Conservative and Non-Conservative forces and Potential Energy Questions in English

(D) force is called conservative if the work done by it in moving a particle between two points is independent of the path taken.
Gravitational force,electrostatic force,and magnetic force are examples of conservative forces because the work done by them depends only on the initial and final positions.
Frictional force is a non-conservative force because the work done by it depends on the path taken,and it results in the dissipation of mechanical energy into heat.

(B) For a particle to be in stable equilibrium,the force $F$ must be zero at the equilibrium position,and if the particle is slightly displaced,the force should act in a direction that brings it back to the equilibrium position.
At $x = x_1$,the force is zero. For $x > x_1$ (slightly to the right),the force $F$ is positive (repulsive),which pushes the particle further away from $x_1$. Thus,$x_1$ is a point of unstable equilibrium.
At $x = x_2$,the force is zero. For $x > x_2$ (slightly to the right),the force $F$ is negative (attractive),which pulls the particle back towards $x_2$. For $x < x_2$ (slightly to the left),the force $F$ is positive (repulsive),which pushes the particle back towards $x_2$. Since the force acts to restore the particle to its original position,$x_2$ is a point of stable equilibrium.

(C) The relationship between force $F$ and potential energy $U$ is given by $F = -\frac{dU}{dx}$.
This implies that the force is zero where the slope of the $U-x$ graph is zero,i.e.,$\frac{dU}{dx} = 0$.
In the given graph,the slope of the curve is zero at the local maximum point $B$ and the local minimum point $C$.
Therefore,the force acting on the particle is zero at points $B$ and $C$.

(D) The force between the atoms is given by $F = -\frac{dU}{dr}$.
Attraction force is negative,so $F = -\frac{dU}{dr} < 0$,which means $\frac{dU}{dr} > 0$. The magnitude of the attractive force is $|F| = |\frac{dU}{dr}|$,which represents the slope of the $U-r$ curve.
At point $S$,the slope of the curve is positive and reaches its maximum value. Thus,the attraction is maximum at point $S$.
At point $T$,the curve becomes horizontal,meaning the slope $\frac{dU}{dr} = 0$. Thus,the attraction force is minimum (zero) at point $T$.
Therefore,the points of maximum and minimum attraction are $S$ and $T$ respectively.

(C) The relationship between force $F$ and potential energy $U$ is given by $F = -\frac{dU}{dx}$.
$1$. For $0 < x < a$,the graph of $U(x)$ is a straight line with a positive constant slope. Let the slope be $k$ $(k > 0)$. Thus,$U(x) = kx$. The force is $F = -\frac{d}{dx}(kx) = -k$. This is a negative constant value.
$2$. For $x > a$,the graph of $U(x)$ is a horizontal line,meaning $U(x)$ is constant. Thus,$\frac{dU}{dx} = 0$,which implies $F = 0$.
Comparing this with the given options,the graph that shows a negative constant force for $0 < x < a$ and zero force for $x > a$ is Graph $C$.

(A) The relationship between a conservative force and potential energy is given by $F_{\text{cons}} = -\frac{dU}{dx}$.
This implies $F_{\text{cons}} \cdot dx = -dU$.
The work done by a conservative force is $W_{\text{cons}} = \int F_{\text{cons}} \cdot dx = -\Delta U$.
Therefore,$\Delta U = -W_{\text{cons}}$.
If work is done by the system against a conservative force,the external work done is positive,which results in an increase in the potential energy of the system $(\Delta U > 0)$.

(A) For a conservative force field,the relationship between force $F$ and potential energy $U$ is given by $F = -\frac{dU}{dx}$.
Given $U = ax^2 - bx$.
Now,differentiate $U$ with respect to $x$:
$\frac{dU}{dx} = \frac{d}{dx}(ax^2 - bx) = 2ax - b$.
Substituting this into the force equation:
$F = -(2ax - b) = -2ax + b$.

(C) The force $\vec{F}$ is related to the potential energy $U$ by the negative gradient: $\vec{F} = -\nabla U = -\left( \hat{i} \frac{\partial U}{\partial x} + \hat{j} \frac{\partial U}{\partial y} + \hat{k} \frac{\partial U}{\partial z} \right)$.
Given $U = K(x^2 + y^2 + z^2)$.
Calculating the partial derivatives:
$\frac{\partial U}{\partial x} = 2Kx$,$\frac{\partial U}{\partial y} = 2Ky$,and $\frac{\partial U}{\partial z} = 2Kz$.
Substituting these into the force equation:
$\vec{F} = -(2Kx\hat{i} + 2Ky\hat{j} + 2Kz\hat{k})$.
Therefore,$\vec{F} = -2K(x\hat{i} + y\hat{j} + z\hat{k})$.

(B) The gravitational force is a conservative force.
For a conservative force,the work done in moving a particle from one point to another is independent of the path taken.
It only depends on the initial and final positions of the particle.
In the given figure,all three paths $(1, 2, 3)$ start from point $A$ and end at point $B$.
Since the initial and final positions are the same for all three paths,the work done along each path must be equal.
Therefore,${W_1} = {W_2} = {W_3}$.

(C) The relationship between force $F$ and potential energy $U$ is given by $F = -\frac{dU}{dx}$.
For the interval $0 < x < a$,the potential energy $U(x)$ is a linear function with a positive slope,i.e.,$U(x) = kx$ where $k > 0$. Thus,$F = -\frac{d}{dx}(kx) = -k$,which is a constant negative value.
For $x > a$,the potential energy $U(x)$ is constant,so $\frac{dU}{dx} = 0$,which implies $F = 0$.
Comparing this with the given options,the graph that shows a constant negative force for $0 < x < a$ and zero force for $x > a$ is graph $C$.

(C) The change in potential energy $U$ of a system is defined by the work done by a conservative force as $\Delta U = -W_{cons}$.
If work is done by a conservative force on the system,the force acts in the direction of displacement,meaning $W_{cons} > 0$.
Consequently,$\Delta U = -W_{cons} < 0$,which implies that the potential energy of the system decreases.
Therefore,the potential energy of a system decreases when work is done upon the system by a conservative force.

(A) For equilibrium,the net force $F$ must be zero: $2x^2 - 3x - 2 = 0$.
Factoring the quadratic equation: $2x^2 - 4x + x - 2 = 0 \Rightarrow 2x(x - 2) + 1(x - 2) = 0 \Rightarrow (x - 2)(2x + 1) = 0$.
The equilibrium positions are $x = 2$ and $x = -1/2$.
The potential energy $U$ is related to force by $F = -dU/dx$,so $dU/dx = -F = -(2x^2 - 3x - 2)$.
To determine stability,we check the second derivative $d^2U/dx^2 = -dF/dx = -(4x - 3)$.
For $x = 2$: $d^2U/dx^2 = -(4(2) - 3) = -5 < 0$. Since the second derivative is negative,$x = 2$ is a position of unstable equilibrium.
For $x = -1/2$: $d^2U/dx^2 = -(4(-1/2) - 3) = -(-2 - 3) = 5 > 0$. Since the second derivative is positive,$x = -1/2$ is a position of stable equilibrium.

(A) The particle is released from rest at the origin $(x = 0)$,so the initial kinetic energy $K_i = 0$ and initial potential energy $U_i = U(0) = 0^2 - 3(0) = 0 \ J$.
By the law of conservation of mechanical energy,the total energy $E = K + U$ remains constant.
$E_i = K_i + U_i = 0 + 0 = 0 \ J$.
At $x = 2$,the potential energy is $U_f = U(2) = 2^2 - 3(2) = 4 - 6 = -2 \ J$.
Since $E_i = E_f$,we have $0 = K_f + U_f$.
$0 = K_f - 2 \ J$.
Therefore,$K_f = 2 \ J$.

(B) The work done $W$ by a conservative force $\vec{F}$ is related to the change in potential energy $V$ by the relation: $W = -\Delta V$.
Given that the work done is positive $(W > 0)$,we have $-\Delta V > 0$,which implies $\Delta V < 0$.
Since $\Delta V = V_{final} - V_{initial} < 0$,it follows that $V_{final} < V_{initial}$.
Therefore,the potential energy of the body decreases when a conservative force does positive work on it.

(C) From the given potential energy graph:
At $x = 2\, m$,the potential energy $U_i = 6\, J$.
At $x = 5\, m$,the potential energy $U_f = 2\, J$.
Since the body is released from rest at $x = 2\, m$,its initial kinetic energy $K_i = 0$.
According to the law of conservation of mechanical energy,the total mechanical energy remains constant:
$K_i + U_i = K_f + U_f$
$0 + 6 = K_f + 2$
$K_f = 6 - 2 = 4\, J$
Since $K_f = \frac{1}{2} m v^2$,where $m = 2\, kg$:
$4 = \frac{1}{2} \times 2 \times v^2$
$4 = v^2$
$v = 2\, m/s$.

(B) The force is given by $\vec{F}(x, y) = x^2 \hat{i} + y^2 \hat{j}$.
For a force to be conservative,the curl of the force must be zero,i.e.,$\nabla \times \vec{F} = 0$.
Here,$\frac{\partial F_y}{\partial x} = \frac{\partial}{\partial x}(y^2) = 0$ and $\frac{\partial F_x}{\partial y} = \frac{\partial}{\partial y}(x^2) = 0$.
Since $\frac{\partial F_y}{\partial x} - \frac{\partial F_x}{\partial y} = 0$,the force $\vec{F}$ is a conservative force.
The work done by a conservative force around any closed path is always zero.
Therefore,the work done by the force $\vec{F}$ around the closed path $OABC$ is zero.

(D) The force $F$ between two molecules is related to the potential energy $U$ by the relation $F = -\frac{dU}{dx}$.
This means the force is the negative of the slope of the $U-x$ graph.
At point $A$ $(x = 0.6 \mathring{A})$,the slope $\frac{dU}{dx}$ is negative,so $F = -(\text{negative}) = \text{positive}$,which corresponds to a repulsive force.
At point $B$ $(x = 1.2 \mathring{A})$,the curve has a minimum,so the slope $\frac{dU}{dx} = 0$,which means the force $F = 0$.
At point $C$ $(x = 1.8 \mathring{A})$,the slope $\frac{dU}{dx}$ is positive,so $F = -(\text{positive}) = \text{negative}$,which corresponds to an attractive force.
Therefore,the forces at $A, B, C$ are repulsive,zero,and attractive respectively.

(B) The force $\vec{F}$ is related to the potential energy $U$ by the relation $\vec{F} = -\nabla U = -(\frac{\partial U}{\partial x}\hat{i} + \frac{\partial U}{\partial y}\hat{j})$.
Given $U(x,y) = \cos(x + y)$.
Calculating the partial derivatives:
$\frac{\partial U}{\partial x} = -\sin(x + y)$
$\frac{\partial U}{\partial y} = -\sin(x + y)$
Thus,the force components are:
$F_x = -(-\sin(x + y)) = \sin(x + y)$
$F_y = -(-\sin(x + y)) = \sin(x + y)$
At the point $(0, \frac{\pi}{4})$:
$F_x = \sin(0 + \frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$
$F_y = \sin(0 + \frac{\pi}{4}) = \sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$
Therefore,the force vector is $\vec{F} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} = \frac{1}{\sqrt{2}}(\hat{i} + \hat{j})$.

(C) For a conservative system, the force $F$ is related to the potential energy $U$ by the negative gradient of the potential energy with respect to position:
$F = -\frac{dU}{dx}$
Given $U = ax^2 - bx$, we differentiate $U$ with respect to $x$:
$\frac{dU}{dx} = \frac{d}{dx}(ax^2 - bx) = 2ax - b$
Substituting this into the force equation:
$F = -(2ax - b)$
$F = b - 2ax$
Therefore, the correct option is $C$.

(D) The force $F$ is given by $F = -\frac{dU}{dx}$. Equilibrium occurs where $F = 0$,which implies $\frac{dU}{dx} = 0$. This corresponds to the points where the slope of the $U-x$ graph is zero,which are points $B$ and $D$.
At point $B$,the potential energy is at a local minimum,so $\frac{d^2U}{dx^2} > 0$,which indicates stable equilibrium.
At point $D$,the potential energy is at a local maximum,so $\frac{d^2U}{dx^2} < 0$,which indicates unstable equilibrium.
At point $C$,the slope $\frac{dU}{dx}$ is at its maximum positive value.
Therefore,the statement 'Position $D$ is stable equilibrium' is incorrect.

(D) The total mechanical energy $E$ of a particle is the sum of its kinetic energy $K$ and potential energy $V(x)$,given by $E = K + V(x)$.
Since $K = \frac{1}{2}mv^2$,we have $E = \frac{1}{2}mv^2 + V(x)$.
The force acting on the particle is given by $F = -\frac{dV(x)}{dx}$.
According to Newton's second law,$F = ma$,so $ma = -\frac{dV(x)}{dx}$.
Acceleration $a = -\frac{1}{m} \frac{dV(x)}{dx}$.
For the acceleration to be zero,we must have $\frac{dV(x)}{dx} = 0$.
Therefore,the particle has zero acceleration where the derivative of the potential energy with respect to position is zero.

(A) The potential energy is given by $U = ax + by$.
The force $\overrightarrow{F}$ acting on the object is the negative gradient of the potential energy: $\overrightarrow{F} = -\nabla U = -\left( \frac{\partial U}{\partial x} \hat{i} + \frac{\partial U}{\partial y} \hat{j} \right)$.
Calculating the partial derivatives: $\frac{\partial U}{\partial x} = a$ and $\frac{\partial U}{\partial y} = b$.
Thus,$\overrightarrow{F} = -a \hat{i} - b \hat{j}$.
The magnitude of the force is $|\overrightarrow{F}| = \sqrt{(-a)^2 + (-b)^2} = \sqrt{a^2 + b^2}$.
Using Newton's second law,$F = ma$,the magnitude of acceleration is $a_{acc} = \frac{|\overrightarrow{F}|}{m} = \frac{\sqrt{a^2 + b^2}}{m}$.
Therefore,the correct option is $A$.

(A) The conservative force $\overrightarrow{F}$ is related to the potential energy $U$ by the relation $\overrightarrow{F} = -\nabla U = -\left( \frac{\partial U}{\partial x} \hat{i} + \frac{\partial U}{\partial y} \hat{j} + \frac{\partial U}{\partial z} \hat{k} \right)$.
Given $U = K(x + y)$,we calculate the partial derivatives:
$\frac{\partial U}{\partial x} = K$ and $\frac{\partial U}{\partial y} = K$.
Thus,the force is $\overrightarrow{F} = -(K \hat{i} + K \hat{j})$.
The displacement vector $\overrightarrow{S}$ from $(1, 1)$ to $(2, 3)$ is $\overrightarrow{S} = (2 - 1) \hat{i} + (3 - 1) \hat{j} = \hat{i} + 2 \hat{j}$.
The work done $W$ by the conservative force is given by $W = \overrightarrow{F} \cdot \overrightarrow{S}$.
$W = -(K \hat{i} + K \hat{j}) \cdot (\hat{i} + 2 \hat{j}) = -(K \times 1 + K \times 2) = -(K + 2K) = -3K$.

(A) The force is given by $F = 2x^2 - 3x - 2$. Equilibrium occurs when $F = 0$.
$2x^2 - 3x - 2 = 0$
$2x^2 - 4x + x - 2 = 0$
$2x(x - 2) + 1(x - 2) = 0$
$(2x + 1)(x - 2) = 0$
Thus,equilibrium positions are $x = -1/2$ and $x = 2$.
To determine stability,we check the derivative of force: $dF/dx = 4x - 3$.
For stable equilibrium,$dF/dx < 0$ at the equilibrium point.
At $x = -1/2$: $dF/dx = 4(-1/2) - 3 = -2 - 3 = -5 < 0$. This is stable equilibrium.
At $x = 2$: $dF/dx = 4(2) - 3 = 8 - 3 = 5 > 0$. This is unstable equilibrium.
Therefore,$x = -1/2$ is a position of stable equilibrium.

(A) The force $\vec{F}$ is related to the potential energy $U$ by the negative gradient: $\vec{F} = -\nabla U = -\left( \frac{\partial U}{\partial x}\hat{i} + \frac{\partial U}{\partial y}\hat{j} + \frac{\partial U}{\partial z}\hat{k} \right)$.
Given $U = \frac{1}{2}(x^2 - z^2)$,we calculate the partial derivatives:
$F_x = -\frac{\partial U}{\partial x} = -\frac{\partial}{\partial x} \left( \frac{1}{2}x^2 - \frac{1}{2}z^2 \right) = -\frac{1}{2}(2x) = -x$.
$F_y = -\frac{\partial U}{\partial y} = 0$ (since there is no $y$ dependence).
$F_z = -\frac{\partial U}{\partial z} = -\frac{\partial}{\partial z} \left( \frac{1}{2}x^2 - \frac{1}{2}z^2 \right) = -\frac{1}{2}(-2z) = z$.
Thus,the force vector is $\vec{F} = F_x\hat{i} + F_y\hat{j} + F_z\hat{k} = -x\hat{i} + z\hat{k}$.

(D) force is called conservative if the work done by or against it in moving a particle between two points is independent of the path taken.
Frictional force is a non-conservative force because the work done against friction depends on the path length.
Furthermore,work done against a non-conservative force like friction is dissipated as heat and cannot be recovered as potential energy.
Therefore,both the $Assertion$ and the $Reason$ are incorrect.

(N/A) force is called a conservative force if the work done by or against it in moving a particle between two points is independent of the path taken.
$1$. When an external force does work in moving a body against a conservative force (like gravitational or spring force),this work is stored as potential energy.
$2$. When the external force is removed,the body moves by converting this stored potential energy back into kinetic energy.
$3$. Throughout this process,the total mechanical energy (sum of kinetic and potential energy) remains constant.
Since the work done depends only on the initial and final positions and not on the path,these forces are classified as conservative forces. Examples include gravitational force,spring force,and electrostatic force.

(C) force is called conservative if the work done by it in moving a particle between two points is independent of the path taken.
Field forces,such as gravitational force and electrostatic force,are conservative because the work done by them depends only on the initial and final positions.
Contact forces,such as friction and air resistance,are non-conservative because the work done by them depends on the path taken,and they often dissipate mechanical energy into heat.
Therefore,field forces are generally conservative,and contact forces are generally non-conservative.

(N/A) Conservative force: $A$ force is said to be conservative if the work done by the force on a body is path-independent and depends only on the initial and final positions. Such a field is known as a conservative field.
For example,gravitational force,electrostatic force,and magnetic force.
Non-conservative force: $A$ force is said to be non-conservative if the work done by the force on a body is path-dependent. The work done by such a force in moving a body around a closed path is not zero. Such a field is known as a non-conservative field.
For example,frictional force and viscous force.

(N/A) For a conservative force $F$,the work done $W$ by the force during a small displacement $\Delta x$ is given by $W = F \Delta x$.
According to the work-energy theorem,the work done by all forces equals the change in kinetic energy,$\Delta K = W$.
Thus,$\Delta K = F \Delta x$.
From the law of conservation of mechanical energy,the sum of changes in kinetic energy and potential energy is zero: $\Delta K + \Delta V = 0$.
Substituting $\Delta K = F \Delta x$ into the equation,we get $F \Delta x + \Delta V = 0$.
Rearranging the terms,we have $F \Delta x = -\Delta V$.
Therefore,$F = -\frac{\Delta V}{\Delta x}$.
In the limit $\Delta x \to 0$,this becomes $F = -\frac{dV}{dx}$.
Thus,for a conservative force,the force is the negative gradient of the potential energy with respect to displacement.

(N/A) Suppose that a body undergoes a small displacement $dx$ under the action of a conservative force $F$.
The work done $dW$ by this conservative force is given by $dW = F dx$.
According to the work-energy theorem,the work done by all forces is equal to the change in kinetic energy,$dW = dK$.
From the law of conservation of mechanical energy,the total mechanical energy $E = K + V$ remains constant for a conservative force,meaning $dE = 0$.
Therefore,$dK + dV = 0$,which implies $dK = -dV$.
Substituting $dW = dK$ and $dW = F dx$,we get $F dx = -dV$.
Thus,$F = -\frac{dV}{dx}$.
Hence,for a conservative force,the force is equal to the negative gradient of the potential energy with respect to displacement.

(N/A) conservative force is a force with the property that the total work done in moving a particle between two points is independent of the path taken. Examples include: $1$. Gravitational force,$2$. Electrostatic force,$3$. Spring force (elastic force).
$A$ non-conservative force is a force for which the work done depends on the path taken. Examples include: $1$. Frictional force,$2$. Viscous force,$3$. Air resistance.

(A) For a conservative force $F(x)$,the potential energy $V(x)$ is defined such that the work done by the force is equal to the negative change in potential energy:
$W = -\Delta V$
Since the work done by a force $F$ over a small displacement $dx$ is $dW = F dx$,we have:
$F dx = -dV$
$F = -\frac{dV}{dx}$
Thus,the negative derivative of potential energy with respect to displacement gives the conservative force.

(N/A) When non-conservative forces (such as friction or air resistance) act on a system,the total mechanical energy is not conserved. Instead,the work done by these non-conservative forces $(W_{nc})$ is equal to the change in the total mechanical energy of the system.
Mathematically,this is expressed as: $W_{nc} = \Delta E = \Delta K + \Delta U$,where $\Delta K$ is the change in kinetic energy and $\Delta U$ is the change in potential energy.
Alternatively,it can be written as: $E_f - E_i = W_{nc}$,where $E_f$ is the final mechanical energy and $E_i$ is the initial mechanical energy.
This indicates that the work done by non-conservative forces results in a change in the mechanical energy of the system,often dissipated as heat or sound.

(N/A) $(1)$ $A$ force $F$ is conservative if the work done by it on a particle moving between two points depends only on the initial and final positions,not on the path taken.
$(2)$ $A$ force $F$ is conservative if the work done by it in moving a particle around any closed path is zero.
$(3)$ $A$ force $F$ is conservative if it can be expressed as the negative gradient of a scalar potential energy function $U$,such that $F = -\nabla U$.

(N/A) $(1)$ Conservative forces are path-independent. The work done by a conservative force depends only on the initial and final positions of the object,not on the path taken.
$(2)$ The work done by a conservative force in a closed loop is always zero. Mathematically,$\oint \vec{F} \cdot d\vec{r} = 0$.
$(3)$ $A$ potential energy function $V$ can be defined for conservative forces such that $\vec{F} = -\nabla V$. This allows for the conservation of mechanical energy,where $K + V = \text{constant}$.
$(4)$ Not all forces are conservative. For example,friction is a non-conservative force. In the presence of non-conservative forces,the total mechanical energy is not conserved,and energy is dissipated as heat $(Q)$. The modified law is $K + V + Q = \text{constant}$.
$(5)$ The reference point for potential energy is arbitrary. We choose the zero level based on convenience (e.g.,$x=0$ for springs,Earth's surface for gravity,or infinity for universal gravitation). Only the change in potential energy $(\Delta V)$ is physically significant.

(A) The spring force is given by Hooke's Law,$F = -kx$,where $k$ is the spring constant and $x$ is the displacement from the equilibrium position.
$A$ force is defined as conservative if the work done by the force in moving a particle between two points is independent of the path taken.
For a spring force,the work done in moving from position $x_1$ to $x_2$ is $W = \int_{x_1}^{x_2} (-kx) dx = -\frac{1}{2}k(x_2^2 - x_1^2)$.
Since the work done depends only on the initial and final positions ($x_1$ and $x_2$) and not on the path taken,the spring force is a conservative force.

(N/A) $1$. For conservative forces: The law of conservation of mechanical energy states that if only conservative forces (like gravitational or electrostatic forces) act on a system,the total mechanical energy $(E = K + U)$ remains constant. Mathematically,$\Delta E = \Delta K + \Delta U = 0$,which implies $E_{initial} = E_{final}$.
$2$. For non-conservative forces: When non-conservative forces (like friction or air resistance) act on a system,the total mechanical energy is not conserved. Instead,the work done by non-conservative forces $(W_{nc})$ is equal to the change in total mechanical energy: $W_{nc} = \Delta E = (K_f + U_f) - (K_i + U_i)$. This work is typically dissipated as thermal energy.

(B) force is said to be conservative if the work done by or against the force in moving a body between two points is independent of the path taken.
Alternatively,a force is conservative if the work done by the force in moving a particle around any closed path is zero.
Properties of conservative forces:
$1$. The work done is independent of the path followed.
$2$. The work done in a closed loop is zero.
$3$. The force is the negative gradient of a potential energy function,i.e.,$F = -\nabla U$.
Examples include gravitational force and electrostatic force.

(A) For a conservative force $F(x)$,the potential energy $U(x)$ is defined such that the work done by the force is equal to the negative change in potential energy: $W = -\Delta U$.
For an infinitesimal displacement $dx$,the work done is $dW = F(x) dx$.
Thus,$F(x) dx = -dU$,which implies $F(x) = -\frac{dU}{dx}$.
Therefore,the negative derivative of potential energy with respect to position gives the force.

(N/A) force is said to be a conservative force if and only if:
$(1)$ The work done by the force is independent of the path taken between the initial and final points.
$(2)$ The total work done by the force in a closed loop or path is zero.

(B) No,the work done by non-conservative forces is not always negative.
$1$. If a body is subjected to a force but does not move,the work done by friction is zero.
$2$. If the motion of a body is caused by friction (e.g.,a block placed on a moving plank),the work done by the friction force on the body can be positive.
$3$. Generally,work done by friction is negative when it opposes the relative motion between surfaces,but it is not a universal rule for all non-conservative forces.

(N/A) No,the total mechanical energy will not be conserved.
During the fall,the body experiences an air resistive force (drag),which is a non-conservative force.
Work done against this resistive force results in the dissipation of some mechanical energy into heat.
Consequently,the gain in kinetic energy $(KE)$ is less than the loss in potential energy $(PE)$,meaning the total mechanical energy decreases over time.

(N/A) No,the work done in moving a body along a closed loop is not necessarily zero.
Work done is only zero when the forces acting on the body are conservative forces.
$A$ conservative force is a force with the property that the total work done in moving a particle between two points is independent of the path taken.
Examples of conservative forces include gravitational force and electrostatic force.
In contrast,non-conservative forces like friction or air resistance dissipate energy,meaning the work done over a closed loop is not zero.

(B) conservative force is a force with the property that the total work done in moving a particle between two points is independent of the path taken. Examples include gravitational force and electrostatic force.
$A$ non-conservative force is a force for which the work done depends on the path taken. Examples include frictional force and air resistance.
Therefore,$(1)$ Conservative force matches with $(b)$ Gravitational force,and $(2)$ Non-conservative force matches with $(a)$ Frictional force.
The correct matching is $(1-b), (2-a)$.

(A) The force $F$ is related to potential energy $U$ by the relation $F = -\frac{dU}{dr}$.
Given $U = \frac{a}{r^2} - \frac{b}{r}$,we have $F = -\frac{d}{dr}(\frac{a}{r^2} - \frac{b}{r}) = -(-\frac{2a}{r^3} + \frac{b}{r^2}) = \frac{2a}{r^3} - \frac{b}{r^2}$.
To find the maximum force,we differentiate $F$ with respect to $r$ and set it to zero: $\frac{dF}{dr} = \frac{d}{dr}(\frac{2a}{r^3} - \frac{b}{r^2}) = -\frac{6a}{r^4} + \frac{2b}{r^3} = 0$.
This gives $\frac{2b}{r^3} = \frac{6a}{r^4}$,so $r = \frac{3a}{b}$.
Substituting $a = 2$ and $b = 4$,we get $r = \frac{3(2)}{4} = 1.5 = \frac{3}{2}$.
Now,substitute $r = \frac{3}{2}$ into the expression for $F$:
$F = \frac{2(2)}{(3/2)^3} - \frac{4}{(3/2)^2} = \frac{4}{27/8} - \frac{4}{9/4} = \frac{32}{27} - \frac{16}{9} = \frac{32 - 48}{27} = -\frac{16}{27} \ N$.

(A) The total mechanical energy is $E_{\text{mech}} = K.E. + U = 8 \, J$.
$(A)$ At $x = x_{3}$,from the graph,$U = 4 \, J$. Therefore,$K.E. = E_{\text{mech}} - U = 8 - 4 = 4 \, J$. The statement says $K.E. = 10 \, J$,which is incorrect.
$(B)$ At $x = x_{2}$,from the graph,$U = 0 \, J$. Therefore,$K.E. = 8 - 0 = 8 \, J$. Since $U$ is minimum,$K.E.$ is maximum,and the particle moves at the fastest speed. This statement is correct.
$(C)$ At $x < x_{1}$,from the graph,$U = 8 \, J$. Therefore,$K.E. = 8 - 8 = 0 \, J$. The particle is at rest,which is the slowest possible speed. This statement is correct.
$(D)$ At $x > x_{4}$,from the graph,$U = 6 \, J$ (constant). Therefore,$K.E. = 8 - 6 = 2 \, J$ (constant). This statement is also correct.

(A) The force acting on the particle is given by $F = -\alpha x^3 - \beta x^4$.
Since $F = -\frac{dU}{dx}$,we have $\frac{dU}{dx} = \alpha x^3 + \beta x^4$.
At $x = 0$,$\frac{dU}{dx} = 0$,which implies the particle is in equilibrium.
To determine the stability,we examine the derivatives of the potential energy $U$ at $x = 0$:
$\frac{d^2U}{dx^2} = 3\alpha x^2 + 4\beta x^3$,which is $0$ at $x = 0$.
$\frac{d^3U}{dx^3} = 6\alpha x + 12\beta x^2$,which is $0$ at $x = 0$.
$\frac{d^4U}{dx^4} = 6\alpha + 24\beta x$,which is $6\alpha$ at $x = 0$.
Since $\alpha > 0$,the first non-zero derivative at $x = 0$ is of an even order ($4^{th}$ order) and is positive. This indicates that $U$ has a local minimum at $x = 0$.
Therefore,the particle is in stable equilibrium.

(A) According to the law of conservation of mechanical energy, the total energy remains constant: $K_i + U_i = K_f + U_f$.
At $x = +\infty$, the potential energy $U_i = V(\infty) = \frac{K}{e^{\infty} + e^{-\infty}} = 0$.
The initial kinetic energy is $K_i = K$.
At $x = 0$, the potential energy is $U_f = V(0) = \frac{K}{e^0 + e^0} = \frac{K}{1 + 1} = \frac{K}{2}$.
Let the speed at $x = 0$ be $v$. Then the final kinetic energy is $K_f = \frac{1}{2}mv^2$.
Applying the conservation of energy: $K + 0 = \frac{1}{2}mv^2 + \frac{K}{2}$.
Rearranging the terms: $\frac{1}{2}mv^2 = K - \frac{K}{2} = \frac{K}{2}$.
Solving for $v$: $mv^2 = K \Rightarrow v^2 = \frac{K}{m} \Rightarrow v = \sqrt{\frac{K}{m}}$.
Thus, the correct option is $A$.

(C) For equilibrium,the force $F$ must be zero.
$F = 2x^2 - 3x - 2 = 0$
Factoring the quadratic equation: $2x^2 - 4x + x - 2 = 0 \Rightarrow 2x(x - 2) + 1(x - 2) = 0 \Rightarrow (2x + 1)(x - 2) = 0$.
Thus,the equilibrium positions are $x = 2$ and $x = -1/2$.
The potential energy $U$ is related to force by $F = -dU/dx$,so $dU/dx = -F$.
The condition for stable equilibrium is that the potential energy $U$ is minimum,which implies $d^2U/dx^2 > 0$.
Since $dU/dx = -F$,we have $d^2U/dx^2 = -dF/dx$.
Calculating the derivative: $dF/dx = d/dx(2x^2 - 3x - 2) = 4x - 3$.
Therefore,$d^2U/dx^2 = -(4x - 3) = 3 - 4x$.
At $x = -1/2$: $d^2U/dx^2 = 3 - 4(-1/2) = 3 + 2 = 5 > 0$. Since the second derivative is positive,$x = -1/2$ is a position of stable equilibrium.
At $x = 2$: $d^2U/dx^2 = 3 - 4(2) = 3 - 8 = -5 < 0$. Since the second derivative is negative,$x = 2$ is a position of unstable equilibrium.