Potential Energy Questions in English

(D) For a particle to be in equilibrium,the net force must be zero,which implies $\frac{dU}{dx} = 0$. This corresponds to the points where the slope of the $U-x$ graph is zero.
At $x_2$,the potential energy $U$ is at a local maximum. For a potential energy maximum,the equilibrium is unstable because any small displacement results in a force that pushes the particle further away from the equilibrium position.
At $x_1$ and $x_3$,the slope is not zero,so the particle is not in equilibrium at these points.
Therefore,none of the given options correctly describe the state of the particle at the points shown.

(A) The force components are derived from the negative gradient of the potential energy:
$F_x = -\frac{\partial U}{\partial x} = -a$
$F_y = -\frac{\partial U}{\partial y} = -b$
The magnitude of the net force $F$ is given by:
$F = \sqrt{F_x^2 + F_y^2} = \sqrt{(-a)^2 + (-b)^2} = \sqrt{a^2 + b^2}$
Using Newton's second law,the acceleration $a_{acc}$ is:
$a_{acc} = \frac{F}{m} = \frac{\sqrt{a^2 + b^2}}{m} = \frac{(a^2 + b^2)^{1/2}}{m}$

(B) Let the total length of the chain be $L = 2\,m$ and the total mass be $M = 4\,kg$.
The length of the hanging part is $l = 60\,cm = 0.6\,m$.
The mass per unit length of the chain is $\lambda = \frac{M}{L} = \frac{4}{2} = 2\,kg/m$.
The mass of the hanging part is $m = \lambda \times l = 2 \times 0.6 = 1.2\,kg$.
The center of mass of the hanging part is at a distance $h = \frac{l}{2} = \frac{0.6}{2} = 0.3\,m$ below the table edge.
The work done to pull the chain onto the table is equal to the change in potential energy of the hanging part,which is $W = mgh$.
Substituting the values: $W = 1.2 \times 10 \times 0.3 = 3.6\,J$.

(C) Let $x$ be the height of the reference point (where potential energy is zero) with respect to the ground.
The potential energy $U$ is given by $U = mgh'$,where $h'$ is the height relative to the reference point.
For the first body: $m_1 = 2\, kg$,$h_1 = 5\, m$,$U_1 = 70\, J$,and $g = 10\, m/s^2$.
$70 = 2 \times 10 \times (5 - x)$
$70 = 20(5 - x)$
$3.5 = 5 - x$
$x = 1.5\, m$.
For the second body: $m_2 = 3\, kg$,$h_2 = 1\, m$,and $g = 10\, m/s^2$.
$U_2 = m_2 \times g \times (h_2 - x)$
$U_2 = 3 \times 10 \times (1 - 1.5)$
$U_2 = 30 \times (-0.5)$
$U_2 = -15\, J$.

(N/A) The total energy of a system is given by $E = P.E. + K.E.$,which implies $K.E. = E - P.E.$. Since kinetic energy $(K.E.)$ must be non-negative,the particle cannot exist in regions where $P.E. > E$.
$(i)$ For the first figure: The particle cannot exist in the region $x > a$ because $V(x) = V_0 > E$. The minimum total energy required is $0$.
(ii) For the second figure: The potential energy $V(x)$ is greater than $E$ in all regions shown. Thus,the particle cannot exist in any of these regions. The minimum total energy required is $V_0$.
(iii) For the third figure: The particle cannot exist in regions where $V(x) > E$. Here,$V(x) = V_0$ for $x < a$ and $x > b$. Thus,the particle is confined to $a < x < b$. The minimum total energy required is $-V_1$.
(iv) For the fourth figure: The particle cannot exist where $V(x) > E$. Based on the graph,this occurs for $x < -b/2$,$-a/2 < x < a/2$,and $x > b/2$. The particle can only exist in the regions $-b/2 < x < -a/2$ and $a/2 < x < b/2$. The minimum total energy required is $-V_1$.

(N/A) The total mechanical energy $(E)$ of a body is the sum of its kinetic energy $(K)$ and potential energy $(U)$.
$E = K + U$
For a body of mass $m$ at height $H$ that is stationary,its velocity $(v)$ is $0$.
Therefore,the kinetic energy $K = \frac{1}{2}mv^2 = \frac{1}{2}m(0)^2 = 0$.
The gravitational potential energy $U$ at height $H$ relative to the ground is given by $U = mgH$.
Thus,the total mechanical energy is $E = 0 + mgH = mgH$.

(B) The negative sign in the equation $W = -mgh$ indicates that the work is being done against the gravitational force. When an object is lifted vertically,the displacement is upward while the gravitational force acts downward. Since the force and displacement are in opposite directions,the work done by the external agent is positive,but the work done by the gravitational force is negative.

(B) When the bowstring is stretched,work is done on the bow,which is stored as elastic potential energy in the bow and the string. When the arrow is released,this stored potential energy is converted into the kinetic energy of the arrow.

(B) The total energy of a particle is given by $E = V + K$,where $V$ is the potential energy and $K$ is the kinetic energy. Since $K = E - V$ and kinetic energy $K$ must always be non-negative $(K \ge 0)$,a particle can exist in a region only if $E - V \ge 0$,which implies $V \le E$.
$1.$ Region $A$ $(V > E)$: Here,$K = E - V < 0$. Since kinetic energy cannot be negative,the particle cannot be found in this region.
$2.$ Region $B$ $(V < E)$: Here,$K = E - V > 0$. Since kinetic energy is positive,the particle can be found in this region.
$3.$ Region $C$ $(K < E)$: Since $K = E - V$,the condition $K < E$ implies $E - V < E$,which simplifies to $V > 0$. As long as the potential energy $V$ is positive,this condition is satisfied. Thus,the particle can be found in this region.
$4.$ Region $D$ $(V > E)$: Similar to region $A$,here $K = E - V < 0$. Since kinetic energy cannot be negative,the particle cannot be found in this region.

(B) In physics,a system is considered to be in a state of stable equilibrium when its potential energy is at a minimum value.
According to the principle of minimum potential energy,systems naturally tend to move towards configurations that minimize their total potential energy to achieve maximum stability.
Therefore,for the stability of a system,its potential energy should be as low (less) as possible.

(C) The potential energy is given by $U(r) = -\frac{A}{r^6} + \frac{B}{r^{12}}$.
At equilibrium,the force $F = -\frac{dU}{dr} = 0$.
Calculating the derivative: $\frac{dU}{dr} = -A(-6r^{-7}) + B(-12r^{-13}) = \frac{6A}{r^7} - \frac{12B}{r^{13}}$.
Setting $\frac{dU}{dr} = 0$: $\frac{6A}{r^7} = \frac{12B}{r^{13}} \Rightarrow r^6 = \frac{12B}{6A} = \frac{2B}{A}$.
Thus,the equilibrium separation is $r = \left(\frac{2B}{A}\right)^{1/6}$.
Substituting this value into the potential energy expression:
$U = -\frac{A}{(2B/A)} + \frac{B}{(2B/A)^2} = -\frac{A^2}{2B} + \frac{B \cdot A^2}{4B^2} = -\frac{A^2}{2B} + \frac{A^2}{4B} = -\frac{A^2}{4B}$.

(B) For a system to be in equilibrium,the net force must be zero,which implies that the derivative of potential energy with respect to distance must be zero: $\frac{dU}{dr} = 0$.
Given $U = \alpha r^{-10} - \beta r^{-5} - 3$.
Differentiating with respect to $r$:
$\frac{dU}{dr} = -10\alpha r^{-11} + 5\beta r^{-6} = 0$.
Rearranging the terms:
$5\beta r^{-6} = 10\alpha r^{-11}$.
Dividing both sides by $5\beta r^{-11}$:
$r^5 = \frac{10\alpha}{5\beta} = \frac{2\alpha}{\beta}$.
$r = \left(\frac{2\alpha}{\beta}\right)^{\frac{1}{5}}$.
Comparing this with the given form $\left(\frac{2\alpha}{\beta}\right)^{\frac{a}{b}}$,we find $\frac{a}{b} = \frac{1}{5}$.
Thus,$a = 1$.

(C) The equilibrium position occurs where the force $F = -\frac{dU}{dr} = 0$.
Given $U = A r^{-10} - B r^{-5}$.
Differentiating with respect to $r$: $\frac{dU}{dr} = -10 A r^{-11} + 5 B r^{-6}$.
Setting $\frac{dU}{dr} = 0$ for equilibrium:
$-10 A r^{-11} + 5 B r^{-6} = 0$.
$5 B r^{-6} = 10 A r^{-11}$.
$r^5 = \frac{10A}{5B} = \frac{2A}{B}$.
Therefore,the equilibrium distance is $r = \left(\frac{2A}{B}\right)^{\frac{1}{5}}$.

(C) The potential energy of the particle is given by $U(x) = \frac{x^4}{4} - \frac{x^2}{2}$.
To find the equilibrium points,we differentiate $U(x)$ with respect to $x$ and set it to zero:
$\frac{dU}{dx} = x^3 - x = x(x^2 - 1) = 0$.
This gives critical points at $x = 0$ and $x = \pm 1$.
Using the second derivative test: $\frac{d^2U}{dx^2} = 3x^2 - 1$.
At $x = 0$,$\frac{d^2U}{dx^2} = -1$ (local maximum).
At $x = \pm 1$,$\frac{d^2U}{dx^2} = 2$ (local minima).
The potential energy graph shows two wells separated by a barrier at $x = 0$. Since the total mechanical energy $E$ is small and negative (specifically,$E < U(0) = 0$),the particle is trapped in one of the potential wells.
Given that at $t = 0 \, s$,the particle is at $x = -0.5 \, m$,it is located in the left potential well. Because the total energy is less than the potential barrier at $x = 0$,the particle cannot cross the barrier and will remain confined between the turning points in the region $x = -1.0 \, m$ to $x = 0.0 \, m$.

(B) The particle is at the origin $(x=0)$ and is projected towards the right. To escape to infinity on the right side,it must overcome the potential barrier of $4 \,J$.
Using the law of conservation of energy: $K_i + V_i = K_f + V_f$.
At the origin,$V_i = 0$ and $K_i = \frac{1}{2}mv^2$.
To just cross the barrier,the final kinetic energy at the peak $(V_f = 4 \,J)$ must be at least $0$.
$\frac{1}{2} \times 0.5 \times v^2 = 4 \Rightarrow 0.25 \times v^2 = 4 \Rightarrow v^2 = 16 \Rightarrow v = 4 \,m/s$.
However,if the particle does not have enough energy to cross the right barrier,it will be reflected back towards the left. On the left side,there is a smaller potential barrier of $1 \,J$.
If the particle is reflected,it moves towards the left and must overcome the barrier of $1 \,J$ to escape to infinity on the left side.
For this,the initial kinetic energy must be at least $1 \,J$:
$\frac{1}{2} \times 0.5 \times v^2 = 1 \Rightarrow 0.25 \times v^2 = 1 \Rightarrow v^2 = 4 \Rightarrow v = 2 \,m/s$.
Since $2 \,m/s < 4 \,m/s$,the minimum speed required for the particle to escape to infinity (by going over the left barrier after reflection) is $2 \,m/s$.

(D) The length of the hanging part is $l = \frac{L}{5}$.
The mass of the hanging part is $m = \frac{M}{L} \times \frac{L}{5} = \frac{M}{5}$.
The center of mass of the hanging part is at a distance $h = \frac{l}{2} = \frac{L/5}{2} = \frac{L}{10}$ below the edge of the table.
The work done to pull the hanging part back onto the table is equal to the change in potential energy of the hanging part.
$W = mgh = \left( \frac{M}{5} \right) g \left( \frac{L}{10} \right) = \frac{M g L}{50}$.

(A) The potential energy of the stick is associated with its center of mass.
Initially,the center of mass of the stick is at a distance $h_1 = \frac{l}{2}$ from the pivot point.
When the stick is displaced by an angle $\theta$ from the vertical,the vertical distance of the center of mass from the pivot point becomes $h_2 = \frac{l}{2} \cos \theta$.
The change in the height of the center of mass is $\Delta h = h_1 - h_2 = \frac{l}{2} - \frac{l}{2} \cos \theta = \frac{l}{2}(1 - \cos \theta)$.
The increase in potential energy is given by $\Delta U = mg \Delta h = mg \frac{l}{2}(1 - \cos \theta)$.

(B) body is in equilibrium when the net force acting on it is zero.
For a conservative force field,the force $F$ is related to the potential energy $U$ by the relation $F = -\frac{dU}{dx}$.
Equilibrium occurs when $F = 0$,which implies $\frac{dU}{dx} = 0$.
Graphically,$\frac{dU}{dx}$ represents the slope of the $U-x$ curve.
At point $B$,the tangent to the curve is horizontal,meaning the slope $\frac{dU}{dx} = 0$.
Therefore,the body is in an equilibrium state at point $B$.

(D) The potential energy is given by $U(x) = \frac{x^3}{3} - \frac{x^2}{2}$.
To find the equilibrium points,we set the force $F = -\frac{dU}{dx} = 0$.
$\frac{dU}{dx} = x^2 - x = 0 \implies x(x - 1) = 0$,so $x = 0$ or $x = 1$.
We check the second derivative $\frac{d^2U}{dx^2} = 2x - 1$.
At $x = 0$,$\frac{d^2U}{dx^2} = -1 < 0$ (unstable equilibrium).
At $x = 1$,$\frac{d^2U}{dx^2} = 1 > 0$ (stable equilibrium,minimum potential energy).
The minimum potential energy is $U_{min} = \frac{1^3}{3} - \frac{1^2}{2} = \frac{1}{3} - \frac{1}{2} = -\frac{1}{6} \, J$.
The total mechanical energy $E = K + U = 4 \, J$.
Maximum kinetic energy $K_{max}$ occurs when $U$ is minimum: $K_{max} = E - U_{min} = 4 - (-\frac{1}{6}) = \frac{25}{6} \, J$.
Using $K_{max} = \frac{1}{2} m v_{max}^2$ with $m = 2 \, kg$:
$\frac{1}{2} (2) v_{max}^2 = \frac{25}{6} \implies v_{max}^2 = \frac{25}{6} \implies v_{max} = \frac{5}{\sqrt{6}} \, m \, s^{-1}$.

(C) The potential energy function is given by $U = 2x^2 + 3y^3 + 2z$.
The $x$-component of the force $F_x$ is related to the potential energy by the relation $F_x = -\frac{\partial U}{\partial x}$.
Differentiating $U$ with respect to $x$ while keeping $y$ and $z$ constant,we get $\frac{\partial U}{\partial x} = \frac{\partial}{\partial x}(2x^2 + 3y^3 + 2z) = 4x$.
Therefore,$F_x = -4x$.
At the point $P(1, 2, 3) \ m$,the value of $x$ is $1 \ m$.
Substituting $x = 1$ into the expression for $F_x$,we get $F_x = -4(1) = -4 \ N$.
The magnitude of the $x$-component of the force is $|F_x| = |-4| = 4 \ N$.

(D) The relationship between force $F$ and potential energy $U$ is given by $F = -\frac{dU}{dx}$.
Therefore,$dU = -F dx$.
Given $F = -2x$,we substitute this into the equation:
$dU = -(-2x) dx = 2x dx$.
Integrating both sides with the condition that $U = 0$ at $x = 0$:
$\int_0^U dU = \int_0^x 2x dx$.
$U = [x^2]_0^x = x^2$.
The potential energy function is $U = x^2$,which represents a parabola opening upwards with its vertex at the origin. This corresponds to the graph shown in option $D$.

(B) The density $\rho = M/V$ of the carpet remains constant.
Let $M_1 = M$ and $R_1 = R$. The volume $V = \pi R^2 l$,where $l$ is the length of the carpet.
When unrolled to radius $R_2 = R/2$,the mass $M_2$ of the carpet remaining in the rolled form is proportional to the volume of the cylinder.
$M_2 = \frac{M}{\pi R^2 l} \times \pi (R/2)^2 l = \frac{M}{4}$.
The potential energy of the initial rolled carpet (assuming center of mass at height $R$) is $U_1 = MgR$.
The potential energy of the final rolled carpet (with radius $R_2 = R/2$ and mass $M_2 = M/4$) is $U_2 = M_2 g R_2 = (M/4) g (R/2) = \frac{1}{8} MgR$.
The change in potential energy is $\Delta U = U_1 - U_2 = MgR - \frac{1}{8} MgR = \frac{7}{8} MgR$.

(B) Mass per unit length of the chain, $\lambda = \frac{M}{L} = \frac{4 \,kg}{2 \,m} = 2 \,kg/m$.
Let the length of the chain hanging from the table be $l = 0.6 \,m$.
The mass of the hanging part is $m = \lambda \times l = 2 \times 0.6 = 1.2 \,kg$.
The center of mass of the hanging part is at a distance $x = \frac{l}{2} = \frac{0.6}{2} = 0.3 \,m$ below the edge of the table.
The work done to pull the chain onto the table is equal to the change in potential energy of the hanging part, which is equivalent to lifting its center of mass to the level of the table.
$W = m \times g \times x = 1.2 \,kg \times 10 \,m/s^2 \times 0.3 \,m = 3.6 \,J$.

(A) The potential energy of the particle is given by $U(x) = 5x(x-4) = 5x^2 - 20x \ J$.
For a particle moving in a conservative field,the force acting on it is given by $F = -\frac{dU}{dx}$.
$F = -\frac{d}{dx}(5x^2 - 20x) = -(10x - 20) = 20 - 10x$.
The speed of the particle is maximum when the net force acting on it is zero (equilibrium position).
Setting $F = 0$,we get $20 - 10x = 0$.
$10x = 20$,which implies $x = 2 \ m$.
At this position,the potential energy is minimum,and by the law of conservation of energy,the kinetic energy is maximum,meaning the speed is maximum.

(C) The potential energy of the body is given by $U = (4x^2 + 2x) \,J$.
We know that the force $F$ acting on a body is related to the potential energy $U$ by the relation $F = -\frac{dU}{dx}$.
Calculating the derivative of $U$ with respect to $x$:
$F = -\frac{d}{dx}(4x^2 + 2x) = -(8x + 2) \,N$.
At the position $x = 2 \,m$, the magnitude of the force is:
$|F| = |-(8(2) + 2)| = |-(16 + 2)| = |-18| = 18 \,N$.
Therefore, the force acting on the body is $18 \,N$.

(D) The work done in lifting a body against gravity is stored as its gravitational potential energy.
Given:
Mass of the body, $m = 2 \,kg$
Height, $h = 4 \,m$
Acceleration due to gravity, $g = 10 \,ms^{-2}$
Formula for work done, $W = mgh$
Substituting the values:
$W = 2 \,kg \times 10 \,ms^{-2} \times 4 \,m$
$W = 80 \,J$
Therefore, the work done is $80 \,J$.

(A) The work done in pulling the hanging part of the chain onto the table is equal to the increase in the gravitational potential energy of that part.
Let the mass of the hanging part be $m' = \frac{m}{6}$ and its length be $l' = \frac{l}{6}$.
The center of mass of the hanging part is at a distance $h = \frac{l'}{2} = \frac{l/6}{2} = \frac{l}{12}$ below the edge of the table.
The work done $W$ is equal to the potential energy required to raise this center of mass to the level of the table:
$W = m' g h$
$W = \left(\frac{m}{6}\right) g \left(\frac{l}{12}\right)$
$W = \frac{m g l}{72}$

(C) The potential energy of the particle is given by $U(x) = 2 - 20x + 5x^2 \text{ J}$.
At the starting position $x_i = -3 \text{ m}$,the initial potential energy is:
$U_i = 2 - 20(-3) + 5(-3)^2 = 2 + 60 + 45 = 107 \text{ J}$.
Since the particle is released from rest,its total mechanical energy $E$ is equal to its initial potential energy:
$E = U_i = 107 \text{ J}$.
The particle will reach its maximum $x$ position when its kinetic energy becomes zero,meaning its potential energy at that point $x_f$ must equal the total mechanical energy $E$:
$U(x_f) = E$
$2 - 20x_f + 5x_f^2 = 107$
$5x_f^2 - 20x_f - 105 = 0$
Dividing by $5$:
$x_f^2 - 4x_f - 21 = 0$
Factoring the quadratic equation:
$(x_f - 7)(x_f + 3) = 0$
This gives two solutions: $x_f = 7 \text{ m}$ or $x_f = -3 \text{ m}$.
Since the particle starts at $x = -3 \text{ m}$ and moves to reach a maximum $x$,the maximum value is $x = 7 \text{ m}$.

(C) The work done in pulling the chain onto the table is equal to the change in the gravitational potential energy of the chain.
Let the table surface be the reference level $(U = 0)$.
The initial potential energy $(U_i)$ is due to the hanging part of the chain,which has mass $M/2$ and its center of mass is at a distance $L/4$ below the table.
$U_i = -(\frac{M}{2}) g (\frac{L}{4}) = -\frac{M g L}{8}$.
After pulling the entire chain onto the table,the final potential energy $(U_f)$ is $0$ because the entire chain is on the table surface.
Work done $W = U_f - U_i = 0 - (-\frac{M g L}{8}) = \frac{M g L}{8}$.

(B) Let the angle subtended by the chain at the center of the hemisphere be $\alpha = l/R$.
Consider an element of the chain of length $dl$ at an angle $\theta$ from the base.
The height of this element from the base is $h = R \sin \theta$.
The length of the element is $dl = R d\theta$.
The mass of this element is $dm = \frac{m}{l} dl = \frac{m}{l} R d\theta$.
The gravitational potential energy of this element is $dU = (dm)gh = \left(\frac{m}{l} R d\theta\right) g (R \sin \theta) = \frac{mgR^2}{l} \sin \theta d\theta$.
The chain extends from the top $(\theta = \pi/2)$ to an angle $\theta = \pi/2 - \alpha = \pi/2 - l/R$.
Integrating $dU$ from $\pi/2 - l/R$ to $\pi/2$:
$U = \int_{\pi/2 - l/R}^{\pi/2} \frac{mgR^2}{l} \sin \theta d\theta = \frac{mgR^2}{l} [-\cos \theta]_{\pi/2 - l/R}^{\pi/2}$
$U = \frac{mgR^2}{l} [-\cos(\pi/2) - (-\cos(\pi/2 - l/R))]$
$U = \frac{mgR^2}{l} [0 + \sin(l/R)] = \frac{mgR^2}{l} \sin \left(\frac{l}{R}\right)$.

(D) When a person climbs stairs,work is done against the gravitational force which acts in the downward direction.
This work against gravity is performed by the chemical energy converted into mechanical work by the internal forces within the person's body (muscular contraction).
Therefore,the source of the increase in gravitational potential energy is the work done by internal forces within the person's body.
Hence,option $D$ is the correct answer.

(A) The relation between central force $F$ and potential energy $U(r)$ is given by $F = -\frac{dU}{dr}$.
Given $U(r) = r^{-2} - r^{-1}$.
Calculating the force: $F = -\frac{d}{dr}(r^{-2} - r^{-1}) = -(-2r^{-3} + r^{-2}) = \frac{2}{r^3} - \frac{1}{r^2}$.
For the maximum attractive force,we find the extremum of $F$ by setting $\frac{dF}{dr} = 0$.
$\frac{dF}{dr} = \frac{d}{dr}(2r^{-3} - r^{-2}) = -6r^{-4} + 2r^{-3} = 0$.
$2r^{-3} = 6r^{-4} \Rightarrow \frac{2}{r^3} = \frac{6}{r^4} \Rightarrow r = 3$.
Substituting $r = 3$ into the force equation:
$F = \frac{2}{(3)^3} - \frac{1}{(3)^2} = \frac{2}{27} - \frac{1}{9} = \frac{2-3}{27} = -\frac{1}{27}$.
The magnitude of the force is $|F| = \frac{1}{27}$.

(C) Given: Mass of the metal chain $m = 2 \,kg$, total length $l = 90 \,cm = 0.9 \,m$.
Length of the hanging part $l' = (90 - 60) \,cm = 30 \,cm = 0.3 \,m$.
The mass of the hanging part $m'$ is proportional to its length: $m' = (l'/l) \times m = (0.3 / 0.9) \times 2 = 2/3 \,kg$.
The center of gravity of the hanging part is at a distance $l_c = l'/2 = 0.3 / 2 = 0.15 \,m$ below the table surface.
The work done to pull the hanging part onto the table is equal to the change in potential energy: $W = m' g l_c$.
Substituting the values: $W = (2/3) \times 10 \times 0.15 = 1 \,J$.

(C) Equilibrium points occur where the force $F = -\frac{dV}{dx} = 0$,which corresponds to the points where the slope of the $V(x)$ versus $x$ graph is zero (i.e.,the peaks and valleys of the curve).
Looking at the graph,there are three such points: two local maxima (peaks) and one local minimum (valley).
$1$. At a local minimum,the potential energy is at a minimum,so $\frac{d^2V}{dx^2} > 0$. This represents a stable equilibrium point.
$2$. At a local maximum,the potential energy is at a maximum,so $\frac{d^2V}{dx^2} < 0$. This represents an unstable equilibrium point.
Since there are two peaks (unstable) and one valley (stable),there are three equilibrium points in total: one stable and two unstable.