Potential Energy Questions in English

(B) Let the total length of the chain be $L = 2\,m$ and 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 edge of the table.
The work done in pulling the chain onto the table is equal to the increase in the potential energy of the hanging part,which is $W = mgh$.
Substituting the values: $W = 1.2 \times 10 \times 0.3 = 3.6\,J$.

(B) The relationship between force $F$ and potential energy $U$ is given by $F = -\frac{dU}{dx}$.
Given the potential energy function $U = A - Bx^2$.
Differentiating $U$ with respect to $x$:
$\frac{dU}{dx} = \frac{d}{dx}(A - Bx^2) = 0 - 2Bx = -2Bx$.
Substituting this into the force equation:
$F = -(-2Bx) = 2Bx$.
The magnitude of the force is $|F| = 2Bx$.
Since $2B$ is a constant,the magnitude of the force is directly proportional to the displacement $x$ $(F \propto x)$.

(D) The condition for stable equilibrium is that the net force $F$ acting on the system must be zero,where $F = -\frac{dU}{dx} = 0$.
Given $U(x) = ax^{-12} - bx^{-6}$.
Differentiating with respect to $x$:
$\frac{dU}{dx} = -12ax^{-13} - (-6bx^{-7}) = -12ax^{-13} + 6bx^{-7}$.
Setting the force to zero:
$F = -(-12ax^{-13} + 6bx^{-7}) = 0$
$12ax^{-13} - 6bx^{-7} = 0$
$\frac{12a}{x^{13}} = \frac{6b}{x^7}$
$\frac{12a}{6b} = \frac{x^{13}}{x^7}$
$\frac{2a}{b} = x^6$
$x = \sqrt[6]{\frac{2a}{b}}$.

(A) body at rest has a velocity of $0 \ m/s$,which implies its speed is also $0 \ m/s$.
Since momentum $p = mv$,if $v = 0$,then momentum $p = 0$.
However,a body at rest can possess potential energy due to its position or configuration in a field (e.g.,gravitational potential energy $U = mgh$).
Therefore,a body at rest may have energy.

(B) The potential energy $(PE)$ of an object is given by the formula $PE = mgh$,where $m$ is the mass of the object,$g$ is the acceleration due to gravity,and $h$ is the height above the reference level (ground).
Since $m$ and $g$ are constants,the potential energy is directly proportional to the height $h$.
Therefore,the potential energy is maximum when the height $h$ is at its maximum value.
This occurs at the maximum height reached by the stone during its flight.

(B) When a watch spring is wound,work is done against the elastic restoring force of the spring. This work is stored in the spring in the form of elastic potential energy. Therefore,the energy stored in a wound watch spring is potential energy $(P.E.)$. The correct option is $(B)$.

(B) The mass of the body is $m = 200\, g = 0.2\, kg$.
The height from which the body falls is $h = 200\, m$.
The acceleration due to gravity is $g = 10\, m/s^2$.
The potential energy $(P.E.)$ at the initial height is given by $U = mgh$.
At the point of contact with the earth's surface,the final height is $0$,so the final potential energy is $0$.
The decrease in $P.E.$ is $\Delta U = U_{initial} - U_{final} = mgh - 0 = mgh$.
Substituting the values: $\Delta U = 0.2\, kg \times 10\, m/s^2 \times 200\, m = 400\, J$.
Therefore,the decrease in $P.E.$ is $400\, J$.

(B) The potential energy $(U)$ of an object is given by the formula $U = mgh$,where $m$ is the mass,$g$ is the acceleration due to gravity,and $h$ is the height.
Given:
Mass $(m)$ = $20 \,kg$
Height $(h)$ = $50 \,cm = 0.5 \,m$
Acceleration due to gravity $(g)$ = $9.8 \,m/s^2$
Decrease in potential energy $(\Delta U)$ = $mgh$
$\Delta U = 20 \times 9.8 \times 0.5$
$\Delta U = 10 \times 9.8 = 98 \,J$.
Therefore,the correct option is $B$.

(D) The mass of the hanging part of the chain is $m = M/3$.
The length of the hanging part is $l = L/3$.
The center of mass of the hanging part is at a distance $h = l/2 = L/6$ below the edge of the table.
The potential energy of the hanging part relative to the table surface is $U = -mgh = -(M/3)g(L/6) = -MgL/18$.
To pull the hanging part onto the table, the work done by an external force must be equal to the change in potential energy, which is $W = \Delta U = U_{final} - U_{initial} = 0 - (-MgL/18) = MgL/18$.

(D) The relationship between force $F(x)$ and potential energy $U(x)$ is given by $F(x) = -\frac{dU}{dx}$.
Integrating this,we get $U(x) = -\int F(x) dx$.
Substituting $F(x) = -kx + ax^3$,we have $U(x) = -\int (-kx + ax^3) dx = \frac{kx^2}{2} - \frac{ax^4}{4} + C$.
Assuming $U(0) = 0$,we get $C = 0$,so $U(x) = \frac{kx^2}{2} - \frac{ax^4}{4} = \frac{x^2}{4}(2k - ax^2)$.
For $x \ge 0$,$U(x) = 0$ at $x = 0$ and $x = \sqrt{\frac{2k}{a}}$.
Between $x = 0$ and $x = \sqrt{\frac{2k}{a}}$,$U(x)$ is positive. For $x > \sqrt{\frac{2k}{a}}$,$U(x)$ becomes negative.
The graph starts at the origin,rises to a maximum,and then decreases,crossing the $x-$axis at $x = \sqrt{\frac{2k}{a}}$. This matches Graph $D$.

(A) For a stable molecule,the system must exist in a state of minimum potential energy at an equilibrium distance $r_0$.
When the distance between atoms is large,the interatomic force is very weak (attractive).
As they come closer,the force of attraction increases until it reaches a point where the potential energy is minimum.
At this equilibrium distance $r_0$,the net force is zero $(F = -dU/dr = 0)$.
If they are brought even closer,the force becomes strongly repulsive due to the overlapping of electron clouds.
This behavior is characteristic of the potential energy curve shown in graph $A$,which exhibits a clear potential well.

(A) The relationship between force $F$ and potential energy $U$ is given by $F = -\frac{dU}{dx}$.
Given $F = kx$,we have $-\frac{dU}{dx} = kx$.
Integrating both sides with respect to $x$:
$\int dU = -\int kx \, dx$
$U(x) = -\frac{1}{2}kx^2 + C$
Given $U(0) = 0$,we find $C = 0$.
Thus,$U(x) = -\frac{1}{2}kx^2$.
This is the equation of a downward-opening parabola symmetric about the $U$-axis,as shown in option $A$.

(B) When a rubber band is stretched,work is done against the internal elastic forces of the material.
This work done is stored within the rubber band in the form of elastic potential energy.
Therefore,a stretched rubber band possesses increased potential energy compared to its unstretched state.

(A) The potential energy $(PE)$ of an object at a height $h$ is given by the formula: $PE = mgh$.
Here,the weight of the body is $W = mg = 1 \, N$.
The potential energy is given as $PE = 1 \, J$.
Substituting these values into the formula:
$1 \, J = (1 \, N) \times h$.
Therefore,$h = 1 \, m$.
Thus,the body is at a height of $1 \, m$.

(D) The relationship between force $F$ and potential energy $U$ is given by $F = -\frac{dU}{dx}$.
Integrating this,we get $dU = -F dx$.
$U(x) = -\int_0^x (-kx + ax^3) dx = \int_0^x (kx - ax^3) dx$.
$U(x) = \left[ \frac{kx^2}{2} - \frac{ax^4}{4} \right]_0^x = \frac{kx^2}{2} - \frac{ax^4}{4}$.
At $x = 0$,$U(0) = 0$.
The slope of the $U-x$ graph is $\frac{dU}{dx} = -F = kx - ax^3$. At $x = 0$,the slope is $0$.
For small $x$,$U(x) \approx \frac{kx^2}{2}$,which is a parabola opening upwards.
Setting $U(x) = 0$,we get $\frac{kx^2}{2} = \frac{ax^4}{4} \Rightarrow x^2 = \frac{2k}{a} \Rightarrow x = \sqrt{\frac{2k}{a}}$.
Beyond this point,$U(x)$ becomes negative. Thus,the graph starts at the origin,increases to a maximum,and then decreases,crossing the $x$-axis at $x = \sqrt{\frac{2k}{a}}$. This matches the shape shown in Graph $D$.

(A) The equilibrium distance $r_0$ occurs when the potential energy $U(r)$ is at a minimum,which implies that the force $F(r) = -dU/dr = 0$.
Given $U(r) = ar^{-12} - br^{-6}$.
Differentiating with respect to $r$:
$dU/dr = a(-12)r^{-13} - b(-6)r^{-7} = -12ar^{-13} + 6br^{-7}$.
Setting $dU/dr = 0$ for equilibrium:
$-12ar_0^{-13} + 6br_0^{-7} = 0$.
$6br_0^{-7} = 12ar_0^{-13}$.
$b/2a = r_0^{-13} / r_0^{-7} = r_0^{-6}$.
$r_0^6 = 2a/b$.
Therefore,$r_0 = (2a/b)^{1/6}$.

(B) The total mass of the chain is $M$ and the total length is $L$. The mass per unit length is $\lambda = M/L$.
The length of the hanging part is $l = L/4$.
The mass of the hanging part is $m = \lambda \times l = (M/L) \times (L/4) = M/4$.
The center of mass of the hanging part is located at a distance $h = l/2 = (L/4)/2 = L/8$ below the edge of the table.
To pull the hanging part back onto the table,the external force must overcome the gravitational potential energy of the hanging part.
The work done $W$ is equal to the change in potential energy of the hanging part,which is $W = mgh$.
Substituting the values: $W = (M/4) \times g \times (L/8) = \frac{MgL}{32}$.
Therefore,the work done by the external force is $\frac{MgL}{32}$.

(A) body at rest has a velocity $v = 0$ and speed $v = 0$.
Since momentum $p = mv$,if $v = 0$,then $p = 0$.
However,a body at rest can possess potential energy due to its position or configuration in a field (e.g.,gravitational potential energy $U = mgh$).
Therefore,a body at rest possesses energy.

(B) The force $F$ acting on a particle is related to the potential energy $U$ by the relation $F = -\frac{dU}{dx}$.
Given $U = A - Bx^2$.
Differentiating $U$ with respect to $x$:
$\frac{dU}{dx} = \frac{d}{dx}(A - Bx^2) = 0 - 2Bx = -2Bx$.
Substituting this into the force equation:
$F = -(-2Bx) = 2Bx$.
Since $A$ and $B$ are constants,$F = 2Bx$ implies that $F \propto x$.
Therefore,the force is proportional to the displacement $x$.

(B) The potential energy $(PE)$ of an object is given by the formula $PE = mgh$,where $m$ is the mass,$g$ is the acceleration due to gravity,and $h$ is the height above the reference level (ground).
Since $m$ and $g$ are constants,the potential energy is directly proportional to the height $h$.
Therefore,the potential energy is maximum when the height $h$ is at its maximum value.
This occurs at the highest point of the stone's trajectory.

(A) Potential energy is defined as the energy possessed by an object due to its position or configuration in a field.
It is always defined relative to a reference point (where potential energy is taken as zero).
Since the choice of the reference point is arbitrary,the value of potential energy depends on the chosen reference frame.
Therefore,potential energy is a relative quantity.

(D) Let the mass per unit length of the chain be $\lambda = M/L$.
The length of the hanging part is $l = L/n$.
The mass of the hanging part is $m = \lambda l = (M/L) \times (L/n) = M/n$.
The center of mass of the hanging part is at a distance $y = l/2 = L/(2n)$ below the edge of the table.
To pull the chain back onto the table,we must lift the center of mass of the hanging part to the level of the table.
The work done $W$ is equal to the change in potential energy of the hanging part:
$W = mgh = (M/n) \times g \times (L/2n) = MgL/(2n^2)$.

(A) The work done to pump out water from a vessel is equal to the change in potential energy of the water.
For a cubical vessel of height $h$ filled with water,the center of mass of the water is at a height $h/2$ from the bottom.
When the water is pumped out to the top level,the effective height through which the mass is lifted is $h/2$.
The mass of water $m = \text{density} \times \text{volume} = \rho \times (L^2 \times h)$. Since it is a cube,$L=h=1 \ m$,so $m = 1000 \ kg/m^3 \times 1 \ m^3 = 1000 \ kg$.
The work done $W = m \times g \times (h/2)$.
$W = 1000 \times 10 \times (1/2) = 5000 \ J$.

(B) The work done in raising the rod to a vertical position is equal to the change in its gravitational potential energy.
When the rod is lying horizontally,its center of mass is at height $h_1 = 0$.
When the rod is in a vertical position,its center of mass is at height $h_2 = \frac{L}{2}$,where $L = 4 \ m$.
So,$h_2 = \frac{4}{2} = 2 \ m$.
The work done $W = \Delta U = mg(h_2 - h_1)$.
Substituting the values: $W = 20 \times 9.8 \times (2 - 0)$.
$W = 20 \times 9.8 \times 2 = 392 \ J$.

(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 $U = a(\frac{b}{2a})^2 - b(\frac{b}{2a}) = \frac{b^2}{4a} - \frac{b^2}{2a} = -\frac{b^2}{4a}$.

(C) The length of the hanging part is $l = L/4$.
The mass of the hanging part is $m = (M/L) \times (L/4) = M/4$.
The center of mass of the hanging part is at a distance $h = l/2 = L/8$ below the edge of the table.
The work done to pull the hanging part onto the table is equal to the change in potential energy of the hanging part,which is $W = mgh$.
Substituting the values: $W = (M/4) \times g \times (L/8) = \frac{MgL}{32}$.

(B) The decrease in potential energy is given by the formula $\Delta U = mgh$.
Given:
Mass $m = 20 \ g = 20 \times 10^{-3} \ kg = 0.02 \ kg$.
Height $h = 200 \ m$.
Acceleration due to gravity $g = 10 \ m/s^2$.
Substituting the values:
$\Delta U = 0.02 \ kg \times 10 \ m/s^2 \times 200 \ m$.
$\Delta U = 0.2 \times 200 = 40 \ J$.
Therefore,the decrease in potential energy is $40 \ J$.

(B) The length of the chain is $L = 2 \ m$ and the total mass is $M = 4 \ kg$. The mass per unit length is $\lambda = M/L = 4/2 = 2 \ kg/m$.
The length of the hanging part is $l = 60 \ cm = 0.6 \ m$.
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 $h = l/2 = 0.6/2 = 0.3 \ m$ below the edge of the table.
To pull the chain onto the table,we must lift the center of mass of the hanging part to the level of the table.
The work done $W$ is equal to the change in potential energy of the hanging part: $W = mgh$.
$W = 1.2 \ kg \times 10 \ m/s^2 \times 0.3 \ m = 3.6 \ J$.

(A) The potential energy is given by $V(x) = \frac{x^4}{4} - \frac{x^2}{2}$.
To find the equilibrium points,we set the derivative to zero: $\frac{dV}{dx} = x^3 - x = 0$.
This gives $x(x^2 - 1) = 0$,so $x = 0, 1, -1$.
The potential energy is minimum at $x = \pm 1$.
$V_{\min} = V(\pm 1) = \frac{(\pm 1)^4}{4} - \frac{(\pm 1)^2}{2} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \ J$.
Since the total mechanical energy $E = K.E. + V$ is constant at $2 \ J$,the kinetic energy is maximum when potential energy is minimum.
$K.E._{\max} = E - V_{\min} = 2 - (-1/4) = 2 + 0.25 = 2.25 \ J$.
Using $K.E._{\max} = \frac{1}{2} m v_{\max}^2$ with $m = 1 \ kg$:
$\frac{1}{2} \times 1 \times v_{\max}^2 = \frac{9}{4}$.
$v_{\max}^2 = \frac{9}{2}$.
$v_{\max} = \frac{3}{\sqrt{2}} \ m/s$.

(B) The potential energy is given by $U(x) = \frac{x^2}{2} - x$.
To find the minimum potential energy,we set the derivative to zero: $\frac{dU}{dx} = x - 1 = 0$,which gives $x = 1 \ m$.
The second derivative $\frac{d^2U}{dx^2} = 1 > 0$,confirming a minimum at $x = 1 \ m$.
The minimum potential energy is $U_{min} = \frac{(1)^2}{2} - 1 = -0.5 \ J$.
The total mechanical energy $E$ is the sum of kinetic energy $K$ and potential energy $U$: $E = K + U$.
Since $E = 2 \ J$,the maximum kinetic energy $K_{max}$ occurs when $U$ is at its minimum: $K_{max} = E - U_{min} = 2 - (-0.5) = 2.5 \ J$.
Using $K_{max} = \frac{1}{2}mv_{max}^2$,we have $2.5 = \frac{1}{2}(1)v_{max}^2$.
$v_{max}^2 = 5$,so $v_{max} = \sqrt{5} \ ms^{-1}$.

(D) The relationship between force $F$ and potential energy $U$ is given by $F = -\frac{dU}{dx}$.
Integrating this,we get $dU = -F \, dx$,so $U(x) = -\int F \, dx$.
Substituting $F(x) = -kx + ax^3$,we have $U(x) = -\int (-kx + ax^3) \, dx = \int (kx - ax^3) \, dx$.
Performing the integration,we get $U(x) = \frac{1}{2}kx^2 - \frac{1}{4}ax^4 + C$.
Assuming $U(0) = 0$,we find $C = 0$,so $U(x) = \frac{1}{2}kx^2 - \frac{1}{4}ax^4 = \frac{1}{4}x^2(2k - ax^2)$.
For $x \ge 0$,$U(x) = 0$ at $x = 0$ and $x = \sqrt{\frac{2k}{a}}$.
For $0 < x < \sqrt{\frac{2k}{a}}$,$U(x)$ is positive.
For $x > \sqrt{\frac{2k}{a}}$,$U(x)$ becomes negative and decreases as $x^4$ dominates.
This corresponds to a curve that starts at the origin,rises to a maximum,and then crosses the $x$-axis to become negative,which matches the graph in option $D$.

(B) The relationship between force $F$ and potential energy $U$ is given by $F = -\frac{dU}{dx}$.
Given $U = A - Bx^2$.
Differentiating with respect to $x$:
$F = -\frac{d}{dx}(A - Bx^2)$
$F = -(0 - 2Bx)$
$F = 2Bx$.
Since $A$ and $B$ are constants,$F \propto x$.

(B) The force $F$ acting on a particle is related to its potential energy $U$ by the relation $F = -\frac{dU}{dx}$.
Given $U = 8x^2 - 4x + 400$.
To find the position where the force is zero,we set $F = 0$,which implies $\frac{dU}{dx} = 0$.
Calculating the derivative: $\frac{dU}{dx} = \frac{d}{dx}(8x^2 - 4x + 400) = 16x - 4$.
Setting the derivative to zero: $16x - 4 = 0$.
Solving for $x$: $16x = 4$,which gives $x = \frac{4}{16} = 0.25 \, m$.

(D) The work done by the pump is equal to the change in the potential energy of the water.
Initial potential energy $(U_i)$: The water is in two columns of height $h$ and cross-sectional area $a$. The center of mass of each column is at $h/2$.
$U_i = (a \cdot h \cdot d \cdot g) \cdot (h/2) + (a \cdot h \cdot d \cdot g) \cdot (h/2) = a \cdot h \cdot d \cdot g \cdot h = dgh^2a$.
Final potential energy $(U_f)$: All water is in one column of height $2h$ and cross-sectional area $a$. The center of mass is at $2h/2 = h$.
$U_f = (a \cdot 2h \cdot d \cdot g) \cdot h = 2dgh^2a$.
Work done $(W)$ = $U_f - U_i = 2dgh^2a - dgh^2a = dgh^2a$.

(B) The work done to lift the rod is equal to the change in its gravitational potential energy.
Initially,the center of mass of the rod is on the table,so the initial potential energy $U_i = 0$.
When the rod is placed vertically,its center of mass is at a height $h = \frac{l}{2}$ from the table.
The final potential energy $U_f = mgh = mg(\frac{l}{2}) = \frac{mgl}{2}$.
Therefore,the work done $W = U_f - U_i = \frac{mgl}{2} - 0 = \frac{mgl}{2}$.

(A) The mass of the meter scale is $m = 400 \, g = 0.4 \, kg$. The length of the meter scale is $l = 1 \, m$.
The center of mass of the scale is at its midpoint,$l/2 = 0.5 \, m$.
When the scale is rotated by an angle $\theta = {60^o}$,the vertical displacement $h$ of the center of mass is given by:
$h = \frac{l}{2} - \frac{l}{2} \cos \theta = \frac{l}{2}(1 - \cos \theta)$.
The work done $W$ against gravity is equal to the change in potential energy:
$W = mgh = mg \times \frac{l}{2}(1 - \cos {60^o})$.
Substituting the values:
$W = 0.4 \times 10 \times \frac{1}{2} \times (1 - 0.5) = 0.4 \times 10 \times 0.5 \times 0.5 = 1 \, J$.

(B) Given potential energy: $U(r) = \frac{A}{r^2} - \frac{B}{r}$.
For equilibrium,the force $F = -\frac{dU}{dr} = 0$,which implies $\frac{dU}{dr} = 0$.
$\frac{dU}{dr} = \frac{d}{dr}(Ar^{-2} - Br^{-1}) = -2Ar^{-3} + Br^{-2} = 0$.
$\frac{B}{r^2} = \frac{2A}{r^3} \implies r = \frac{2A}{B}$.
For stable equilibrium,the second derivative must be positive: $\frac{d^2U}{dr^2} > 0$.
$\frac{d^2U}{dr^2} = \frac{d}{dr}(-2Ar^{-3} + Br^{-2}) = 6Ar^{-4} - 2Br^{-3}$.
Substituting $r = \frac{2A}{B}$:
$\frac{d^2U}{dr^2} = 6A(\frac{B}{2A})^4 - 2B(\frac{B}{2A})^3 = 6A(\frac{B^4}{16A^4}) - 2B(\frac{B^3}{8A^3}) = \frac{3B^4}{8A^3} - \frac{B^4}{4A^3} = \frac{3B^4 - 2B^4}{8A^3} = \frac{B^4}{8A^3}$.
Since $A, B > 0$,$\frac{B^4}{8A^3} > 0$. Thus,the condition for stable equilibrium is satisfied at $r = \frac{2A}{B}$.

(B) The force $F$ acting on the particle is given by the negative gradient of potential energy: $F = -\frac{dU}{dx}$.
Given $U = x^2 - 2x$,we find the derivative: $\frac{dU}{dx} = 2x - 2$.
For equilibrium,the net force must be zero: $F = -(2x - 2) = 0$,which gives $x = 1$.
To check for stability,we examine the second derivative: $\frac{d^2U}{dx^2} = \frac{d}{dx}(2x - 2) = 2$.
Since $\frac{d^2U}{dx^2} > 0$ at $x = 1$,the potential energy is at a minimum,which corresponds to a state of stable equilibrium.

(A) The total mechanical energy $E$ of a particle is the sum of its kinetic energy $K$ and potential energy $U(x)$,given by $E = K + U(x)$.
Since kinetic energy $K = \frac{1}{2}mv^2$,where $m$ is the mass and $v$ is the speed of the particle.
If the speed of the particle is zero $(v = 0)$,then the kinetic energy $K$ must be zero $(K = 0)$.
Substituting $K = 0$ into the total energy equation,we get $E = 0 + U(x)$,which simplifies to $U(x) = E$.
Therefore,the speed of the particle is zero at the points where the potential energy equals the total energy.

(B) The force acting on the particle is $\vec{F} = -\nabla V = -(\frac{\partial V}{\partial x} \hat{i} + \frac{\partial V}{\partial y} \hat{j}) = -(-7 \hat{i} + 24 \hat{j}) = 7 \hat{i} - 24 \hat{j} \ N$.
The acceleration is $\vec{a} = \frac{\vec{F}}{m} = \frac{7 \hat{i} - 24 \hat{j}}{5} = 1.4 \hat{i} - 4.8 \hat{j} \ m/s^2$.
The magnitude of acceleration is $|\vec{a}| = \sqrt{1.4^2 + (-4.8)^2} = \sqrt{1.96 + 23.04} = \sqrt{25} = 5 \ m/s^2$. (Option $B$ is correct).
At $t = 4 \ s$,the velocity is $\vec{v} = \vec{v}_0 + \vec{a}t = (1.44 \hat{i} + 4.2 \hat{j}) + (1.4 \hat{i} - 4.8 \hat{j}) \times 4 = (1.44 + 5.6) \hat{i} + (4.2 - 19.2) \hat{j} = 7.04 \hat{i} - 15 \hat{j} \ m/s$.
The magnitude is $|\vec{v}| = \sqrt{7.04^2 + (-15)^2} \approx \sqrt{49.56 + 225} \approx 16.57 \ m/s$. (Option $A$ is incorrect).
For option $C$,check the dot product $\vec{v}_0 \cdot \vec{a} = (1.44)(1.4) + (4.2)(-4.8) = 2.016 - 20.16 \neq 0$. (Option $C$ is incorrect).
Since only $B$ is correct,the answer is $B$.

(D) Let the length of the string be $\ell$. When the mass is displaced by an angle $\theta = 45^\circ$,the vertical height gained by the mass is $h = \ell - \ell \cos 45^\circ = \ell(1 - \frac{1}{\sqrt{2}})$.
The horizontal displacement of the mass is $x = \ell \sin 45^\circ = \frac{\ell}{\sqrt{2}}$.
Using the work-energy theorem,the work done by the external horizontal force $F$ must equal the change in potential energy of the mass (assuming it starts and ends at rest,so $\Delta K = 0$):
$W_F + W_g = 0$
$F \cdot x - Mg \cdot h = 0$
$F \cdot (\frac{\ell}{\sqrt{2}}) = Mg \cdot \ell(1 - \frac{1}{\sqrt{2}})$
$F = Mg \cdot \sqrt{2}(1 - \frac{1}{\sqrt{2}})$
$F = Mg(\sqrt{2} - 1)$

(C) At equilibrium,the force is zero,which means the derivative of the potential energy with respect to distance is zero: $\frac{dU(x)}{dx} = 0$.
Given $U(x) = ax^{-12} - bx^{-6}$,we differentiate:
$\frac{dU}{dx} = -12ax^{-13} + 6bx^{-7} = 0$.
$12ax^{-13} = 6bx^{-7} \Rightarrow \frac{2a}{x^6} = b \Rightarrow x^6 = \frac{2a}{b}$.
Now,substitute $x^6 = \frac{2a}{b}$ into the potential energy function to find $U_{\text{at equilibrium}}$:
$U_{\text{at equilibrium}} = \frac{a}{(x^6)^2} - \frac{b}{x^6} = \frac{a}{(2a/b)^2} - \frac{b}{(2a/b)} = \frac{a}{4a^2/b^2} - \frac{b^2}{2a} = \frac{b^2}{4a} - \frac{2b^2}{4a} = -\frac{b^2}{4a}$.
Since $U(x = \infty) = 0$,the dissociation energy $D$ is:
$D = U(\infty) - U_{\text{at equilibrium}} = 0 - (-\frac{b^2}{4a}) = \frac{b^2}{4a}$.

(B) At equilibrium,the potential energy is at a minimum,so the derivative of potential energy with respect to distance $r$ must be zero: $\frac{dU}{dr} = 0$.
Given $U = Ar^{-12} - Br^{-6}$.
Differentiating with respect to $r$: $\frac{dU}{dr} = -12Ar^{-13} + 6Br^{-7} = 0$.
Rearranging the equation: $\frac{6B}{r^7} = \frac{12A}{r^{13}}$.
Multiplying both sides by $r^{13}$: $6B r^6 = 12A$.
Solving for $r^6$: $r^6 = \frac{12A}{6B} = \frac{2A}{B}$.
Therefore,the equilibrium distance is $r = (\frac{2A}{B})^{1/6}$.

(B) The total number of blocks is $N = n + 1$. Initially, all blocks are on the table, so their center of mass is at height $h_i = a/2$.
After stacking them into a column of height $H = (n + 1)a$, the center of mass of the column is at $h_f = H/2 = (n + 1)a/2$.
The total mass of the system is $M = (n + 1)m$.
The work done is equal to the change in gravitational potential energy: $W = \Delta U = M g (h_f - h_i)$.
$W = (n + 1)m g \left[ \frac{(n + 1)a}{2} - \frac{a}{2} \right]$.
$W = (n + 1)m g \left[ \frac{na + a - a}{2} \right] = (n + 1)m g \left( \frac{na}{2} \right)$.
$W = \frac{1}{2} m a g (n^2 + n)$.

(B) The total mechanical energy $E$ of the particle is conserved. Since the particle is released from rest at $x = a$,its kinetic energy $K = 0$ at $x = a$,so its total energy is $E = U(a)$.
As the particle moves,$E = K + U(x) = U(a)$. Since $K \ge 0$,the particle can only move in regions where $U(x) \le U(a)$.
Looking at the graph,$U(x) = U(a)$ at $x = a$ and $x = e$. Between $x = a$ and $x = e$,$U(x) \le U(a)$.
Therefore,the particle will move from $x = a$ towards the right,pass through $x = b$,$x = c$,and $x = d$,and reach $x = e$. At $x = e$,its kinetic energy becomes zero $(K = E - U(e) = 0)$,so it stops and then reverses its direction to move back to the left.

(C) In diagram $(I)$,the center of mass of the chain is at a distance of $\frac{l}{2}$ below the edge $A$. The potential energy is $U_I = -mg(\frac{l}{2})$.
In diagram $(II)$,the chain is folded such that it hangs to a depth of $\frac{l}{2}$. The center of mass of this folded chain is at a distance of $\frac{l}{4}$ below the edge $A$. The potential energy is $U_{II} = -mg(\frac{l}{4})$.
The work done $W$ is equal to the change in potential energy: $W = \Delta U = U_{II} - U_I$.
$W = -mg(\frac{l}{4}) - (-mg(\frac{l}{2})) = -mg\frac{l}{4} + mg\frac{l}{2} = mg\frac{l}{4}$.

(B) The force $F$ acting on the particle is given by the negative gradient of the potential energy: $F = -\frac{dU}{dr}$.
Given $U = \alpha r^{-4} - \beta r^{-5}$.
Differentiating with respect to $r$:
$F = -\frac{d}{dr}(\alpha r^{-4} - \beta r^{-5}) = -[\alpha(-4r^{-5}) - \beta(-5r^{-6})] = 4\alpha r^{-5} - 5\beta r^{-6}$.
For the particle to be in equilibrium,the net force must be zero: $F = 0$.
$4\alpha r^{-5} - 5\beta r^{-6} = 0$.
$4\alpha r^{-5} = 5\beta r^{-6}$.
Multiplying both sides by $r^6$:
$4\alpha r = 5\beta$.
Therefore,$r = \frac{5\beta}{4\alpha}$.

(C) The potential energy function is $U(x) = 20 + (x - 2)^2$.
The minimum potential energy $(U_{min})$ occurs when $(x - 2)^2 = 0$,which is at $x = 2\,m$.
Thus,$U_{min} = 20 + 0 = 20\,J$.
The total mechanical energy $(TE)$ of the particle is given as $36\,J$.
According to the law of conservation of mechanical energy,$TE = U + KE$.
To find the maximum kinetic energy $(KE_{max})$,we use the relation $KE_{max} = TE - U_{min}$.
Substituting the values,$KE_{max} = 36\,J - 20\,J = 16\,J$.

(D) The length of the hanging part is $l = \frac{L}{3}$.
The mass of the hanging part is $m = \frac{M}{L} \times \frac{L}{3} = \frac{M}{3}$.
The center of mass of the hanging part is at a distance $h = \frac{l}{2} = \frac{L/3}{2} = \frac{L}{6}$ below the edge of the table.
The work done to pull the hanging part onto the table is equal to the change in potential energy of the hanging part.
$W = mgh = \left(\frac{M}{3}\right) g \left(\frac{L}{6}\right) = \frac{MgL}{18}$.

(B) The length of the hanging part is $l = L/n$.
Since the cable is uniform,the mass of the hanging part is $m = (M/L) \times l = M/n$.
The center of mass of the hanging part is at a distance $h_{COM} = l/2 = L/(2n)$ below the edge of the surface.
To lift the hanging part onto the surface,we must raise its center of mass to the level of the surface.
The work done $W$ is equal to the change in potential energy of the hanging part:
$W = m \times g \times h_{COM}$
$W = (M/n) \times g \times (L/(2n))$
$W = \frac{MgL}{2n^2}$