Work Energy Theorem and Conservation of Mechanical Energy Questions in English

(C) Let the point of suspension be $O$. The pendulum is released from $A$,where the string is horizontal. The vertical distance of $A$ from $O$ is $0$.
The bob moves to position $B$,where the string makes an angle of $30^{\circ}$ with the vertical.
The vertical distance of $B$ below the point of suspension $O$ is $h' = l \cos 30^{\circ} = l \frac{\sqrt{3}}{2}$.
The vertical distance fallen by the bob from $A$ to $B$ is $h = l - h' = l - l \cos 30^{\circ} = l(1 - \frac{\sqrt{3}}{2})$.
By the law of conservation of energy,the gain in kinetic energy is equal to the loss in potential energy.
Gain in kinetic energy $= mgh = mgl(1 - \frac{\sqrt{3}}{2})$.
Note: The provided options seem to assume the height $h$ is measured from the lowest point or relative to the vertical projection. Given the options,the intended calculation is likely based on the vertical drop from the horizontal level to the position $B$ relative to the lowest point,or a misinterpretation of the angle. Re-evaluating the standard interpretation for this specific problem type: The potential energy at $A$ (relative to the lowest point) is $mgl$. The potential energy at $B$ (relative to the lowest point) is $mgl(1 - \cos 30^{\circ})$. The gain in kinetic energy is $PE_A - PE_B = mgl - mgl(1 - \cos 30^{\circ}) = mgl \cos 30^{\circ} = mgl \frac{\sqrt{3}}{2}$.

(C) According to the law of conservation of mechanical energy,the total mechanical energy at point $P$ is equal to the total mechanical energy at point $Q$.
Total energy at $P$ = $PE_P + KE_P = mg(5R) + 0 = 5mgR$.
Total energy at $Q$ = $PE_Q + KE_Q = mgR + \frac{1}{2}mv^2$.
Equating the two,we get:
$5mgR = mgR + \frac{1}{2}mv^2$
$4mgR = \frac{1}{2}mv^2$
$8gR = v^2$
$v = \sqrt{8gR} = 2\sqrt{2gR}$.

(C) According to the work-energy theorem,the total work done on the body is equal to the change in its kinetic energy.
$W_{\text{gravity}} + W_{\text{air}} = \Delta KE$
Given:
Mass $m = 10\, kg$,Height $h = 20\, m$,Final velocity $v = 10\, m/s$,Initial velocity $u = 0\, m/s$,$g = 10\, m/s^2$.
Work done by gravity $W_{\text{gravity}} = mgh = 10 \times 10 \times 20 = 2000\, J$.
Change in kinetic energy $\Delta KE = \frac{1}{2}mv^2 - 0 = \frac{1}{2} \times 10 \times (10)^2 = 500\, J$.
Substituting these values into the work-energy theorem:
$2000 + W_{\text{air}} = 500$
$W_{\text{air}} = 500 - 2000 = -1500\, J$.
Thus,the work done by the air resistance is $-1500\, J$.

(C) The position of the particle is given by $x = 3t - 4t^2 + t^3$.
The velocity $v$ is the derivative of position with respect to time: $v = \frac{dx}{dt} = 3 - 8t + 3t^2$.
At $t = 0$,the initial velocity $v_1 = 3 - 8(0) + 3(0)^2 = 3\, m/s$.
At $t = 4$,the final velocity $v_2 = 3 - 8(4) + 3(4)^2 = 3 - 32 + 48 = 19\, m/s$.
According to the work-energy theorem,the work done $W$ is equal to the change in kinetic energy: $W = \Delta K = \frac{1}{2} m(v_2^2 - v_1^2)$.
Given mass $m = 3\, g = 3 \times 10^{-3}\, kg$.
$W = \frac{1}{2} \times 3 \times 10^{-3} \times (19^2 - 3^2) = \frac{1}{2} \times 3 \times 10^{-3} \times (361 - 9) = \frac{1}{2} \times 3 \times 10^{-3} \times 352$.
$W = 3 \times 10^{-3} \times 176 = 528 \times 10^{-3}\, J = 528\, mJ$.

(B) Since the track is frictionless,the total mechanical energy of the particle is conserved throughout its motion.
Let the potential energy be zero at the ground level.
At point $A$,the particle starts from rest,so its initial kinetic energy $K_A = 0$ and potential energy $U_A = mgh_A = 1 \times 10 \times 2 = 20 \; J$.
At the highest point $P$ of the projectile motion,the particle still has a horizontal velocity component. Let the kinetic energy at $P$ be $K_P$ and potential energy be $U_P = mgh_P = 1 \times 10 \times 1 = 10 \; J$.
By the law of conservation of mechanical energy:
$K_A + U_A = K_P + U_P$
$0 + 20 = K_P + 10$
$K_P = 20 - 10 = 10 \; J$.

(A) Given: Mass of the bullet $m = 0.04\; kg$,initial velocity $u = 90\; m/s$,final velocity $v = 0\; m/s$,and distance $s = 60\; cm = 0.6\; m$.
Using the third equation of motion,$v^2 = u^2 + 2as$,we find the acceleration $a$:
$0^2 = (90)^2 + 2 \times a \times 0.6$
$0 = 8100 + 1.2a$
$a = -\frac{8100}{1.2} = -6750\; m/s^2$.
The negative sign indicates retardation.
According to Newton's second law of motion,the average resistive force $F$ is given by $F = m \times |a|$.
$F = 0.04\; kg \times 6750\; m/s^2 = 270\; N$.
Thus,the average resistive force exerted by the block on the bullet is $270\; N$.

(A) Mass of the body,$m = 0.5\; kg$.
Velocity of the body is given by $v = a x^{3/2}$ with $a = 5\; m^{-1/2} s^{-1}$.
Initial velocity $u$ at $x = 0$ is $v = 5(0)^{3/2} = 0\; m/s$.
Final velocity $v$ at $x = 2\; m$ is $v = 5(2)^{3/2} = 5 \times 2\sqrt{2} = 10\sqrt{2}\; m/s$.
According to the Work-Energy Theorem,the work done $W$ is equal to the change in kinetic energy:
$W = \Delta K = \frac{1}{2} m (v^2 - u^2)$
$W = \frac{1}{2} \times 0.5 \times [(10\sqrt{2})^2 - 0^2]$
$W = 0.25 \times [100 \times 2]$
$W = 0.25 \times 200 = 50\; J$.

(C) The stopping distance $s$ of a vehicle is given by the work-energy theorem,where the work done by the braking force $F$ equals the change in kinetic energy: $F \cdot s = \frac{1}{2}mv^2$.
Since the braking force $F$ and mass $m$ are constant,the stopping distance is proportional to the square of the velocity: $s \propto v^2$.
If the initial velocity is $v_1 = v$ and the stopping distance is $s_1 = d$,then for a new velocity $v_2 = 3v$,the new stopping distance $s_2$ is:
$\frac{s_2}{s_1} = \left(\frac{v_2}{v_1}\right)^2 = \left(\frac{3v}{v}\right)^2 = 3^2 = 9$.
Therefore,$s_2 = 9d$.

(N/A) For straight-line motion: If a body is moving with initial velocity $u$ and constant acceleration $a$, it acquires velocity $v$ after displacement $s$. From the third equation of motion, $v^{2} - u^{2} = 2as$.
Multiplying by $\frac{m}{2}$ on both sides of the equation:
$\frac{1}{2}mv^{2} - \frac{1}{2}mu^{2} = mas$
Since $F = ma$, we get:
$K_{f} - K_{i} = Fs = W$
For motion in three dimensions:
$v^{2} - u^{2} = 2\vec{a} \cdot \vec{d}$
where $v$ is final velocity, $u$ is initial velocity, $\vec{a}$ is constant acceleration, and $\vec{d}$ is displacement.
Multiplying by $\frac{m}{2}$ on both sides:
$\frac{1}{2}mv^{2} - \frac{1}{2}mu^{2} = m\vec{a} \cdot \vec{d}$
$= \vec{F} \cdot \vec{d} = W$
The left-hand side represents the change in kinetic energy $(\Delta K)$ of the body, while the right-hand side represents the work done $(W)$ by the force.
Therefore, $\Delta K = W$. This relation is known as the work-energy theorem.

(A) Work done on a body by an external force is defined as $W = \int \vec{F} \cdot d\vec{r}$. According to the work-energy theorem,if work is done on a body,energy is transferred to the body,resulting in an increase in its kinetic energy.
Conversely,when a body does work against an external force,it loses energy,resulting in a decrease in its kinetic energy.
Therefore,work done on a body increases its energy,and work done by a body decreases its energy.

(N/A) Consider a body of mass $m$ moving in one dimension along the $X$-axis under the influence of a variable force $F(x)$. The kinetic energy $K$ of the body is given by $K = \frac{1}{2}mv^2$.
Differentiating $K$ with respect to time $t$:
$\frac{dK}{dt} = \frac{d}{dt} \left( \frac{1}{2}mv^2 \right) = \frac{1}{2}m \cdot 2v \cdot \frac{dv}{dt} = mv \cdot a$
Using Newton's second law,$F = ma$,we substitute $ma$ with $F$:
$\frac{dK}{dt} = Fv$
Since velocity $v = \frac{dx}{dt}$,we have:
$\frac{dK}{dt} = F \frac{dx}{dt}$
Multiplying both sides by $dt$:
$dK = F dx$
Integrating both sides from the initial position $x_i$ (with kinetic energy $K_i$) to the final position $x_f$ (with kinetic energy $K_f$):
$\int_{K_i}^{K_f} dK = \int_{x_i}^{x_f} F(x) dx$
$K_f - K_i = \int_{x_i}^{x_f} F(x) dx$
Since the work done $W$ by a variable force is defined as $W = \int_{x_i}^{x_f} F(x) dx$,we get:
$W = K_f - K_i = \Delta K$
This is the work-energy theorem for a variable force. It states that the work done by the net force on a body equals the change in its kinetic energy.

(N/A) The work-energy theorem states that the net work done on an object equals the change in its kinetic energy,expressed as $W_{net} = \Delta K = K_f - K_i$.
Importance:
$(1)$ If the change in kinetic energy $\Delta K$ of a body is zero,the net work done on the body is zero,meaning its kinetic energy and speed remain constant. For example,a particle in uniform circular motion has constant speed and zero net work done.
$(2)$ If the displacement is in the same direction as the net force (or its component),the work done is positive,causing the kinetic energy of the body to increase. For example,a body in free fall.
$(3)$ If the displacement is in the opposite direction to the net force (or its component),the work done is negative,causing the kinetic energy of the body to decrease.
Nature:
The work-energy theorem is a scalar relation because work $(W)$ and kinetic energy $(K)$ are both scalar quantities.

(A) According to the Work-Energy Theorem,the net work done on an object is equal to the change in its kinetic energy.
Mathematically,$W = \Delta K = K_f - K_i$.
If the work done $(W)$ is positive,then $\Delta K > 0$,which implies $K_f - K_i > 0$ or $K_f > K_i$.
Therefore,the kinetic energy of the object increases.

(N/A) The principle of conservation of mechanical energy states that if only conservative forces (like gravitational force) do work on a system,the total mechanical energy (sum of kinetic energy $K$ and potential energy $U$) remains constant.
Proof:
Consider an object of mass $m$ falling freely under gravity from a height $h$ above the ground.
$1$. At height $h$ (point $A$): The velocity $v = 0$. Kinetic energy $K = 0$. Potential energy $U = mgh$. Total energy $E = K + U = mgh$.
$2$. At a distance $x$ below $A$ (point $B$): The object has fallen a distance $x$,so its height is $h - x$. Using $v^2 = u^2 + 2as$,$v^2 = 0 + 2gx = 2gx$. Kinetic energy $K = \frac{1}{2}mv^2 = mgx$. Potential energy $U = mg(h - x)$. Total energy $E = K + U = mgx + mgh - mgx = mgh$.
$3$. At the ground (point $C$): The object has fallen a distance $h$. Using $v^2 = u^2 + 2as$,$v^2 = 0 + 2gh = 2gh$. Kinetic energy $K = \frac{1}{2}mv^2 = mgh$. Potential energy $U = 0$. Total energy $E = K + U = mgh + 0 = mgh$.
Since the total mechanical energy is $mgh$ at all points,the principle is established.

(N/A) Consider a ball of mass $m$ falling from the top of a cliff of height $H$,as shown in the figure.
For heights $H$,$h$,and $0$ (at the ground),the total mechanical energies $E_H$,$E_h$,and $E_0$ are calculated as follows:
At height $H$,the total mechanical energy is:
$E_H = mgH + \frac{1}{2}mv^2$
Since the velocity at the maximum height is $v = 0$,
$\therefore E_H = mgH$
At height $h$,the total energy is:
$E_h = \text{Potential Energy} + \text{Kinetic Energy}$
$E_h = mgh + \frac{1}{2}mv_h^2$ (where $v_h$ is the velocity of the ball at height $h$)
At the ground (height $0$),the total energy is:
$E_0 = \frac{1}{2}mv_f^2$
where $v_f$ is the final velocity of the ball when it hits the ground.
By the law of conservation of energy,$E_H = E_0$:
$mgH = \frac{1}{2}mv_f^2$
$\therefore v_f = \sqrt{2gH}$
Also,from the conservation of mechanical energy between height $H$ and height $h$:
$E_H = E_h$
$mgH = mgh + \frac{1}{2}mv_h^2$

(N/A) The principle of conservation of mechanical energy states that if only conservative forces act on a system, the total mechanical energy of the system remains constant.
Mathematically, the total mechanical energy $E$ is the sum of kinetic energy $K$ and potential energy $U$.
$E = K + U = \text{constant}$.
This implies that any change in kinetic energy is equal to the negative change in potential energy:
$\Delta K + \Delta U = 0$ or $\Delta K = -\Delta U$.

(N/A) The total mechanical energy $(E)$ of a body is the sum of its kinetic energy $(K)$ and potential energy $(U)$.
For a body of mass $m$ falling freely from a height $H$,at any height $h$ (where $0 \le h \le H$),the velocity $v$ of the body is given by the equation of motion $v^2 = u^2 + 2g(H-h)$. Since the body falls from rest,$u = 0$,so $v^2 = 2g(H-h)$.
The kinetic energy is $K = \frac{1}{2}mv^2 = \frac{1}{2}m(2g(H-h)) = mg(H-h)$.
The potential energy at height $h$ is $U = mgh$.
The total mechanical energy is $E = K + U = mg(H-h) + mgh = mgH - mgh + mgh = mgH$.
Thus,the equation for total mechanical energy is $E = mgH$.

(A) Let the total energy be $E = mgh$. At any height $y$ from the ground,Potential Energy $U = mgy$ and Kinetic Energy $K = mg(h-y)$.
For $(1)$,$U = K \implies mgy = mg(h-y) \implies y = h-y \implies 2y = h \implies y = \frac{h}{2}$. Thus,$(1) \rightarrow (c)$.
For $(2)$,$U = 2K \implies mgy = 2mg(h-y) \implies y = 2h - 2y \implies 3y = 2h \implies y = \frac{2h}{3}$. Thus,$(2) \rightarrow (a)$.
Therefore,the correct matching is $(1)-(c)$ and $(2)-(a)$.

(A) Let $h$ be the height at which the kinetic energy becomes half of the initial kinetic energy.
Initial kinetic energy $K_0 = \frac{1}{2} m v_0^2$.
At height $h$,kinetic energy $K = \frac{K_0}{2} = \frac{1}{4} m v_0^2$.
Using the work-energy theorem or conservation of energy,the loss in kinetic energy equals the gain in potential energy:
$K_0 - K = mgh$
$\frac{1}{2} m v_0^2 - \frac{1}{4} m v_0^2 = mgh$
$\frac{1}{4} m v_0^2 = mgh$
$h = \frac{v_0^2}{4g}$
Given $v_0 = 7 \, m/s$ and $g = 9.8 \, m/s^2$:
$h = \frac{7^2}{4 \times 9.8} = \frac{49}{39.2} = 1.25 \, m$.

(C) According to the Work-Energy Theorem,the net work done on an object is equal to the change in its kinetic energy.
$W_{net} = \Delta K = K_f - K_i$
Since the object is moving with a constant speed,its kinetic energy $K = \frac{1}{2}mv^2$ remains constant.
Therefore,the change in kinetic energy $\Delta K = 0$.
Hence,the net work done on the object is $0 \ J$.

(A) According to the work-energy theorem,the net work done on an object is equal to the change in its kinetic energy $(W = \Delta K)$.
If positive mechanical work is done on an object,its kinetic energy increases.
If negative mechanical work is done on an object,its kinetic energy decreases.
Assuming the question refers to work done by an external force in the direction of motion,the kinetic energy increases.

(C) According to the work-energy theorem,the change in kinetic energy is equal to the work done by the force: $\Delta K = W$.
Since the bodies are brought to rest,the work done by the retarding force $F$ over a distance $d$ is $W = F \cdot d$.
Given that the initial kinetic energy $K$ is the same for both bodies and the final kinetic energy is $0$,we have $\Delta K = K - 0 = K$.
Thus,$K = F \cdot d$,which implies $d = K / F$.
Since both the kinetic energy $K$ and the retarding force $F$ are the same for both bodies,the distance $d$ moved by them before coming to rest will be equal.

(N/A) The mechanical energy of the balloon alone is not conserved because an external buoyant force acts on it. The work done by the buoyant force is responsible for the increase in the total mechanical energy (kinetic energy + potential energy) of the balloon.
Let $m$ be the mass of the balloon,$V$ be its volume,$\rho_{He}$ be the density of helium,and $\rho_{air}$ be the density of air.
The net upward force on the balloon is $F_{net} = V(\rho_{air} - \rho_{He})g - mg = ma$.
As the balloon rises to a height $h$,the work done by the buoyant force is $W_b = V \rho_{air} g h$.
The change in mechanical energy is $\Delta E = \Delta K + \Delta U = \frac{1}{2}mv^2 + mgh$.
Using the work-energy theorem,the work done by the net force is equal to the change in kinetic energy: $W_{net} = [V(\rho_{air} - \rho_{He})g - mg]h = \frac{1}{2}mv^2$.
Rearranging this,we get $V \rho_{air} g h = \frac{1}{2}mv^2 + mgh$.
This shows that the work done by the buoyant force $(V \rho_{air} g h)$ is exactly equal to the increase in the total mechanical energy of the balloon.

(N/A) The correct statement is: '$A$ system can possess mechanical energy but cannot possess work.'
Reasoning:
$1$. Energy is a state function,meaning it is a property of the system's state. Therefore,a system can possess internal energy or mechanical energy.
$2$. Work is a path function,not a state function. It represents energy in transit across the boundary of a system due to a force acting through a distance. $A$ system does not 'have' or 'possess' work; it performs work or has work done on it.

(B) According to the Work-Energy Theorem,the work done by the force $F$ on the ball is equal to the change in its kinetic energy,which is equal to the potential energy gained at the maximum height.
Work done by the machine,$W = F \times S$,where $S = 0.2 \, m$.
Potential energy gained at maximum height,$U = mgh$,where $m = 0.15 \, kg$,$g = 10 \, m/s^2$,and $h = 20 \, m$.
Equating the work done to the potential energy: $F \times S = mgh$.
$F \times 0.2 = 0.15 \times 10 \times 20$.
$F \times 0.2 = 30$.
$F = \frac{30}{0.2} = 150 \, N$.

(D) According to the law of conservation of mechanical energy,the total mechanical energy at point $A$ is equal to the total mechanical energy at point $B$.
$E_A = E_B$
$PE_A + KE_A = PE_B + KE_B$
Given that the particle starts from rest at $A$,$KE_A = 0$.
$mgh_A + 0 = mgh_B + \frac{1}{2}mv^2$
$gh_A = gh_B + \frac{1}{2}v^2$
$v^2 = 2g(h_A - h_B)$
Substituting the given values: $g = 10 \ m/s^2$,$h_A = 10 \ m$,$h_B = 5 \ m$.
$v^2 = 2 \times 10 \times (10 - 5)$
$v^2 = 20 \times 5 = 100$
$v = \sqrt{100} = 10 \ m/s$
Therefore,the value of $x$ is $10$.

(A) Let the total length of the chain be $L = 3\, m$ and total mass be $M = 3\, kg$. The linear mass density is $\lambda = M/L = 3/3 = 1\, kg/m$.
Initially,$2\, m$ of the chain is on the table and $1\, m$ is hanging. The center of mass of the hanging part is at a distance of $0.5\, m$ below the table level. Taking the table surface as the reference level $(U = 0)$,the initial potential energy is $U_i = -(m_{hanging})g(h_{cm}) = -(1\, kg)(10\, m/s^2)(0.5\, m) = -5\, J$.
The initial kinetic energy is $K_i = 0$.
Finally,when the chain completely slips off,the entire chain of length $3\, m$ hangs vertically. Its center of mass is at $1.5\, m$ below the table level. The final potential energy is $U_f = -(M)g(h_{cm}) = -(3\, kg)(10\, m/s^2)(1.5\, m) = -45\, J$.
By the law of conservation of mechanical energy,$K_i + U_i = K_f + U_f$.
$0 + (-5\, J) = K_f + (-45\, J)$.
$K_f = 45 - 5 = 40\, J$.
Thus,the value of $k$ is $40$.

(A) According to the Work-Energy Theorem,the total work done on the body is equal to the change in its kinetic energy.
$W_{total} = \Delta K$
$W_{gravity} + W_{air\ friction} = K_f - K_i$
Given that the body is dropped,initial velocity $u = 0$,so $K_i = 0$.
The work done by gravity is $W_{gravity} = mgh$.
The final kinetic energy is $K_f = \frac{1}{2} m v^2 = \frac{1}{2} m (0.8 \sqrt{gh})^2$.
$K_f = \frac{1}{2} m (0.64 gh) = 0.32 mgh$.
Substituting these values into the theorem:
$mgh + W_{air\ friction} = 0.32 mgh - 0$
$W_{air\ friction} = 0.32 mgh - mgh = -0.68 mgh$.
Thus,the work done by air friction is $-0.68 mgh$.

(D) Let the mass of the bob be $m$, length of the pendulum be $l = 0.5 \, m$, and initial speed at the lowest point be $u = 3 \, m/s$.
Using the law of conservation of mechanical energy between the lowest point $(A)$ and the point where the string makes an angle $\theta = 60^{\circ}$ with the vertical $(B)$:
Total energy at $A$ = Total energy at $B$
$\frac{1}{2} mu^2 = \frac{1}{2} mv^2 + mgh$
Where $h$ is the vertical height gained by the bob, given by $h = l(1 - \cos \theta)$.
Substituting the values:
$\frac{1}{2} m(3)^2 = \frac{1}{2} mv^2 + mg(0.5)(1 - \cos 60^{\circ})$
$9 = v^2 + 2(10)(0.5)(1 - 0.5)$
$9 = v^2 + 10(0.5)$
$9 = v^2 + 5$
$v^2 = 4$
$v = 2 \, m/s$.

(D) According to the work-energy theorem, the work done by all forces acting on an object is equal to the change in its kinetic energy.
Here, the only force doing work on the block is gravity.
Work done by gravity $(W_g)$ = Change in kinetic energy $(\Delta K)$
$W_g = K_B - K_A$
Since the block is dropped from point $A$, its initial velocity is zero, so $K_A = 0$.
The vertical displacement of the block from point $A$ to point $B$ is $y_0$.
Therefore, the work done by gravity is $W_g = m g y_0$.
Substituting these values into the work-energy theorem:
$m g y_0 = K_B - 0$
$K_B = m g y_0$

(D) According to the work-energy theorem,the work done by the net force is equal to the change in kinetic energy: $W = \Delta K.E. = K_f - K_i$.
Given mass $m = 500 \,g = 0.5 \,kg$ and velocity $v = b x^{5/2}$.
At $x_i = 0$,$v_i = b(0)^{5/2} = 0$,so $K_i = 0$.
At $x_f = 4 \,m$,$v_f = b(4)^{5/2} = 0.25 \times (2^2)^{5/2} = 0.25 \times 2^5 = 0.25 \times 32 = 8 \,m/s$.
The work done is $W = \frac{1}{2} m v_f^2 - 0 = \frac{1}{2} \times 0.5 \times (8)^2$.
$W = 0.25 \times 64 = 16 \,J$.

(B) According to the Work-Energy Theorem,the net work done is equal to the change in kinetic energy: $W = \Delta K = K_f - K_i$.
Given mass $m = 0.5\, kg$ and velocity $v = (3x^2 + 4)\, m/s$.
At initial position $x_i = 0\, m$,the velocity is $v_i = 3(0)^2 + 4 = 4\, m/s$.
At final position $x_f = 2\, m$,the velocity is $v_f = 3(2)^2 + 4 = 3(4) + 4 = 16\, m/s$.
The change in kinetic energy is $\Delta K = \frac{1}{2} m (v_f^2 - v_i^2)$.
Substituting the values: $W = \frac{1}{2} \times 0.5 \times (16^2 - 4^2)$.
$W = 0.25 \times (256 - 16) = 0.25 \times 240$.
$W = 60\, J$.

(D) According to the work-energy theorem,the work done by the spring force is equal to the change in kinetic energy of the block.
$W_{\text{net}} = K_f - K_i$
Given that the initial kinetic energy is $K_i = E = \frac{1}{2}mv^2$.
When the speed is halved,the final speed is $v' = \frac{v}{2}$.
The final kinetic energy is $K_f = \frac{1}{2}m(\frac{v}{2})^2 = \frac{1}{4}(\frac{1}{2}mv^2) = \frac{E}{4}$.
The work done by the spring in compressing it by a distance $x = 25\,cm = 0.25\,m = \frac{1}{4}\,m$ is $W = -\frac{1}{2}Kx^2$.
Applying the work-energy theorem:
$-\frac{1}{2}Kx^2 = K_f - K_i$
$-\frac{1}{2}K(\frac{1}{4})^2 = \frac{E}{4} - E$
$-\frac{1}{2}K(\frac{1}{16}) = -\frac{3E}{4}$
$\frac{K}{32} = \frac{3E}{4}$
$K = \frac{3E \times 32}{4} = 24E$
Comparing this with $nE$,we get $n = 24$.

(A) Let $v$ be the speed of the particle when it separates from the hemisphere.
Since the surface is smooth,there is no friction,and mechanical energy is conserved.
The particle starts from rest at the top of the hemisphere (height $R$ from the horizontal surface).
The potential energy at the top is $U_i = mgR$.
The potential energy at height $h$ is $U_f = mgh$.
The kinetic energy at height $h$ is $K_f = \frac{1}{2}mv^2$.
By the law of conservation of energy:
$U_i = U_f + K_f$
$mgR = mgh + \frac{1}{2}mv^2$
$mg(R-h) = \frac{1}{2}mv^2$
$v^2 = 2g(R-h)$
$v = \sqrt{2g(R-h)}$
Thus,the correct option is $A$.

(C) Since the block is sliding down the inclined plane with a constant speed,its acceleration is zero. Therefore,the change in kinetic energy $(\Delta KE)$ of the block as it moves from height $h$ to the ground is zero.
According to the work-energy theorem,the total work done on the block is equal to the change in its kinetic energy:
$W_{\text{total}} = \Delta KE = 0$
The total work done is the sum of the work done by gravity $(W_{\text{gravity}})$ and the work done by friction $(W_{\text{friction}})$:
$W_{\text{gravity}} + W_{\text{friction}} = 0$
The work done by gravity as the block descends a vertical height $h$ is $W_{\text{gravity}} = mgh$.
Substituting this into the equation:
$mgh + W_{\text{friction}} = 0$
$W_{\text{friction}} = -mgh$
The energy dissipated by friction is the magnitude of the work done by the friction force:
$\text{Energy dissipated} = |W_{\text{friction}}| = mgh$.

(B) Let the initial position of the center of mass be $h_i = 1.0 \,m$.
When she bends her knees,the center of mass is lowered by $0.2 \,m$,so the new position is $h_{new} = 1.0 - 0.2 = 0.8 \,m$.
She pushes the ground with a constant force $F$ over a distance of $d = 0.2 \,m$.
At the maximum height,the feet are $0.3 \,m$ above the ground. Since the center of mass is $1.0 \,m$ above the feet when standing straight,the final height of the center of mass is $h_f = 0.3 + 1.0 = 1.3 \,m$.
Applying the work-energy theorem from the position where she starts pushing to the maximum height:
Work done by force $F$ + Work done by gravity = Change in kinetic energy.
$F \times 0.2 - Mg(h_f - h_{new}) = 0 - 0$.
$F \times 0.2 - Mg(1.3 - 0.8) = 0$.
$F \times 0.2 = Mg \times 0.5$.
$\frac{F}{Mg} = \frac{0.5}{0.2} = 2.5$.

(D) According to the Work-Energy Theorem,the work done by all forces acting on an object is equal to the change in its kinetic energy.
$W = \Delta K = K_f - K_i$
Given:
Mass $m = 20 \,g = 0.02 \,kg$
Initial velocity $v_i = 100 \,m/s$
Final velocity $v_f = 50 \,m/s$
$W = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 = \frac{1}{2} m (v_f^2 - v_i^2)$
$W = \frac{1}{2} \times 0.02 \times (50^2 - 100^2)$
$W = 0.01 \times (2500 - 10000)$
$W = 0.01 \times (-7500) = -75 \,J$
The work done by the air resistance is $-75 \,J$. Since the question asks for the amount of work done by the resistance of air,we consider the magnitude,which is $75 \,J$.

(B) According to the work-energy theorem,the net work done on the object is equal to the change in its kinetic energy: $W_{\text{net}} = \Delta K$.
During the upward motion,both the gravitational force $(w = mg)$ and the air drag force $(f)$ act downwards,opposing the motion.
The work done by gravity is $W_g = -wh = -mgh$.
The work done by air drag is $W_f = -fh$.
The change in kinetic energy is $\Delta K = K_f - K_i = 0 - \frac{1}{2}mv_0^2$.
Applying the theorem: $W_g + W_f = \Delta K$.
$-mgh - fh = -\frac{1}{2}mv_0^2$.
Dividing by $m$: $-gh - \frac{f}{m}h = -\frac{1}{2}v_0^2$.
Since $w = mg$,we have $m = \frac{w}{g}$. Substituting this into the drag term: $-gh - \frac{fg}{w}h = -\frac{1}{2}v_0^2$.
$h(g + \frac{fg}{w}) = \frac{1}{2}v_0^2$.
$h g (1 + \frac{f}{w}) = \frac{1}{2}v_0^2$.
$h = \frac{v_0^2}{2g(1 + \frac{f}{w})}$.

(C) According to the Work-Energy Theorem,the total work done on an object is equal to the change in its kinetic energy: $W_{\text{all}} = \Delta K$.
Here,the forces acting on the knife are gravity and the resistive force exerted by the wooden block. Since the knife starts from rest and comes to rest,the change in kinetic energy $\Delta K = 0$.
Thus,$W_{\text{gravity}} + W_{\text{block}} = 0$.
The total vertical displacement of the knife is $(x + y)$. Therefore,the work done by gravity is $W_{\text{gravity}} = m g (x + y)$.
Substituting this into the equation: $m g (x + y) + W_{\text{block}} = 0$.
Solving for the work done by the block: $W_{\text{block}} = -m g (x + y)$.

(D) Let the total distance travelled by the block along the incline be $d = 1 \, m$.
According to the law of conservation of mechanical energy,the loss in gravitational potential energy of the block is equal to the gain in elastic potential energy of the spring.
Let $d$ be the total distance travelled by the block along the incline. The vertical height dropped by the block is $h = d \sin 45^{\circ}$.
The block comes to rest after compressing the spring by $x = 1 \, m$.
Thus,the total distance travelled by the block is $d = (d_{initial} + x)$.
However,the problem asks for the total distance travelled along the incline. Let this distance be $d$.
Initial potential energy $U_i = mgd \sin 45^{\circ}$.
Final potential energy $U_f = \frac{1}{2} k x^2$.
Equating $U_i = U_f$:
$mgd \sin 45^{\circ} = \frac{1}{2} k x^2$
$(\sqrt{2}) \times 10 \times d \times \frac{1}{\sqrt{2}} = \frac{1}{2} \times 100 \times (1)^2$
$10d = 50$
$d = 5 \, m$.

(D) Given: Mass $m = 2 \, kg$,position $x = \frac{t^2}{3}$.
First,find the velocity $v$ by differentiating $x$ with respect to $t$: $v = \frac{dx}{dt} = \frac{d}{dt}(\frac{t^2}{3}) = \frac{2t}{3}$.
At $t = 0 \, s$,velocity $v_i = \frac{2(0)}{3} = 0 \, m/s$.
At $t = 2 \, s$,velocity $v_f = \frac{2(2)}{3} = \frac{4}{3} \, m/s$.
According to the Work-Energy Theorem,the work done $W$ is equal to the change in kinetic energy $\Delta K.E. = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$.
$W = \frac{1}{2} \times 2 \times (\frac{4}{3})^2 - 0 = \frac{16}{9} \, J$.

(C) According to the Work-Energy Theorem,the work done by the net force is equal to the change in kinetic energy of the particle.
$W = \Delta K = K_f - K_i$
Given $v = a \sqrt{x}$ and mass $m = 2 \,kg$.
At $x = 0$,$v_i = a \sqrt{0} = 0$.
At $x = 4 \,m$,$v_f = a \sqrt{4} = 2a$.
Initial kinetic energy $K_i = \frac{1}{2} m v_i^2 = \frac{1}{2} (2) (0)^2 = 0$.
Final kinetic energy $K_f = \frac{1}{2} m v_f^2 = \frac{1}{2} (2) (2a)^2 = 4a^2$.
Therefore,work done $W = 4a^2 - 0 = 4a^2$.

(A) The Work-Energy Theorem states that the net work done by all forces (conservative,non-conservative,internal,and external) acting on a particle is equal to the change in its kinetic energy.
Mathematically,$W_{\text{net}} = \Delta K = K_f - K_i$.
This theorem is derived from Newton's second law of motion and is valid in all inertial frames of reference. When applied correctly by including pseudo forces in non-inertial frames,it remains valid there as well.
Therefore,the statement is always applicable.

(B) According to the Work-Energy Theorem,the work done by the braking force is equal to the change in kinetic energy of the bus.
For the unloaded bus of mass $m$:
$W = F \cdot x = \frac{1}{2} m v^2$ --- (Equation $1$)
For the loaded bus,the new mass is $m' = m + 0.5 m = 1.5 m$. Let the new stopping distance be $x'$.
$W' = F \cdot x' = \frac{1}{2} (1.5 m) v^2$ --- (Equation $2$)
Dividing Equation $2$ by Equation $1$:
$\frac{F \cdot x'}{F \cdot x} = \frac{\frac{1}{2} (1.5 m) v^2}{\frac{1}{2} m v^2}$
$\frac{x'}{x} = 1.5$
$x' = 1.5 x$
Therefore,the bus will travel a distance of $1.5 x$.

(D) According to the work-energy theorem,the total work done on the block is equal to the change in its kinetic energy.
$W_{\text{total}} = \Delta KE = KE_f - KE_i$
Here,$W_{\text{total}} = W_{\text{friction}} + W_{\text{gravity}}$.
The block starts at the top and stops at the bottom,so the net change in height is $h = 2r = 2 \times 0.15\,m = 0.3\,m$.
The work done by gravity is $W_g = mg \times h = 1 \times 10 \times 0.3 = 3\,J$.
The initial kinetic energy is $KE_i = \frac{1}{2}mv^2 = \frac{1}{2} \times 1 \times (22)^2 = \frac{484}{2} = 242\,J$.
The final kinetic energy is $KE_f = 0\,J$ (since it stops).
Applying the work-energy theorem:
$W_f + W_g = KE_f - KE_i$
$W_f + 3 = 0 - 242$
$W_f = -242 - 3 = -245\,J$.
Therefore,the work done by the tube (friction) on the block is $-245\,J$.

(B) According to the work-energy theorem,the work done (energy spent) is equal to the change in kinetic energy.
For the first process (from rest to $u \ m/s$):
$E_1 = \frac{1}{2} m u^2 - 0 = \frac{1}{2} m u^2 = E$
For the second process (from $u \ m/s$ to $2u \ m/s$):
$E_2 = \frac{1}{2} m (2u)^2 - \frac{1}{2} m u^2$
$E_2 = \frac{1}{2} m (4u^2) - \frac{1}{2} m u^2 = 2 m u^2 - 0.5 m u^2 = 1.5 m u^2 = 3 \left( \frac{1}{2} m u^2 \right)$
Since $E = \frac{1}{2} m u^2$,we have $E_2 = 3E$.
Comparing this with $nE$,we get $n = 3$.

(A) Given,length of the pendulum $\ell = 10 \ m$ and acceleration due to gravity $g = 10 \ ms^{-2}$.
Initially,the bob is at a horizontal position,so its initial potential energy with respect to the lowest point is $U_i = mg\ell$.
The bob is released from rest,so its initial kinetic energy is $K_i = 0$.
Total initial energy $E_i = mg\ell$.
As the bob moves to the lowest point,it dissipates $10 \%$ of its initial energy against air resistance.
Energy lost = $0.10 \times mg\ell$.
Remaining energy at the lowest point = $E_f = E_i - 0.10 \times E_i = 0.90 \times mg\ell$.
At the lowest point,the potential energy is $0$,so the entire remaining energy is kinetic energy $K_f = \frac{1}{2}mv^2$.
Equating the energies: $\frac{1}{2}mv^2 = 0.90 \times mg\ell$.
$v^2 = 2 \times 0.90 \times g \times \ell = 1.8 \times 10 \times 10 = 180$.
$v = \sqrt{180} = \sqrt{36 \times 5} = 6\sqrt{5} \ ms^{-1}$.

(B) Since the track is frictionless,the total mechanical energy of the particle is conserved.
Applying the Law of Conservation of Mechanical Energy between points $A$ and $B$:
$KE_A + PE_A = KE_B + PE_B$
Here,$KE_A = 0$ (as it is gently pushed,initial velocity is negligible),$PE_A = mgh_A$,$KE_B = \frac{1}{2}mv^2$,and $PE_B = mgh_B$.
Given: $h_A = 1 \ m$,$h_B = 0.5 \ m$,and $g = 10 \ m/s^2$.
Substituting the values:
$0 + mg(1) = \frac{1}{2}mv^2 + mg(0.5)$
$mg(1 - 0.5) = \frac{1}{2}mv^2$
$mg(0.5) = \frac{1}{2}mv^2$
$g = v^2$
$v = \sqrt{g} = \sqrt{10} \ m/s$.

(C) According to the Work-Energy Theorem,the total work done by all forces is equal to the change in kinetic energy of the particle.
$W = \Delta K = K_f - K_i$
Given the velocity equation $v = \alpha \sqrt{x}$:
At $x = 0$,the initial velocity $v_i = \alpha \sqrt{0} = 0$.
At $x = d$,the final velocity $v_f = \alpha \sqrt{d}$.
The initial kinetic energy $K_i = \frac{1}{2} m v_i^2 = \frac{1}{2} m (0)^2 = 0$.
The final kinetic energy $K_f = \frac{1}{2} m v_f^2 = \frac{1}{2} m (\alpha \sqrt{d})^2 = \frac{1}{2} m \alpha^2 d$.
Therefore,the total work done is $W = \frac{1}{2} m \alpha^2 d - 0 = \frac{m \alpha^2 d}{2}$.

(B) According to the work-energy theorem,the total work done on the block is equal to the change in its kinetic energy:
$W_{total} = W_{force} + W_{gravity} = K_f - K_i$
Given that the block starts from rest,$K_i = 0$. The force $F = 18 \ N$ is applied parallel to the displacement vector $\vec{PQ}$. The length of the displacement $PQ$ is calculated using the Pythagorean theorem:
$PQ = \sqrt{OP^2 + OQ^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = 5 \ m$
The work done by the force is:
$W_{force} = F \times PQ = 18 \ N \times 5 \ m = 90 \ J$
The work done by gravity is:
$W_{gravity} = -mgh = -1 \ kg \times 10 \ m/s^2 \times 4 \ m = -40 \ J$
Therefore,the final kinetic energy $K_f$ is:
$K_f = 90 \ J - 40 \ J = 50 \ J$
We are given $K_f = (n \times 10) \ J$,so:
$n \times 10 = 50 \implies n = 5$