3 x)J . The equilibrium position of the particle is at:

Question

The potential energy of a particle under a conservative force is given by U(x)=(x2 - 3 x)J . The equilibrium position of the particle is at: (A) x=1.5 m (B) x=2m (C) x=2.5m (D) x=3m

Tardigrade Admin · Accepted Answer

U(x)=(x2 - 3 x)J For a conservative field, Force, F=-(d U/d x) ∴ F=-(d/d x)(x2 - 3 x)=-(2 x - 3)=-2x+3 At equilibrium position, F=0 -2x+3=0⇒ x=(3/2)m=1.5m

Q. The potential energy of a particle under a conservative force is given by $U (x) = (x^{2} - 3 x) J$ . The equilibrium position of the particle is at:

$x = 1.5 m$

$x = 2 m$

$x = 2.5 m$

$x = 3 m$

$U (x) = (x^{2} - 3 x) J$
For a conservative field, Force, $F = - \frac{d U}{d x}$
$∴ F = - \frac{d}{d x} (x^{2} - 3 x) = - (2 x - 3) = - 2 x + 3$
At equilibrium position, $F = 0$
$- 2 x + 3 = 0 \Rightarrow x = \frac{3}{2} m = 1.5 m$

Q. The potential energy of a particle under a conservative force is given by U(x)=(x2−3x)J . The equilibrium position of the particle is at:

x=1.5m

x=2m

x=2.5m

x=3m