Work Energy Theorem and Conservation of Mechanical Energy Questions in English

(B) According to the work-energy theorem,the loss in potential energy is equal to the work done against friction on the horizontal surface.
Let $h = 1.5 \ m$ be the initial height and $\mu = 0.2$ be the coefficient of friction.
The potential energy at the start is $PE = Mgh$.
The work done by friction over a distance $s$ on the horizontal surface is $W_f = f \cdot s = (\mu Mg) \cdot s$.
Equating the two: $Mgh = \mu Mgs$.
Solving for $s$: $s = \frac{h}{\mu} = \frac{1.5}{0.2} = 7.5 \ m$.
Since $7.5 \ m < 15 \ m$,the body comes to rest at a distance of $7.5 \ m$ from point $P$.

(A) Initial kinetic energy $K_i = \frac{1}{2} m u^2 = \frac{1}{2} \times 2 \times (4)^2 = 16 \, J$.
At the required height $h$,the kinetic energy $K_f = \frac{1}{2} K_i = 8 \, J$.
Using the law of conservation of energy: $K_i + P_i = K_f + P_f$.
Taking the ground as the reference level $(P_i = 0)$,we have $16 + 0 = 8 + mgh$.
$16 = 8 + 2 \times 10 \times h$.
$8 = 20h$.
$h = \frac{8}{20} = 0.4 \, m$.

(B) According to the work-energy theorem,the work done by all forces equals the change in kinetic energy.
Initial potential energy at $A$ is $U_A = mgh = 2 \times 10 \times 1 = 20 \, J$.
At point $C$,the body comes to rest,so the final kinetic energy is $0$.
The total initial energy is $20 \, J$ (assuming $g = 10 \, m/s^2$).
Since the body comes to rest at $C$,all the initial potential energy is dissipated as work done against friction.
Therefore,the work done against friction is $W_f = 20 \, J$.

(B) Let the total mechanical energy of the body be $E = mgh$,where $m$ is the mass,$g$ is the acceleration due to gravity,and $h$ is the maximum height.
At a height of $y = \frac{3h}{4}$,the potential energy $U$ is given by $U = mgy = mg(\frac{3h}{4}) = \frac{3}{4}mgh$.
According to the law of conservation of energy,the total energy $E$ remains constant.
Therefore,the kinetic energy $K$ at this height is $K = E - U = mgh - \frac{3}{4}mgh = \frac{1}{4}mgh$.
The ratio of kinetic energy to potential energy is $\frac{K}{U} = \frac{\frac{1}{4}mgh}{\frac{3}{4}mgh} = \frac{1}{3}$.

(A) According to the work-energy theorem,the work done by friction is equal to the change in kinetic energy of the block.
The heat produced is equal to the magnitude of the work done by friction.
Given: mass $m = 100 \, g = 0.1 \, kg$,initial velocity $u = 10 \, m/s$,final velocity $v = 5 \, m/s$.
Change in kinetic energy $\Delta K = \frac{1}{2} m (u^2 - v^2)$.
$\Delta K = \frac{1}{2} \times 0.1 \times (10^2 - 5^2)$.
$\Delta K = 0.05 \times (100 - 25) = 0.05 \times 75 = 3.75 \, J$.
Thus,the heat produced is $3.75 \, J$.

(A) According to the work-energy theorem,the net work done on the system is equal to the change in kinetic energy of the ball.
Let the initial position of the ball be at height $h$ above the spring. The final position is when the spring is compressed by distance $d$.
The total vertical displacement of the ball is $(h + d)$.
The forces acting on the ball are gravity (downward) and the spring force (upward).
The work done by gravity is $W_g = mg(h + d)$.
The work done by the spring force is $W_s = -\int_0^d kx \, dx = -\frac{1}{2}kd^2$.
The net work done on the ball is $W_{net} = W_g + W_s = mg(h + d) - \frac{1}{2}kd^2$.
Since the ball starts from rest and comes to rest momentarily at the maximum compression $d$,the change in kinetic energy is zero,which implies $W_{net} = 0$ for the entire process. However,the question asks for the net work done by the external forces (gravity and spring) during the displacement $d$,which is $mg(h + d) - \frac{1}{2}kd^2$.

(C) According to the law of conservation of energy,the loss in gravitational potential energy is equal to the gain in kinetic energy.
Given that the loss in gravitational potential energy is $U$.
The gain in kinetic energy is $\frac{1}{2}mv^2$.
Equating the two: $U = \frac{1}{2}mv^2$.
Solving for mass $m$: $m = \frac{2U}{v^2}$.

(B) Let the equilibrium position be at a distance $x$ from the natural length.
At equilibrium,the spring force equals the gravitational force: $mg = kx$.
Given $k = mg/a$,we have $mg = (mg/a)x$,which gives $x = a$.
Using the principle of conservation of energy,the loss in gravitational potential energy equals the gain in elastic potential energy plus the kinetic energy at the equilibrium position.
$mgx = \frac{1}{2}kx^2 + KE$.
Substituting $x = a$ and $k = mg/a$:
$mg(a) = \frac{1}{2}(mg/a)(a^2) + KE$.
$mga = \frac{1}{2}mga + KE$.
$KE = mga - \frac{1}{2}mga = \frac{1}{2}mga$.

(C) According to the work-energy theorem,the net work done on an object is equal to the change in its kinetic energy.
For observer $A$ (who is in an inertial frame),the initial kinetic energy of the block is $K_i = \frac{1}{2}mv_0^2$.
Since the block comes to rest,the final kinetic energy is $K_f = 0$.
The net work done on the block is $W_{net} = K_f - K_i = 0 - \frac{1}{2}mv_0^2 = -\frac{1}{2}mv_0^2$.

(C) $1$. The system consists of a block of mass $m$ attached to a spring of constant $K$ on a rough surface with coefficient of friction $\mu$.
$2$. Since the surface is rough,there is a non-conservative force (friction) acting on the system. Therefore,mechanical energy is not conserved.
$3$. The net external force on the system is not zero (due to the wall and friction),so linear momentum is not conserved.
$4$. The work-energy principle states that the work done by all forces (conservative and non-conservative) is equal to the change in kinetic energy of the system. This principle is universally applicable to any system,regardless of whether forces are conservative or non-conservative.
$5$. Thus,the work-energy principle can be applied to this system.

(D) The correct answer is $mgh = \frac{1}{2}mv^2$. This equation is an application of the law of conservation of mechanical energy. In this situation,the loss in potential energy,$mgh$,is equal to the gain in kinetic energy,$\frac{1}{2}mv^2$.
The other equations are kinematic equations that are only valid when the acceleration is constant. Since the track is curved,the component of the gravitational force on the box that is tangent to the track surface changes as the box slides. This component of the gravitational force is what accelerates the box. According to Newton's second law,$F = ma$,as the force changes,the acceleration also changes. Since the acceleration is not constant,the other equations are not valid.

(A) The mass of the particle is $m = 100 \ g = 0.1 \ kg$.
The initial velocity is $u = 5 \ m/s$.
At the highest point,the final velocity is $v = 0 \ m/s$.
According to the work-energy theorem,the work done by all forces is equal to the change in kinetic energy: $W_{total} = \Delta K.E$.
Here,the only force acting during the upward motion is gravity $(W_g)$.
$W_g = K.E_{final} - K.E_{initial} = 0 - \frac{1}{2} m u^2$.
$W_g = -\frac{1}{2} \times 0.1 \times (5)^2 = -0.5 \times 0.1 \times 25 = -1.25 \ J$.

(A) The total mechanical energy $E$ is the sum of kinetic energy $K$ and potential energy $U$: $E = K + U = 2 \ J$.
To find the maximum speed,we need to maximize the kinetic energy $K = E - U$.
This occurs when the potential energy $U(x)$ is at its minimum.
To find the minimum of $U(x)$,we set the derivative $\frac{dU}{dx} = 0$:
$\frac{dU}{dx} = \frac{4x^3}{4} - \frac{2x}{2} = x^3 - x = 0$.
$x(x^2 - 1) = 0$,which gives $x = 0, 1, -1$.
Evaluating $U(x)$ at these points:
$U(0) = 0 \ J$.
$U(1) = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \ J$.
$U(-1) = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \ J$.
The minimum potential energy is $U_{\min} = -\frac{1}{4} \ J$.
Thus,$K_{\max} = E - U_{\min} = 2 - (-0.25) = 2.25 \ J = \frac{9}{4} \ J$.
Using $K_{\max} = \frac{1}{2}mv_{\max}^2$ with $m = 1 \ kg$:
$\frac{1}{2} \times 1 \times v_{\max}^2 = \frac{9}{4}$.
$v_{\max}^2 = \frac{9}{2}$.
$v_{\max} = \frac{3}{\sqrt{2}} \ m/s$.

(C) Let the mass per unit length of the chain be $\lambda = \frac{m}{L}$.
Initially,a length $b$ of the chain is on the incline. The center of mass of this part is at a vertical depth of $\frac{b}{2} \sin \theta$ below the horizontal surface.
The potential energy of the chain initially is $PE_i = -(\lambda b) g (\frac{b}{2} \sin \theta) = -\frac{m g b^2}{2L} \sin \theta$.
Finally,the entire chain of length $L$ is on the incline. The center of mass of the entire chain is at a vertical depth of $\frac{L}{2} \sin \theta$ below the horizontal surface.
The potential energy of the chain finally is $PE_f = -(m) g (\frac{L}{2} \sin \theta) = -\frac{m g L}{2} \sin \theta$.
By the law of conservation of mechanical energy,the loss in potential energy equals the gain in kinetic energy:
$PE_i - PE_f = KE_f - KE_i$
$-\frac{m g b^2}{2L} \sin \theta - (-\frac{m g L}{2} \sin \theta) = \frac{1}{2} m v^2 - 0$
$\frac{m g}{2L} \sin \theta (L^2 - b^2) = \frac{1}{2} m v^2$
$v^2 = \frac{g \sin \theta}{L} (L^2 - b^2)$
$v = \sqrt{\frac{g \sin \theta}{L} (L^2 - b^2)}$

(C) Let $R$ be the radius of the cylinder. The vertical height fallen by the block from point $A$ to point $B$ is $h = R \cos \theta_1 - R \cos \theta_2 = R(\cos \theta_1 - \cos \theta_2)$.
By the law of conservation of energy,the kinetic energy gained is equal to the potential energy lost: $\frac{1}{2}mv^2 = mgh = mgR(\cos \theta_1 - \cos \theta_2)$.
At point $B$,the normal force $N$ becomes zero as the block leaves the surface. The radial equation of motion is $mg \cos \theta_2 - N = \frac{mv^2}{R}$.
Setting $N = 0$,we get $v^2 = Rg \cos \theta_2$.
Substituting this into the energy equation: $\frac{1}{2}m(Rg \cos \theta_2) = mgR(\cos \theta_1 - \cos \theta_2)$.
Dividing by $mgR$: $\frac{1}{2} \cos \theta_2 = \cos \theta_1 - \cos \theta_2$.
Rearranging gives: $\frac{3}{2} \cos \theta_2 = \cos \theta_1$,or $\cos \theta_2 = \frac{2}{3} \cos \theta_1$.

(C) Given: Mass $m = 0.5\ g = 0.5 \times 10^{-3}\ kg$, Spring constant $k = 1.25\ N/cm = 125\ N/m$, Compression $x = 2.0\ cm = 0.02\ m$, Distance $d = 10\ cm = 0.1\ m$, Friction force $f = 0.0475\ N$.
According to the Work-Energy Theorem, the work done by the spring minus the work done against friction equals the change in kinetic energy:
$W_{spring} - W_{friction} = \Delta K$
$\frac{1}{2} k x^2 - f d = \frac{1}{2} m v^2$
Substituting the values:
$\frac{1}{2} \times 125 \times (0.02)^2 - 0.0475 \times 0.1 = \frac{1}{2} \times 0.5 \times 10^{-3} \times v^2$
$0.5 \times 125 \times 4 \times 10^{-4} - 0.00475 = 0.25 \times 10^{-3} \times v^2$
$0.025 - 0.00475 = 0.25 \times 10^{-3} \times v^2$
$0.02025 = 0.25 \times 10^{-3} \times v^2$
$v^2 = \frac{0.02025}{0.25 \times 10^{-3}} = \frac{20.25}{0.25} = 81$
$v = 9\ m/s$.

(C) Let the mass of the chain be $M$. The length of the chain is $L = \pi R$. The mass per unit length is $\lambda = M/(\pi R)$.
Initially,the center of mass of the semicircular chain is at a height $h_i = \frac{2R}{\pi}$ from the horizontal diameter.
Initial potential energy $PE_i = Mgh_i = Mg \frac{2R}{\pi}$. Initial kinetic energy $KE_i = 0$.
When the chain emerges from the tube,it hangs vertically. The center of mass of the chain of length $\pi R$ is at a distance $L/2 = \pi R / 2$ below the horizontal diameter.
Final potential energy $PE_f = -Mg \frac{\pi R}{2}$. Final kinetic energy $KE_f = \frac{1}{2} Mv^2$.
By the law of conservation of mechanical energy: $KE_i + PE_i = KE_f + PE_f$.
$0 + Mg \frac{2R}{\pi} = \frac{1}{2} Mv^2 - Mg \frac{\pi R}{2}$.
$MgR \left( \frac{2}{\pi} + \frac{\pi}{2} \right) = \frac{1}{2} Mv^2$.
$v^2 = 2gR \left( \frac{2}{\pi} + \frac{\pi}{2} \right)$.
$v = \sqrt{2gR \left( \frac{2}{\pi} + \frac{\pi}{2} \right)}$.

(C) According to the Work-Energy Theorem,the total work done on an object is equal to the change in its kinetic energy: $W = \Delta K = K_f - K_i$.
Since both monkeys start from rest at the same height $h$,their initial kinetic energy $K_i = 0$ is the same.
Both monkeys hit the floor at the same final height (zero),so their final kinetic energy $K_f$ depends only on the change in potential energy,which is $\Delta U = mgh$ for both.
For monkey $B$,the only force doing work is gravity,so $W_B = \Delta K_B = mgh$.
For monkey $A$,the forces acting are gravity and the tension in the rope. The tension acts perpendicular to the displacement during the swing,so it does zero work. When monkey $A$ lets go,it is in free fall. Thus,the total work done on monkey $A$ is also $W_A = mgh$.
Since $W_A = W_B$,both monkeys have the same change in kinetic energy,and since they start from rest,they have the same final kinetic energy $K_f = mgh$.
Therefore,their final speeds $v = \sqrt{2gh}$ are also equal: $v_A = v_B$.

(B) The marble rolls without slipping. The force of static friction acts on the marble,but it does no work because the point of contact is instantaneously at rest relative to the track. Since we are neglecting air resistance and any other non-conservative forces (like rolling resistance),the total mechanical energy of the system is conserved.
Initially,the marble is at rest at height $h$,so its total mechanical energy is $E_i = mgh$.
At the highest point on the other side,the marble briefly stops at height $h_1$. Its velocity is zero,so its kinetic energy (both translational and rotational) is zero. Its total mechanical energy is $E_f = mgh_1$.
By the law of conservation of mechanical energy,$E_i = E_f$,which implies $mgh = mgh_1$,or $h_1 = h$.
Therefore,the height reached on the other side is equal to the initial height,regardless of the angles of inclination $\theta$ and $\phi$.

(D) According to the work-energy theorem, the work done by all forces is equal to the change in kinetic energy: $W = \Delta K = \frac{1}{2} m(v_f^2 - v_i^2)$.
Since the slope of the displacement-time $(s-t)$ graph represents velocity $(v = ds/dt)$, we analyze the slope in each region:
$1$. In region $OA$, the slope is constant, so velocity is constant. Thus, $\Delta K = 0$ and work done is zero.
$2$. In region $AB$, the slope is increasing, which means the velocity is increasing $(v_f > v_i)$. Therefore, the change in kinetic energy is positive, and the work done by all forces is positive.
$3$. In region $BC$, the slope is decreasing, which means the velocity is decreasing $(v_f < v_i)$. Therefore, the change in kinetic energy is negative, and the work done by all forces is negative.
Thus, the correct statement is that the work done by all forces in region $AB$ is positive.

(C) According to the work-energy theorem,the work done by friction is equal to the change in kinetic energy of the block.
Since the surface is rough,the loss in kinetic energy is dissipated as thermal energy $(H)$.
$H = |\Delta K| = |K_f - K_i| = |\frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2|$
Given: $m = 1\,kg$,$v_i = 10\,m/s$,$v_f = 8\,m/s$.
$H = \frac{1}{2} \times 1 \times (10^2 - 8^2)$
$H = \frac{1}{2} \times (100 - 64)$
$H = \frac{1}{2} \times 36 = 18\,J$.
Therefore,the thermal energy developed is $18\,J$.

(B) According to the work-energy theorem,the total work done on the body is equal to the change in its kinetic energy.
$W_{\text{gravity}} + W_{\text{air friction}} = \Delta K.E.$
$W_{\text{gravity}} = mgH$
Final kinetic energy $K.E. = \frac{1}{2} m v^2 = \frac{1}{2} m (1.1 \sqrt{gH})^2 = \frac{1}{2} m (1.21 gH) = 0.605 mgH$
Since the body starts from rest,initial kinetic energy is $0$.
So,$mgH + W_{\text{air friction}} = 0.605 mgH$
$W_{\text{air friction}} = 0.605 mgH - mgH = -0.395 mgH$
Therefore,the work done by air friction is $-0.395 mgH$.

(A) According to the law of conservation of mechanical energy, the total mechanical energy at the top of the tower must be equal to the total mechanical energy at the ground level.
Let $m$ be the mass of each particle, $h$ be the height of the tower, and $u$ be the initial speed.
At the top: $E_i = mgh + \frac{1}{2}mu^2$
At the ground: $E_f = 0 + \frac{1}{2}mV^2$
Since $E_i = E_f$, we have $mgh + \frac{1}{2}mu^2 = \frac{1}{2}mV^2$.
This simplifies to $V^2 = u^2 + 2gh$, or $V = \sqrt{u^2 + 2gh}$.
Since $u, g$, and $h$ are the same for all three particles, the final speed $V$ will be the same for all three particles regardless of the direction of projection.
Therefore, $V_A = V_B = V_C$.

(B) For the $2\, kg$ block to leave the ground,the spring force must equal the weight of the $2\, kg$ block: $kx = m_2g \Rightarrow 40x = 2 \times 10 \Rightarrow x = 0.5\, m$.
Initially,the spring is at its natural length (as the $5\, kg$ block is released from rest). When the $5\, kg$ block descends by $x = 0.5\, m$,the spring stretches by $x = 0.5\, m$.
Applying the Conservation of Mechanical Energy $(COME)$ for the system (including the $5\, kg$ block and the spring):
Loss in gravitational potential energy of $5\, kg$ block = Gain in kinetic energy of $5\, kg$ block + Gain in elastic potential energy of the spring.
$m_1gx = \frac{1}{2}m_1v^2 + \frac{1}{2}kx^2$
$5 \times 10 \times 0.5 = \frac{1}{2} \times 5 \times v^2 + \frac{1}{2} \times 40 \times (0.5)^2$
$25 = 2.5v^2 + 20 \times 0.25$
$25 = 2.5v^2 + 5$
$2.5v^2 = 20$
$v^2 = 8$
$v = \sqrt{8} = 2\sqrt{2}\, m/s$.

(B) Let the body strike the sand floor with velocity $v$. By the law of conservation of energy,the potential energy at height $h$ is converted into kinetic energy just before impact: $Mgh = \frac{1}{2} Mv^2$.
When the body enters the sand,it travels a distance $x$ before coming to rest. Let $F$ be the average resistive force of the sand.
Applying the work-energy theorem: The net work done on the body is equal to the change in its kinetic energy.
The forces acting on the body inside the sand are the weight $Mg$ (downwards) and the resistive force $F$ (upwards).
Work done by net force = $(F - Mg)x = \Delta K = 0 - \frac{1}{2} Mv^2$.
Since $\frac{1}{2} Mv^2 = Mgh$,we have $(F - Mg)x = Mgh$.
$Fx - Mgx = Mgh \implies Fx = Mgh + Mgx$.
$F = Mg \left( 1 + \frac{h}{x} \right)$.

(A) Let the ball be dropped from an initial height $h$.
Initial mechanical energy $E_i = mgh$.
After the bounce,the ball reaches a height $h' = 80\% \text{ of } h = 0.8h = \frac{4}{5}h$.
The mechanical energy after the bounce is $E_f = mgh' = mg(\frac{4}{5}h) = \frac{4}{5}mgh$.
The energy lost in each bounce is $\Delta E = E_i - E_f = mgh - \frac{4}{5}mgh = \frac{1}{5}mgh$.
The fraction of mechanical energy lost is $\frac{\Delta E}{E_i} = \frac{\frac{1}{5}mgh}{mgh} = \frac{1}{5} = 0.20$.

(A) Let the mass of the bullet be $m$ and the resistive force be $F$. Using the work-energy theorem for the first case:
$W = \Delta K$
$-F \times 1 = 0 - \frac{1}{2} m (100)^2$
$F = 5000m$
Now,for the second case where the thickness is $0.5\,m$,let the final velocity be $v'$.
$-F \times 0.5 = \frac{1}{2} m (v')^2 - \frac{1}{2} m (100)^2$
Substituting $F = 5000m$:
$-(5000m) \times 0.5 = \frac{1}{2} m (v')^2 - \frac{1}{2} m (10000)$
$-2500m = \frac{1}{2} m (v')^2 - 5000m$
$2500m = \frac{1}{2} m (v')^2$
$(v')^2 = 5000$
$v' = \sqrt{5000} = 50\sqrt{2}\,m/s$.

(A) Given the velocity $v = a\sqrt{s}$.
Since $v = \frac{ds}{dt}$,we have $\frac{ds}{dt} = a\sqrt{s}$.
Separating variables,we get $\int s^{-1/2} ds = \int a dt$.
Integrating both sides,$2\sqrt{s} = at + C$. Since the body starts from rest at $t=0$,$s=0$,so $C=0$.
Thus,$\sqrt{s} = \frac{at}{2}$,which implies $s = \frac{a^2 t^2}{4}$.
Differentiating $s$ with respect to $t$,we get velocity $v = \frac{ds}{dt} = \frac{a^2 t}{2}$.
Differentiating $v$ with respect to $t$,we get acceleration $a_{acc} = \frac{dv}{dt} = \frac{a^2}{2}$.
According to the work-energy theorem,the total work done is equal to the change in kinetic energy: $W = \Delta K = \frac{1}{2} m v^2$.
Substituting $v = \frac{a^2 t}{2}$,we get $W = \frac{1}{2} m \left( \frac{a^2 t}{2} \right)^2 = \frac{1}{2} m \left( \frac{a^4 t^2}{4} \right) = \frac{1}{8} m a^4 t^2$.

(C) From the given velocity-time graph,the initial velocity at $t = 0$ is $u = 50\, m/s$ and the final velocity at $t = 10\, s$ is $v = 0\, m/s$.
The acceleration $a$ is the slope of the graph:
$a = \frac{v_f - v_i}{t_f - t_i} = \frac{0 - 50}{10 - 0} = -5\, m/s^2$.
The velocity at $t = 2\, s$ is given by $v(t) = u + at$:
$v(2) = 50 + (-5)(2) = 50 - 10 = 40\, m/s$.
According to the work-energy theorem,the work done $W$ is equal to the change in kinetic energy $\Delta K.E.$:
$W = \Delta K.E. = \frac{1}{2} m (v_f^2 - v_i^2)$
$W = \frac{1}{2} \times 10\, kg \times ((40\, m/s)^2 - (50\, m/s)^2)$
$W = 5 \times (1600 - 2500)$
$W = 5 \times (-900) = -4500\, J$.

(D) The block reaches its maximum speed when its acceleration becomes zero.
At this point,the spring force balances the applied force $F$,so $kx = F$,which gives $x = \frac{F}{k}$.
Applying the work-energy theorem,the total work done on the block equals the change in its kinetic energy:
$W_{\text{spring}} + W_F = \Delta K.E.$
$-\frac{1}{2}kx^2 + Fx = \frac{1}{2}mv_{\max}^2$
Substituting $x = \frac{F}{k}$ into the equation:
$-\frac{1}{2}k\left(\frac{F}{k}\right)^2 + F\left(\frac{F}{k}\right) = \frac{1}{2}mv_{\max}^2$
$-\frac{F^2}{2k} + \frac{F^2}{k} = \frac{1}{2}mv_{\max}^2$
$\frac{F^2}{2k} = \frac{1}{2}mv_{\max}^2$
$v_{\max}^2 = \frac{F^2}{mk}$
$v_{\max} = \frac{F}{\sqrt{mk}}$

(D) Given: Mass $m = 2\,kg$, Position $x = 3t^2 + 5$.
First, find the velocity $v$ by differentiating position with respect to time: $v = \frac{dx}{dt} = \frac{d}{dt}(3t^2 + 5) = 6t\,m/s$.
At $t = 0\,s$, velocity $v_i = 6(0) = 0\,m/s$.
At $t = 5\,s$, velocity $v_f = 6(5) = 30\,m/s$.
According to the Work-Energy Theorem, the work done $W$ is equal to the change in kinetic energy $\Delta K$: $W = \Delta K = K_f - K_i$.
$W = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$.
$W = \frac{1}{2}(2)(30)^2 - \frac{1}{2}(2)(0)^2$.
$W = 900 - 0 = 900\,J$.

(D) According to the Work-Energy Theorem,the work done by the net force on a particle is equal to the change in its kinetic energy.
$W = \Delta K.E. = K.E._{final} - K.E._{initial}$
Given:
Force $\vec F = 3\hat i - 12\hat j$
Displacement $\vec d = 4\hat i$
Initial Kinetic Energy $K.E._{initial} = 3\, J$
Work done $W = \vec F \cdot \vec d = (3\hat i - 12\hat j) \cdot (4\hat i) = (3 \times 4) + (-12 \times 0) = 12\, J$
Using the theorem:
$12 = K.E._{final} - 3$
$K.E._{final} = 12 + 3 = 15\, J$

(D) Using the law of conservation of energy between point $A$ and point $B$:
$E_A = E_B$
$\frac{1}{2}mv_A^2 + mgh = \frac{1}{2}mv_B^2$
$v_B = \sqrt{v_A^2 + 2gh}$
Given $v_A = 5\,m/s$,$g = 10\,m/s^2$,and $h = 10\,m$:
$v_B = \sqrt{5^2 + 2 \times 10 \times 10} = \sqrt{25 + 200} = \sqrt{225} = 15\,m/s$
At point $B$,the particle is at a distance $r = a = 10\,m$ from point $O$. The velocity vector at $B$ is perpendicular to the radius $OB$.
The angular momentum $L$ about $O$ is given by:
$L = m \cdot v_B \cdot r$
$L = (20 \times 10^{-3}\,kg) \times (15\,m/s) \times (10\,m)$
$L = 0.02 \times 150 = 3\,kg \cdot m^2/s$.

(A) Given:
Mass of the bullet,$m = 20\,g = 20 \times 10^{-3}\,kg = 0.02\,kg$
Initial velocity,$u = 1\,ms^{-1}$
Thickness of the wall,$s = 20\,cm = 0.2\,m$
Mean resistance force,$F = 2.5 \times 10^{-2}\,N$
According to the Work-Energy Theorem,the work done by the resistive force is equal to the change in kinetic energy:
$W = \Delta K$
$-F \times s = \frac{1}{2}m(v^2 - u^2)$
Substituting the values:
$-(2.5 \times 10^{-2}) \times 0.2 = \frac{1}{2} \times (20 \times 10^{-3}) \times (v^2 - 1^2)$
$-0.5 \times 10^{-2} = 10 \times 10^{-3} \times (v^2 - 1)$
$-0.005 = 0.01 \times (v^2 - 1)$
$-0.5 = v^2 - 1$
$v^2 = 1 - 0.5 = 0.5$
$v = \sqrt{0.5} \approx 0.707\,ms^{-1}$
Thus,the speed is close to $0.7\,ms^{-1}$.

(A) According to the law of conservation of mechanical energy,the loss in potential energy is equal to the gain in kinetic energy at the lowest point.
Let $m$ be the mass of the child,$g$ be the acceleration due to gravity $(g = 10\,m/s^2)$,$h_{max} = 2\,m$,and $h_{min} = 0.75\,m$.
Loss in potential energy = $mg(h_{max} - h_{min})$
Gain in kinetic energy = $\frac{1}{2}mv^2$
Equating the two:
$mg(2 - 0.75) = \frac{1}{2}mv^2$
$g(1.25) = \frac{1}{2}v^2$
$v^2 = 2 \times 10 \times 1.25$
$v^2 = 25$
$v = 5\,m/s$.

(D) The work-energy theorem states that the net work done on an object by all forces (external,internal,conservative,and non-conservative) is equal to the change in its kinetic energy,i.e.,$W_{net} = \Delta K$.
This theorem is derived from Newton's second law of motion and is a fundamental principle in mechanics.
It does not depend on the nature of the forces involved,whether they are conservative or non-conservative.
Therefore,the work-energy theorem is valid in the presence of all types of forces.

(B) Let the chain have mass $m$ and length $2l$. Initially,it hangs symmetrically over the pin,so $l$ length is on each side. The center of mass of each half is at a distance $l/2$ below the pin.
When the chain is displaced,let it slip by a distance $x$. The new lengths on the two sides are $(l+x)$ and $(l-x)$.
Using the Conservation of Mechanical Energy $(COME)$: The decrease in Potential Energy $(P.E.)$ equals the increase in Kinetic Energy $(K.E.)$.
Initially,the center of mass of each half is at $h_i = -l/2$ relative to the pin. Total $P.E._i = 2 \times (m/2)g(-l/2) = -mgl/2$.
Finally,when the chain leaves the pin,one end is at $2l$ and the other is at $0$. The center of mass is at $h_f = -l$. Total $P.E._f = mg(-l) = -mgl$.
Change in $P.E. = P.E._i - P.E._f = (-mgl/2) - (-mgl) = mgl/2$.
Equating to $K.E. = \frac{1}{2}mv^2$,we get $\frac{1}{2}mv^2 = mgl/2$,which simplifies to $v = \sqrt{gl}$.

(C) According to the Work-Energy Theorem,the work done by an external force on a body is equal to the change in its kinetic energy $(W = \Delta K)$.
Additionally,work done by an external force can also change the potential energy of a system (e.g.,lifting an object against gravity).
Therefore,depending on the nature of the force and the system,an external force can increase the kinetic energy,the potential energy,or both simultaneously.
Thus,the correct statement is that both kinetic and potential energies may increase.

(A) According to the law of conservation of mechanical energy,the potential energy lost by the body is equal to the kinetic energy gained by it.
Let $m$ be the mass of the body,$g$ be the acceleration due to gravity,$h$ be the height,and $v$ be the final velocity.
Potential Energy lost = $mgh$
Kinetic Energy gained = $\frac{1}{2}mv^2$
Equating the two: $mgh = \frac{1}{2}mv^2$
$v^2 = 2gh$
Given $h = 1 \, m$ and $g = 9.8 \, m/s^2$:
$v = \sqrt{2 \times 9.8 \times 1} = \sqrt{19.6} \approx 4.43 \, m/s$.

(D) To reach the height of the suspension point,the sphere must gain a potential energy equal to the work done against gravity.
Let the mass of the sphere be $m$ and the initial velocity be $v$.
At the lowest point,the kinetic energy is $K.E. = \frac{1}{2}mv^2$.
At the height of the suspension point,the potential energy is $P.E. = mg\ell$.
By the law of conservation of mechanical energy:
$\frac{1}{2}mv^2 = mg\ell$
$v^2 = 2g\ell$
$v = \sqrt{2g\ell}$

(A) According to the Work-Energy Theorem,the work done by the resistive force is equal to the change in kinetic energy of the bullet.
$W = \Delta K$
$F \cdot s = \frac{1}{2} m u^2$
Given:
Mass $m = 25\,g = 25 \times 10^{-3}\,kg$
Velocity $u = 200\,m/s$
Distance $s = 5\,cm = 5 \times 10^{-2}\,m$
Substituting the values:
$F = \frac{m u^2}{2 s} = \frac{25 \times 10^{-3} \times (200)^2}{2 \times 5 \times 10^{-2}}$
$F = \frac{25 \times 10^{-3} \times 40000}{10 \times 10^{-2}}$
$F = \frac{1000}{0.1} = 10000\,N$
$F = 10\,kN$.

(A) According to the work-energy theorem,the work done on an object is equal to the change in its kinetic energy.
$W = \Delta K$
Given that the initial kinetic energy is zero,the work done is equal to the final kinetic energy:
$W = \frac{1}{2}mv^2$
Substituting the given values ($W = 25\,J$,$m = 2\,kg$):
$25 = \frac{1}{2} \times 2 \times v^2$
$25 = v^2$
$v = \sqrt{25} = 5\,m/s$
Therefore,the velocity acquired by the mass is $5\,m/s$.

(A) 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 $0$.
At point $B$,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 of $A$ below the point of suspension $O$ is $0$. Therefore,the vertical distance of $A$ above $B$ is $h = l - h' = l - l \cos 30^{\circ} = l(1 - \frac{\sqrt{3}}{2})$.
By the law of conservation of mechanical energy,the gain in kinetic energy at $B$ is equal to the loss in potential energy as the bob moves from $A$ to $B$.
Loss in potential energy $= mg h = mg l(1 - \cos 30^{\circ}) = mgl(1 - \frac{\sqrt{3}}{2})$.
Wait,looking at the provided image,$h$ is defined as the vertical distance from the horizontal level of $A$ to the level of $B$. The vertical distance of $B$ from the horizontal level of $A$ is $h = l \sin 30^{\circ} = l/2$.
Loss in potential energy $= mgh = mg(l \sin 30^{\circ}) = mg(l/2) = \frac{mgl}{2}$.
Therefore,the gain in kinetic energy at $B$ is $\frac{mgl}{2}$.

(C) Mechanical energy $E$ is defined as the sum of kinetic energy $K$ and potential energy $U$,so $E = K + U$. Thus,statement $(A)$ is correct.
Mechanical energy is conserved only when only conservative forces do work. If a non-conservative force (like friction) acts on the system,it typically changes the mechanical energy (often dissipating it as heat). Therefore,statement $(B)$ is incorrect.
The work done by a conservative force $W_c$ is equal to the negative change in potential energy,i.e.,$W_c = -\Delta U$. Since $\Delta E = \Delta K + \Delta U$,if only conservative forces act,$\Delta E = 0$. Statement $(C)$ claims $W_c = -\Delta E$,which is only true if $\Delta K = 0$. In general,$W_c = -\Delta U$,not $-\Delta E$. Thus,statement $(C)$ is also incorrect.
Since both $(B)$ and $(C)$ are incorrect,and the question asks for the incorrect statement,there is an ambiguity in the provided options. However,based on standard physics definitions,$(C)$ is fundamentally incorrect as a general principle.

(A) According to the work-energy theorem,the total work done on the bead is equal to the change in its kinetic energy.
$W_{F} + W_{mg} + W_{N} = \Delta K$
Here,$W_{F}$ is the work done by the constant horizontal force $F$,$W_{mg}$ is the work done by gravity,and $W_{N}$ is the work done by the normal force.
Since the ring is smooth,the normal force $N$ is always perpendicular to the displacement,so $W_{N} = 0$.
The horizontal force $F$ acts in the direction of the horizontal displacement,which is $R = 5 \, m$. Thus,$W_{F} = F \times R = 5 \, N \times 5 \, m = 25 \, J$.
The bead moves from $A$ to $B$,which is a vertical downward displacement of $R = 5 \, m$. Thus,$W_{mg} = mgR = (\frac{1}{2} \, kg) \times (10 \, m/s^2) \times (5 \, m) = 25 \, J$.
The change in kinetic energy is $\Delta K = \frac{1}{2}mv^2 - 0 = \frac{1}{2} \times (\frac{1}{2} \, kg) \times v^2 = \frac{1}{4}v^2$.
Equating the work done to the change in kinetic energy:
$25 + 25 = \frac{1}{4}v^2$
$50 = \frac{1}{4}v^2$
$v^2 = 200$
$v = \sqrt{200} = 10\sqrt{2} \approx 14.14 \, m/s$.

(B) Let the chain have mass $M$ and length $L$. Initially,a length $L/2$ hangs off the table. The mass of the hanging part is $M/2$. The center of mass of this hanging part is at a depth $L/4$ below the table surface. Taking the table surface as the reference level $(U=0)$,the initial potential energy is $U_i = -(M/2)g(L/4) = -MgL/8$.
When the chain completely leaves the table,its center of mass is at a depth $L/2$ below the table surface. The final potential energy is $U_f = -Mg(L/2) = -MgL/2$.
The initial kinetic energy is $K_i = 0$ (released from rest).
By the law of conservation of mechanical energy,$K_i + U_i = K_f + U_f$.
$0 - MgL/8 = (1/2)MV^2 - MgL/2$.
$(1/2)MV^2 = MgL/2 - MgL/8 = 3MgL/8$.
$V^2 = 3gL/4$.
$V = \sqrt{3gL/4}$.

(B) Let the particle be at height $h$ from the ground. The distance fallen by the particle is $(H - h)$.
The potential energy at height $h$ is $PE = mgh$.
The kinetic energy at height $h$ is $KE = \frac{1}{2}mv^2$. Since the particle falls from rest,$v^2 = 2g(H - h)$,so $KE = mg(H - h)$.
According to the problem,$KE = 2PE$.
Substituting the expressions: $mg(H - h) = 2(mgh)$.
Dividing by $mg$: $H - h = 2h$,which gives $3h = H$,or $h = \frac{H}{3}$.
Now,calculate the speed $v$ at this height:
$v = \sqrt{2g(H - h)} = \sqrt{2g(H - \frac{H}{3})} = \sqrt{2g(\frac{2H}{3})} = \sqrt{\frac{4gH}{3}} = 2\sqrt{\frac{gH}{3}}$.
Thus,the height is $\frac{H}{3}$ and the speed is $2\sqrt{\frac{gH}{3}}$.

(B) According to the law of conservation of mechanical energy,since the surface is frictionless,the total mechanical energy remains constant.
The total energy at position $A$ is equal to the total energy at position $C$.
$E_A = E_C$
At position $A$,the block is released from rest,so its kinetic energy $K_A = 0$. The potential energy is $U_A = mgh_A$,where $h_A = 14\, m$.
At position $C$,the kinetic energy is $K_C$ and the potential energy is $U_C = mgh_C$,where $h_C = 7\, m$.
$0 + mgh_A = K_C + mgh_C$
$K_C = mg(h_A - h_C)$
Given:
$m = 2\, kg$
$g = 10\, m/s^2$
$h_A = 14\, m$
$h_C = 7\, m$
$K_C = 2 \times 10 \times (14 - 7)$
$K_C = 20 \times 7$
$K_C = 140\, J$

(B) Let $x$ be the extension in the spring when the $2 \, kg$ block leaves contact with the ground. At this instant,the tension in the string equals the weight of the $2 \, kg$ block.
$kx = m_2 g$
$x = \frac{m_2 g}{k} = \frac{2 \times 10}{40} = 0.5 \, m$
Now,apply the principle of conservation of mechanical energy for the $5 \, kg$ block as it moves down by a distance $x = 0.5 \, m$:
Loss in gravitational potential energy = Gain in spring potential energy + Gain in kinetic energy
$m_1 g x = \frac{1}{2} k x^2 + \frac{1}{2} m_1 v^2$
$5 \times 10 \times 0.5 = \frac{1}{2} \times 40 \times (0.5)^2 + \frac{1}{2} \times 5 \times v^2$
$25 = 20 \times 0.25 + 2.5 v^2$
$25 = 5 + 2.5 v^2$
$20 = 2.5 v^2$
$v^2 = \frac{20}{2.5} = 8$
$v = \sqrt{8} = 2\sqrt{2} \, m/s$