(B) For a conservative force,$F = -\frac{dU}{dx} = -(2ax - b) = -2ax + b$. At the equilibrium position,$F = 0$,so $-2ax + b = 0$,which gives $x = \fra

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$\frac{b}{2a}, -\frac{b^2}{4a}$

(B) For a conservative force,$F = -\frac{dU}{dx} = -(2ax - b) = -2ax + b$. At the equilibrium position,$F = 0$,so $-2ax + b = 0$,which gives $x = \frac{b}{2a}$. The equilibrium potential energy is $U = a(\frac{b}{2a})^2 - b(\frac{b}{2a}) = \frac{b^2}{4a} - \frac{b^2}{2a} = -\frac{b^2}{4a}$.

Class 11 Physics Work, Energy, Power and Collision Potential Energy

Medium

For a system of conservative forces,the potential energy is given by the formula $U = ax^2 - bx$,where $a$ and $b$ are constants. The equilibrium position and the equilibrium potential energy will be,respectively:

A
$\frac{b}{2a}, -\frac{b}{2a}$
B
$\frac{b}{2a}, -\frac{b^2}{4a}$
C
$\frac{b^2}{2a}, -\frac{b^2}{2a}$
D
$-\frac{b}{4a}, -\frac{b}{4a}$

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Class 11

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Work, Energy, Power and Collision

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Potential Energy

$A$ particle is placed at the origin and a force $F = kx$ is acting on it (where $k$ is a positive constant). If $U(0) = 0$,the graph of $U(x)$ versus $x$ will be (where $U$ is the potential energy function):

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If the potential energy of a particle is given by $U = A - Bx^2$,then the force is proportional to which of the following?

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The potential energy of a particle in a force field is $U(r) = \frac{A}{r^2} - \frac{B}{r}$,where $A$ and $B$ are positive constants and $r$ is the distance of the particle from the centre of the field. For stable equilibrium,the distance of the particle is:

MediumAIPMT 2012

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If a stone is thrown vertically upward and returns to the ground,its potential energy is maximum:

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The potential energy of a particle of mass $5 \ kg$ moving in the $XY$ plane is given by $V = -7x + 24y$ joules,where $x$ and $y$ are in metres. Initially at $t = 0$,the particle is at the origin $(0, 0)$ moving with a velocity of $\vec{v}_0 = 6[0.24 \hat{i} + 0.7 \hat{j}] \ m/s = [1.44 \hat{i} + 4.2 \hat{j}] \ m/s$. Then:

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For a system of conservative forces,the potential energy is given by the formula $U = ax^2 - bx$,where $a$ and $b$ are constants. The equilibrium position and the equilibrium potential energy will be,respectively:

Similar Questions

$A$ particle is placed at the origin and a force $F = kx$ is acting on it (where $k$ is a positive constant). If $U(0) = 0$,the graph of $U(x)$ versus $x$ will be (where $U$ is the potential energy function):

If the potential energy of a particle is given by $U = A - Bx^2$,then the force is proportional to which of the following?

The potential energy of a particle in a force field is $U(r) = \frac{A}{r^2} - \frac{B}{r}$,where $A$ and $B$ are positive constants and $r$ is the distance of the particle from the centre of the field. For stable equilibrium,the distance of the particle is:

If a stone is thrown vertically upward and returns to the ground,its potential energy is maximum:

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