Work Energy Theorem and Conservation of Mechanical Energy Questions in English

(D) Let the height from the surface be $x$. The distance fallen by the particle is $(S - x)$.
Using the equation of motion $v^2 = u^2 + 2as$,where $u = 0$ and $a = g$,we get $v^2 = 2g(S - x)$.
The potential energy at height $x$ is $PE = mgx$.
The kinetic energy at that instant is $KE = \frac{1}{2}mv^2 = \frac{1}{2}m(2g(S - x)) = mg(S - x)$.
According to the problem,$KE = 3 \times PE$.
Substituting the expressions: $mg(S - x) = 3(mgx)$.
Dividing both sides by $mg$: $S - x = 3x$.
$S = 4x \Rightarrow x = \frac{S}{4}$.
Now,substitute $x$ into the velocity equation: $v^2 = 2g(S - \frac{S}{4}) = 2g(\frac{3S}{4}) = \frac{3gS}{2}$.
Therefore,$v = \sqrt{\frac{3gS}{2}}$.
The height is $\frac{S}{4}$ and the speed is $\sqrt{\frac{3gS}{2}}$.

(B) According to the Work-Energy Theorem,the work done by the braking force is equal to the change in kinetic energy of the car.
$W = \Delta KE$
Since the cars come to a stop,the work done by the braking force is $W = -F \cdot S$,where $F$ is the braking force and $S$ is the stopping distance.
For car $A$: $F_{A} \cdot S_{A} = KE_{A} \implies F_{A} \cdot 1000 = 100 \implies F_{A} = 0.1 \ N$.
For car $B$: $F_{B} \cdot S_{B} = KE_{B} \implies F_{B} \cdot 1500 = 225 \implies F_{B} = 225 / 1500 = 0.15 \ N$.
Now,calculating the ratio $F_{A} / F_{B} = 0.1 / 0.15 = 10 / 15 = 2/3$.

(D) According to the Work-Energy Theorem,the total work done on the ball is equal to the change in its kinetic energy.
$W_{\text{ext}} + W_{\text{gravity}} = \Delta KE$
Here,the ball starts from rest and ends at rest at the maximum height,so $\Delta KE = 0$.
The force $F$ is applied over a distance $d_1 = 0.5 \ m$.
The total height reached by the ball from the point where the force application ends is $h = 5 \ m$.
The total displacement against gravity is $d_1 + h = 0.5 + 5 = 5.5 \ m$.
$W_{\text{ext}} = F \times 0.5$
$W_{\text{gravity}} = -mg(d_1 + h) = -(0.4 \times 10 \times 5.5) = -22 \ J$
Substituting these into the theorem:
$F \times 0.5 - 22 = 0$
$0.5F = 22$
$F = 44 \ N$.

(D) Let the total energy of the particle be $E = mgs$.
At the given instant,let the potential energy be $PE = x$.
According to the problem,the kinetic energy is $KE = 3x$.
By the law of conservation of energy,$KE + PE = E$.
$3x + x = mgs \Rightarrow 4x = mgs \Rightarrow x = \frac{mgs}{4}$.
Since $PE = mgh = x$,we have $mgh = \frac{mgs}{4}$,which gives the height $h = \frac{s}{4}$.
Now,for the speed $v$,we use $KE = \frac{1}{2}mv^2 = 3x$.
$\frac{1}{2}mv^2 = 3 \left( \frac{mgs}{4} \right) = \frac{3mgs}{4}$.
$v^2 = \frac{3gs}{2} \Rightarrow v = \sqrt{\frac{3gs}{2}}$.
Thus,the height is $\frac{s}{4}$ and the speed is $\sqrt{\frac{3gs}{2}}$.

(B) According to the work-energy theorem,the total work done on the block is equal to the change in its kinetic energy.
$W_{\text{gravity}} + W_{\text{friction}} = KE_f - KE_i$
Since the block starts from rest and comes to rest after moving a distance $x$ on the rough surface,$KE_i = 0$ and $KE_f = 0$.
The work done by gravity is $W_{\text{gravity}} = mgh = (mg)h = 20 \ N \times 1.0 \ m = 20 \ J$.
The work done by friction is $W_{\text{friction}} = -f_k \cdot x = -(\mu \cdot mg) \cdot x = -(0.40 \times 20 \ N) \cdot x = -8x \ J$.
Applying the work-energy theorem:
$20 - 8x = 0$
$8x = 20$
$x = \frac{20}{8} = 2.5 \ m$.
Thus,the distance the block will move on the rough surface is $2.5 \ m$.

(A) According to the law of conservation of mechanical energy,the total mechanical energy remains constant if only conservative forces act on the body.
Total energy at $x = 2 \ m$ is equal to the total energy at $x = 5 \ m$.
$(KE + PE)_{x=2} = (KE + PE)_{x=5}$
Given that the body is released from rest at $x = 2 \ m$,its kinetic energy $(KE)$ at $x = 2 \ m$ is $0$.
From the graph,the potential energy $(PE)$ at $x = 2 \ m$ is $6 \ J$ and at $x = 5 \ m$ is $2 \ J$.
Substituting these values:
$0 + 6 = \frac{1}{2} mv^2 + 2$
Given mass $m = 1 \ kg$:
$6 = \frac{1}{2} (1) v^2 + 2$
$4 = \frac{1}{2} v^2$
$v^2 = 8$
$v = \sqrt{8} = 2 \sqrt{2} \ m/s$.

(A) According to the work-energy theorem,the total work done by all forces (internal and external) acting on a system is equal to the change in its kinetic energy: $W_{\text{ext}} + W_{\text{int}} = \Delta K$.
If no external force is applied,$W_{\text{ext}} = 0$,so $W_{\text{int}} = \Delta K$.
Internal forces can perform work (e.g.,in an explosion or a spring-mass system),which leads to a change in the kinetic energy of the system.
Thus,the kinetic energy of a system can increase due to internal forces even in the absence of external forces.
Therefore,both Assertion $(A)$ and Reason $(R)$ are true,and $(R)$ is the correct explanation of $(A).$

(A) According to the Work-Energy Theorem,the total work done on the object is equal to the change in its kinetic energy.
$W_{mg} + W_{\text{resistive}} = \Delta KE$
Since the knife starts from rest and comes to rest after penetration,$\Delta KE = 0 - 0 = 0$.
The work done by gravity is $W_{mg} = mg(h + x)$,where $h = 2 \ m$ and $x = 10 \ cm = 0.1 \ m$.
The work done by the resistive force $F$ is $W_{\text{resistive}} = -F \cdot x$.
Thus,$mg(h + x) - Fx = 0$.
$F = \frac{mg(h + x)}{x}$.
Substituting the values: $m = 50 \ g = 0.05 \ kg$,$g = 10 \ m/s^2$,$h = 2 \ m$,$x = 0.1 \ m$.
$F = \frac{0.05 \times 10 \times (2 + 0.1)}{0.1} = \frac{0.5 \times 2.1}{0.1} = 5 \times 2.1 = 10.5 \ N$.

(C) Let the initial velocity be $u = V$ and the final velocity after penetrating $s_1 = 30 \text{ cm}$ be $v = V/2$. Assuming constant retardation $a$, we use the third equation of motion: $v^2 = u^2 + 2as_1$.
$(V/2)^2 = V^2 + 2a(30)$
$V^2/4 = V^2 + 60a$
$60a = -3V^2/4$
$a = -V^2/80$.
Now, for the further penetration until the bullet comes to rest, the initial velocity is $u' = V/2$ and the final velocity is $v' = 0$. Let the additional distance be $s_2$.
$(v')^2 = (u')^2 + 2as_2$
$0 = (V/2)^2 + 2(-V^2/80)s_2$
$V^2/4 = (V^2/40)s_2$
$s_2 = (V^2/4) \times (40/V^2) = 10 \text{ cm}$.

(C) Let the initial velocity be $V$ and the final velocity be $V/2$ after penetrating a distance $s_1 = 30 \ cm$. Assuming constant deceleration $a$,we use the third equation of motion: $v^2 = u^2 + 2as$.
$(V/2)^2 = V^2 + 2a(30)$
$V^2/4 = V^2 + 60a$
$60a = -3V^2/4$
$a = -V^2/80$.
Now,let the bullet penetrate an additional distance $s_2$ before coming to rest. The initial velocity for this phase is $V/2$ and the final velocity is $0$.
$0^2 = (V/2)^2 + 2(-V^2/80)s_2$
$0 = V^2/4 - (V^2/40)s_2$
$V^2/4 = (V^2/40)s_2$
$s_2 = 40/4 = 10 \ cm$.

(C) Let the length of the pendulum be $L = 1 \,m$ and the initial speed at the lowest point be $v_0 = 4 \,m/s$.
At the lowest point, the potential energy is taken as $0$. The total energy is $E = \frac{1}{2}mv_0^2$.
When the string makes an angle $\theta = 60^{\circ}$ with the vertical, the height of the bob is $h = L(1 - \cos \theta)$.
Substituting the values, $h = 1(1 - \cos 60^{\circ}) = 1(1 - 0.5) = 0.5 \,m$.
By the law of conservation of energy, the total energy at the lowest point equals the total energy at the angle $\theta$:
$\frac{1}{2}mv_0^2 = \frac{1}{2}mv^2 + mgh$
$v^2 = v_0^2 - 2gh$
$v^2 = (4)^2 - 2(10)(0.5)$
$v^2 = 16 - 10 = 6$
$v = \sqrt{6} \,m/s$.

(A) When the bob of a pendulum is raised to a height '$h$',the potential energy at the extreme position is converted into kinetic energy at the mean position.
By the law of conservation of energy:
$mgh = \frac{1}{2}mv^2$
$v = \sqrt{2gh}$
From the geometry of the pendulum,the vertical distance from the point of suspension to the bob at the extreme position is '$l \cos \theta$'.
Therefore,the vertical height '$h$' raised by the bob is:
$h = l - l \cos \theta = l(1 - \cos \theta)$
Substituting the value of '$h$' in the velocity equation:
$v = \sqrt{2gl(1 - \cos \theta)}$

(D) According to the Work-Energy Theorem,the net work done on a body is equal to the change in its kinetic energy.
$W_{\text{net}} = \Delta KE = KE_f - KE_i$
Given mass $m = 0.25 \ kg$,speed $v = k x^{3/2}$,and $k = 2$.
At $x = 0$,$v_i = 2(0)^{3/2} = 0 \ m/s$,so $KE_i = 0 \ J$.
At $x = 2 \ m$,$v_f = 2(2)^{3/2} = 2 \times 2 \sqrt{2} = 4\sqrt{2} \ m/s$.
$KE_f = \frac{1}{2} m v_f^2 = \frac{1}{2} \times 0.25 \times (4\sqrt{2})^2$
$KE_f = \frac{1}{2} \times 0.25 \times 16 \times 2 = 0.25 \times 16 = 4 \ J$.
Therefore,$W_{\text{net}} = 4 \ J - 0 \ J = 4 \ J$.

(A) According to the work-energy theorem,the work done is equal to the change in kinetic energy.
Work done to accelerate from $0$ to $v$ is $W_1 = \frac{1}{2}mv^2 - 0 = \frac{1}{2}mv^2$.
Work done to accelerate from $v$ to $2v$ is $W_2 = \Delta K = \frac{1}{2}m(2v)^2 - \frac{1}{2}mv^2$.
$W_2 = \frac{1}{2}m(4v^2) - \frac{1}{2}mv^2 = 2mv^2 - 0.5mv^2 = 1.5mv^2$.
Comparing $W_2$ with $W_1$:
$W_2 = 3 \times (\frac{1}{2}mv^2) = 3W_1$.
Therefore,the work done is three times the work done in accelerating it from rest to $v$.

(A) According to the work-energy theorem,the work done by the retarding force is equal to the change in kinetic energy of the body.
For the first case,the body comes to rest from velocity $u$ after covering distance $s_{1}$:
$F s_{1} = \frac{1}{2} m u^{2} \quad (i)$
For the second case,the body comes to rest from velocity $2u$ after covering distance $s_{2}$:
$F s_{2} = \frac{1}{2} m (2u)^{2} = \frac{1}{2} m (4u^{2}) = 4 \left( \frac{1}{2} m u^{2} \right) \quad (ii)$
Dividing equation $(ii)$ by equation $(i)$:
$\frac{F s_{2}}{F s_{1}} = \frac{4 (\frac{1}{2} m u^{2})}{\frac{1}{2} m u^{2}}$
$\frac{s_{2}}{s_{1}} = 4$
$s_{2} = 4s_{1}$

(A) Given: Mass $m = 5 \,kg$,Initial Kinetic Energy $KE_i = 490 \,J$,Acceleration due to gravity $g = 9.8 \,ms^{-2}$.
At the required height $h$,the kinetic energy $KE_f$ becomes half of the initial value:
$KE_f = \frac{1}{2} KE_i = \frac{490}{2} = 245 \,J$.
According to the law of conservation of energy,the total mechanical energy remains constant:
$KE_i = KE_f + PE_f$
$KE_i = KE_f + mgh$
Substituting the values:
$490 = 245 + 5 \times 9.8 \times h$
$490 - 245 = 49 \times h$
$245 = 49h$
$h = \frac{245}{49} = 5 \,m$.
Therefore,the height is $5 \,m$.

(A) According to the work-energy theorem,the work done by the retarding force is equal to the change in kinetic energy of the car.
Work done $W = F \cdot s = \Delta K = \frac{1}{2}mv^2$.
Since the initial kinetic energy and final kinetic energy (which is $0$) remain the same,the work done to stop the car must be constant.
$F \cdot s = F' \cdot s'$
Given $F' = 3F$,we have:
$F \cdot s = (3F) \cdot s'$
$s' = s/3$.
Therefore,the car will be stopped in a distance of $s/3$.

(A) The stopping distance $S$ is given by the formula $S = \frac{u^2}{2a}$,where $u$ is the initial velocity and $a$ is the deceleration (which is constant for a constant braking force).
From this relation,we can see that $S \propto u^2$.
Given: $u_1 = 80 \ km/h$,$S_1 = 60 \ m$,and $u_2 = 160 \ km/h$.
Using the ratio: $\frac{S_2}{S_1} = \left(\frac{u_2}{u_1}\right)^2$.
Substituting the values: $\frac{S_2}{60} = \left(\frac{160}{80}\right)^2 = (2)^2 = 4$.
Therefore,$S_2 = 4 \times 60 \ m = 240 \ m$.

(C) From the work-energy theorem,the work done by the braking force is equal to the change in kinetic energy of the train.
$W = F \cdot d = \Delta K = \frac{1}{2}mv^2$
Since the initial velocity $v$ and mass $m$ are constant,the work done to stop the train must be constant.
$F_1 d_1 = F_2 d_2$
Given $F_1 = F$ and $d_1 = x$.
Given $F_2 = \frac{F}{4}$.
Substituting these values: $F \cdot x = (\frac{F}{4}) \cdot d_2$
$d_2 = 4x$.
Therefore,the train will travel four times the original distance.

(B) Given: Length of pendulum,$l = 0.5 \ m$. Speed at lowest point,$v_1 = 6 \ ms^{-1}$. Angle made by string with vertical,$\theta = 60^{\circ}$.
In $\triangle OBC$,$\cos \theta = \frac{OC}{OB} = \frac{l-h}{l} = \cos 60^{\circ}$.
$\Rightarrow 1 - \frac{h}{l} = 0.5 \Rightarrow \frac{h}{l} = 0.5 \Rightarrow h = 0.5 \times 0.5 = 0.25 \ m$.
Using the law of conservation of mechanical energy:
$K_i + U_i = K_f + U_f$
$\frac{1}{2}mv_1^2 + 0 = \frac{1}{2}mv_2^2 + mgh$
$v_2^2 = v_1^2 - 2gh$
$v_2^2 = (6)^2 - 2 \times 10 \times 0.25 = 36 - 5 = 31$
$v_2 = \sqrt{31} \ ms^{-1}$.
Hence,the final velocity at an angle $60^{\circ}$ is $\sqrt{31} \ ms^{-1}$.

(A) The total mechanical energy of the body is conserved.
Initial kinetic energy at the ground $(h=0)$ is $K_i = \frac{1}{2} m v^2 = \frac{1}{2} m (20)^2 = 200m \,J$.
Initial potential energy at the ground is $U_i = 0 \,J$.
Total energy $E = K_i + U_i = 200m \,J$.
At a height of $5 \,m$, potential energy $U_5 = mgh = m(10)(5) = 50m \,J$.
Given $U_5 = 100 \,J$, so $50m = 100$, which gives $m = 2 \,kg$.
Total energy $E = 200(2) = 400 \,J$.
At a height of $10 \,m$, potential energy $U_{10} = mgh = 2(10)(10) = 200 \,J$.
By conservation of energy, $E = K_{10} + U_{10}$.
$400 = K_{10} + 200$.
$K_{10} = 200 \,J$.

(B) The mass of the ball is $m = 300 g = 0.3 kg$. The total height fallen by the ball is $H = 10 m + 1.5 m = 11.5 m$.
According to the work-energy theorem,the total work done by all forces is equal to the change in kinetic energy.
Since the ball starts from rest and comes to rest,the change in kinetic energy is $\Delta K = 0$.
The forces acting on the ball are gravity (downward) and the resistance force of the sand (upward).
Work done by gravity $W_g = mgH = 0.3 \times 10 \times 11.5 = 34.5 J$.
Work done by sand resistance $W_R = -F_R \times d$,where $d = 1.5 m$.
Applying the theorem: $W_g + W_R = 0
\Rightarrow 34.5 - F_R \times 1.5 = 0
\Rightarrow F_R = \frac{34.5}{1.5} = 23 N$.

(B) According to the work-energy theorem, the work done by the net force on a body is equal to the change in its kinetic energy.
$W = \Delta K = K_f - K_i$
Given:
Force $\overrightarrow{F} = (4 \hat{i} - 15 \hat{j}) \text{ N}$
Displacement $\overrightarrow{S} = 6 \hat{i} \text{ m}$
Initial kinetic energy $K_i = 7 \text{ J}$
Work done $W = \overrightarrow{F} \cdot \overrightarrow{S} = (4 \hat{i} - 15 \hat{j}) \cdot (6 \hat{i}) = (4 \times 6) + (-15 \times 0) = 24 \text{ J}$
Using the work-energy theorem:
$K_f - K_i = W$
$K_f - 7 = 24$
$K_f = 24 + 7 = 31 \text{ J}$
Therefore, the kinetic energy at the end of the displacement is $31 \text{ J}$.

(A) According to the law of conservation of mechanical energy,the total mechanical energy of the system remains constant in the absence of non-conservative forces like friction.
At point $A$,the particle is at rest,so its kinetic energy is $0$ and its potential energy is $mgy$ (taking the reference level at $BC$).
At point $C$,the particle is at height $0$,so its potential energy is $0$ and its kinetic energy is $KE_C$.
Applying the conservation of energy principle:
$PE_A + KE_A = PE_C + KE_C$
$mgy + 0 = 0 + KE_C$
Therefore,the kinetic energy at point $C$ is $KE_C = mgy$.

(A) The total mechanical energy $E$ of a system is the sum of its potential energy $U$ and kinetic energy $KE$. That is,$E = U + KE$.
Given,$E = 2 \ J$ and $m = 1 \ kg$.
To find the maximum speed,we need the maximum kinetic energy,which occurs when the potential energy is at its minimum.
$U(x) = \frac{x^2}{2} - x$.
To find the minimum potential energy,we set $\frac{dU}{dx} = 0$:
$\frac{d}{dx}(\frac{x^2}{2} - x) = x - 1 = 0 \implies x = 1 \ m$.
$U_{\min} = U(1) = \frac{1^2}{2} - 1 = -0.5 \ J$.
Since $E = U + KE$,we have $KE_{\max} = E - U_{\min} = 2 - (-0.5) = 2.5 \ J$.
Using $KE = \frac{1}{2}mv^2$:
$2.5 = \frac{1}{2}(1)v^2 \implies v^2 = 5 \implies v = \sqrt{5} \ ms^{-1}$.

(B) Let the total height be $H = 30 \ m$ and the velocity of the body be $v$ at a point where potential energy $(PE)$ is twice the kinetic energy $(KE)$.
According to the law of conservation of energy,the total energy at any point is equal to the initial potential energy:
$m g H = PE + KE$
Given that $PE = 2 \times KE$,we substitute this into the equation:
$m g H = 2 \times KE + KE = 3 \times KE$
Since $KE = \frac{1}{2} m v^2$,we have:
$m g H = 3 \times (\frac{1}{2} m v^2)$
$g H = \frac{3}{2} v^2$
$v^2 = \frac{2}{3} g H$
Substituting the values $g = 10 \ m \ s^{-2}$ and $H = 30 \ m$:
$v^2 = \frac{2}{3} \times 10 \times 30$
$v^2 = 2 \times 10 \times 10 = 200$
$v = \sqrt{200} = 10 \sqrt{2} \ m \ s^{-1}$

(A) Let the mass per unit length be $\lambda = \frac{m}{l}$.
Initial state: The length hanging is $h_1 = \frac{l}{n}$. The mass of this part is $m_1 = \lambda \cdot \frac{l}{n} = \frac{m}{n}$. The center of mass of the hanging part is at a distance $\frac{l}{2n}$ below the table surface. Taking the table surface as the reference level $(U=0)$,the initial potential energy is $U_i = -m_1 g \left(\frac{l}{2n}\right) = -\left(\frac{m}{n}\right) g \left(\frac{l}{2n}\right) = -\frac{mgl}{2n^2}$.
Final state: When the chain completely slips off,its center of mass is at a distance $\frac{l}{2}$ below the table surface. The final potential energy is $U_f = -mg \left(\frac{l}{2}\right) = -\frac{mgl}{2}$.
By the law of conservation of energy,the loss in potential energy equals the gain in kinetic energy:
$K_f - K_i = U_i - U_f$
$\frac{1}{2}mv^2 - 0 = -\frac{mgl}{2n^2} - \left(-\frac{mgl}{2}\right)$
$\frac{1}{2}mv^2 = \frac{mgl}{2} \left(1 - \frac{1}{n^2}\right)$
$v^2 = gl \left(1 - \frac{1}{n^2}\right)$
$v = \sqrt{gl \left(1 - \frac{1}{n^2}\right)}$
Wait,checking the calculation: $\frac{1}{2}mv^2 = \frac{mgl}{2}(1 - 1/n^2) \implies v = \sqrt{gl(1 - 1/n^2)}$. Re-evaluating the options,the correct expression is $\sqrt{gl(1 - 1/n^2)}$. However,if the question implies the standard result,let's re-check the factor of $2$. The kinetic energy is $\frac{1}{2}mv^2$. The potential energy change is $\frac{mgl}{2}(1 - 1/n^2)$. Thus $v = \sqrt{gl(1 - 1/n^2)}$. Option $A$ matches this.

(D) The total mechanical energy of the particle is conserved,so at any height $h$ from the ground,the sum of potential energy $(PE)$ and kinetic energy $(KE)$ is equal to the initial potential energy at height $H$:
$PE + KE = m g H$ ... $(i)$
Given that at a certain height,the kinetic energy is half of its potential energy:
$KE = \frac{1}{2} PE \implies PE = 2 KE$
Substituting this into equation $(i)$:
$2 KE + KE = m g H$
$3 KE = m g H$
$KE = \frac{m g H}{3}$
Since $PE = m g h$,we have $PE = 2 KE = 2 \left( \frac{m g H}{3} \right) = \frac{2}{3} m g H$.
Equating $m g h = \frac{2}{3} m g H$,we get the height $h = \frac{2 H}{3}$.
The speed $v$ of the particle at this height can be found using the work-energy theorem or kinematics. Since the particle falls from rest from height $H$,its speed at height $h$ is given by $v = \sqrt{2 g (H - h)}$.
Substituting $h = \frac{2 H}{3}$:
$v = \sqrt{2 g (H - \frac{2 H}{3})} = \sqrt{2 g (\frac{H}{3})} = \sqrt{\frac{2 g H}{3}}$.
Thus,the height is $\frac{2 H}{3}$ and the speed is $\sqrt{\frac{2 g H}{3}}$.

(A) The force acting on the body is given by $\vec{F} = -\nabla U = -(\frac{\partial U}{\partial x} \hat{i} + \frac{\partial U}{\partial y} \hat{j}) = -(6 \hat{i} + 8 \hat{j}) \,N$.
The acceleration of the body is $\vec{a} = \frac{\vec{F}}{m} = \frac{-(6 \hat{i} + 8 \hat{j})}{2} = -(3 \hat{i} + 4 \hat{j}) \,m/s^2$.
Since the body starts from rest at $t = 0$, its velocity at $t = 2 \,s$ is $\vec{v} = \vec{u} + \vec{a}t = 0 + (-(3 \hat{i} + 4 \hat{j})) \times 2 = -(6 \hat{i} + 8 \hat{j}) \,m/s$.
The kinetic energy of the body at $t = 2 \,s$ is $K = \frac{1}{2} m v^2 = \frac{1}{2} \times 2 \times ((-6)^2 + (-8)^2) = 36 + 64 = 100 \,J$.
According to the work-energy theorem, the work done by the external force is equal to the change in kinetic energy. Since the body starts from rest $(K_i = 0)$, the work done by the force is $W = \Delta K = 100 - 0 = 100 \,J$.

(D) Given, displacement of the block, $x = 5t^2$.
Velocity $v = \frac{dx}{dt} = \frac{d}{dt}(5t^2) = 10t \,m/s$.
At $t = 0 \,s$, $v_0 = 10(0) = 0 \,m/s$.
At $t = 3 \,s$, $v_1 = 10(3) = 30 \,m/s$.
At $t = 5 \,s$, $v_2 = 10(5) = 50 \,m/s$.
By the Work-Energy Theorem, the work done $W$ is equal to the change in kinetic energy: $W = \Delta K = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$.
Work done in first $3 \,s$: $W_1 = \frac{1}{2}m(v_1^2 - v_0^2) = \frac{1}{2}m(30^2 - 0^2) = \frac{1}{2}m(900)$.
Work done in first $5 \,s$: $W_2 = \frac{1}{2}m(v_2^2 - v_0^2) = \frac{1}{2}m(50^2 - 0^2) = \frac{1}{2}m(2500)$.
Ratio $\frac{W_1}{W_2} = \frac{900}{2500} = \frac{9}{25}$.

(A) Given: Mass of the bullet $= m$,initial velocity $= v_1$,final velocity $= v_2$,and distance traveled $= L$.
Since the bullet experiences a resistive force $F$,it undergoes uniform deceleration $a = F/m$.
Using the third equation of motion,$v^2 - u^2 = 2as$,where $v = v_2$,$u = v_1$,$a = -F/m$,and $s = L$:
$v_2^2 - v_1^2 = 2(-F/m)L$
$v_2^2 - v_1^2 = -\frac{2FL}{m}$
$F = \frac{m(v_1^2 - v_2^2)}{2L}$
Since $v_1 > v_2$,the force $F$ is positive.

(A) According to the principle of conservation of mechanical energy,the sum of kinetic energy and potential energy at the starting point '$X$' must be equal to the sum of kinetic energy and potential energy at the peak '$Z$'.
$mgh + \frac{1}{2}mv^2 = mgh' + \frac{1}{2}mv'^2$
Dividing by mass $m$ on both sides:
$gh + \frac{1}{2}v^2 = gh' + \frac{1}{2}v'^2$
Rearranging for $v'^2$:
$v'^2 = 2g(h - h') + v^2$
Given: $h = 100 \ m$,$h' = 60 \ m$,$v = 10 \ m \ s^{-1}$,and $g = 10 \ m \ s^{-2}$.
$v'^2 = 2 \times 10 \times (100 - 60) + (10)^2$
$v'^2 = 20 \times 40 + 100$
$v'^2 = 800 + 100 = 900$
$v' = \sqrt{900} = 30 \ m \ s^{-1}$

(A) According to the work-energy theorem,the work done by the retarding force is equal to the change in kinetic energy of the bus.
$W = \Delta K$
$F \cdot s = \frac{1}{2} m v^2$
Since the retarding force $F$ and the initial velocity $v$ are constant,we have:
$s = \frac{m v^2}{2 F} \implies s \propto m$
Let the initial mass be $m_1 = m$ and the initial stopping distance be $s_1 = x$.
The new mass after increasing by $25\%$ is $m_2 = m + 0.25m = 1.25m$.
Using the proportionality $s \propto m$:
$\frac{s_2}{s_1} = \frac{m_2}{m_1}$
$s_2 = s_1 \cdot \frac{1.25m}{m}$
$s_2 = 1.25 x$.

(D) Given initial speed $u = 40 \,ms^{-1}$ and angle $\theta = 30^{\circ}$.
The final speed $v = 1.25 \times u = 1.25 \times 40 = 50 \,ms^{-1}$.
Using the principle of conservation of energy, the total mechanical energy at the point of projection and at the point of striking the ground remains constant.
Taking the ground as the reference level for potential energy $(PE = 0)$:
$E_{initial} = E_{final}$
$\frac{1}{2} m u^2 + mgh = \frac{1}{2} m v^2$
Dividing by $m$ and multiplying by $2$:
$u^2 + 2gh = v^2$
Substitute the given values:
$(40)^2 + 2 \times 10 \times h = (50)^2$
$1600 + 20h = 2500$
$20h = 2500 - 1600$
$20h = 900$
$h = \frac{900}{20} = 45 \,m$.
Thus, the value of $h$ is $45 \,m$.

(B) Let the mass of the body be $m$ and the initial velocity be $u$. At maximum height $h$,the total energy is $E = mgh$.
At a height $y = 0.4h$ from the ground,the potential energy is $PE = mgy = mg(0.4h) = 0.4mgh$.
According to the law of conservation of energy,the total energy remains constant.
Therefore,$KE + PE = E$.
$KE = E - PE = mgh - 0.4mgh = 0.6mgh$.
The ratio of kinetic energy to potential energy is $\frac{KE}{PE} = \frac{0.6mgh}{0.4mgh} = \frac{0.6}{0.4} = \frac{3}{2}$.
Thus,the ratio is $3: 2$.

(C) According to the Work-Energy Theorem,the work done by the net force is equal to the change in kinetic energy:
$W = \Delta K = K_f - K_i$
Given $m = 1 \ kg$,$c = 1$,$v = x^\alpha$,$x_i = 0 \ m$,$x_f = 4 \ m$,and $W = 128 \ J$.
At $x = 0$,$v_i = 1 \times (0)^\alpha = 0$.
At $x = 4$,$v_f = 1 \times (4)^\alpha = 4^\alpha$.
Substituting these into the Work-Energy equation:
$128 = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$
$128 = \frac{1}{2} (1) (4^\alpha)^2 - 0$
$128 = \frac{1}{2} (4^{2\alpha})$
$256 = 4^{2\alpha}$
Since $256 = 2^8 = (2^2)^4 = 4^4$,we have:
$4^4 = 4^{2\alpha}$
$2\alpha = 4$
$\alpha = 2$.

(C) Let the compression of the spring be $x$ (in meters).
According to the law of conservation of mechanical energy,the total potential energy lost by the mass equals the potential energy gained by the spring.
The total height fallen by the mass is $(h + x)$,where $h = 1.2 \ m$.
So,$mg(h + x) = \frac{1}{2} kx^2$.
Substituting the given values: $2 \times 10 \times (1.2 + x) = \frac{1}{2} \times (2 \times 10^4) \times x^2$.
$20(1.2 + x) = 10^4 x^2$.
$24 + 20x = 10000x^2$.
$10000x^2 - 20x - 24 = 0$.
Dividing by $4$: $2500x^2 - 5x - 6 = 0$.
Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$x = \frac{5 \pm \sqrt{(-5)^2 - 4(2500)(-6)}}{2(2500)} = \frac{5 \pm \sqrt{25 + 60000}}{5000} = \frac{5 \pm \sqrt{60025}}{5000} = \frac{5 \pm 245}{5000}$.
Since $x > 0$,$x = \frac{250}{5000} = 0.05 \ m$.
Converting to $mm$: $0.05 \ m = 50 \ mm$.

(A) Given,tangential acceleration $a_t = \alpha x^2$.
Tangential force $F = m a_t = m \alpha x^2$.
According to the work-energy theorem,the work done by the tangential force is equal to the change in kinetic energy: $W = \Delta K$.
Assuming the particle starts from rest at $x = 0$,$K = \int_0^x F dx$.
$K = \int_0^x (m \alpha x^2) dx$.
$K = m \alpha \left[ \frac{x^3}{3} \right]_0^x = \frac{m \alpha}{3} x^3$.
Comparing this with the given expression $K = \beta x^c$,we get $\beta = \frac{m \alpha}{3}$ and $c = 3$.

(B) According to the work-energy theorem,the total work done by all forces is equal to the change in kinetic energy of the particle.
Work done $W = \Delta KE = KE_f - KE_i$.
The velocity of the particle is given by $v = k x^\beta$.
At $x = 0$,the initial velocity $v_i = k(0)^\beta = 0$.
At $x = d$,the final velocity $v_f = k d^\beta$.
The initial kinetic energy $KE_i = \frac{1}{2} m v_i^2 = \frac{1}{2} m (0)^2 = 0$.
The final kinetic energy $KE_f = \frac{1}{2} m v_f^2 = \frac{1}{2} m (k d^\beta)^2 = \frac{1}{2} m k^2 d^{2\beta}$.
Therefore,the total work done $W = \frac{1}{2} m k^2 d^{2\beta} - 0 = \frac{m k^2}{2} d^{2\beta}$.

(C) Let the initial height be $H = 12 \,m$. The potential energy at this height is $PE_1 = mgH$.
When the ball strikes the ground, its kinetic energy just before impact is $KE_1 = mgH$.
The ball loses $25 \%$ of its kinetic energy, so the remaining kinetic energy is $KE_2 = KE_1 - 0.25 KE_1 = 0.75 KE_1$.
The ball bounces back to a height '$h$', so its potential energy at the maximum height is $PE_2 = mgh$.
By the law of conservation of energy, the kinetic energy after the loss is equal to the potential energy at the new height: $mgh = 0.75 mgH$.
Therefore, $h = 0.75 H$.
Substituting $H = 12 \,m$: $h = 0.75 \times 12 = 9 \,m$.

(D) Given: Mass $m = 3 \text{ kg}$, Displacement $x = \frac{t^3}{3} \text{ m}$.
First, find the velocity $v$ by differentiating displacement with respect to time:
$v = \frac{dx}{dt} = \frac{d}{dt}\left(\frac{t^3}{3}\right) = t^2 \text{ m/s}$.
At $t = 0$, $v_i = 0^2 = 0 \text{ m/s}$.
At $t = 2 \text{ s}$, $v_f = 2^2 = 4 \text{ m/s}$.
According to the Work-Energy Theorem, the work done $W$ is equal to the change in kinetic energy:
$W = \Delta KE = KE_f - KE_i = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2$.
Substituting the values:
$W = \frac{1}{2} \times 3 \times (4)^2 - \frac{1}{2} \times 3 \times (0)^2 = \frac{1}{2} \times 3 \times 16 = 24 \text{ J}$.

(C) According to the Work-Energy Theorem,the work done by the net (resultant) force acting on a body is equal to the change in its kinetic energy.
Mathematically,$W_{net} = \Delta KE = KE_f - KE_i$.
Therefore,the correct option is $C$.

(A) Given: $m = 2 \,kg$, $x = \frac{\alpha t^2}{2}$, and $\alpha = 1 \,m/s^2$.
First, we find the velocity $v$ by differentiating the position $x$ with respect to time $t$:
$v = \frac{dx}{dt} = \frac{d}{dt} \left( \frac{\alpha t^2}{2} \right) = \alpha t$.
At $t = 2 \,s$, the velocity $v$ is:
$v = 1 \times 2 = 2 \,m/s$.
According to the work-energy theorem, the work done $W$ is equal to the change in kinetic energy:
$W = \Delta K = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$.
At $t = 0$, $v_i = \alpha(0) = 0$.
At $t = 2 \,s$, $v_f = 2 \,m/s$.
Therefore, $W = \frac{1}{2} \times 2 \,kg \times (2 \,m/s)^2 - 0 = 4 \,J$.

(A) Since the surface of the bowl is smooth,there is no friction to cause the sphere to roll. The sphere will simply slide down the surface.
By the law of conservation of mechanical energy,the loss in potential energy is equal to the gain in kinetic energy.
The change in height of the center of mass of the sphere is $h = R - r$.
Potential energy lost = $mgh = mg(R - r)$.
Since the sphere is sliding (not rolling),its total kinetic energy is purely translational: $K = \frac{1}{2} mv^2$.
Therefore,the total kinetic energy at the lowest point is $K = mg(R - r)$.

(C) From the given velocity-time graph,the initial velocity at $t = 0 \ s$ is $u = 50 \ m/s$. The velocity becomes $0 \ m/s$ at $t = 10 \ s$.
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 $v = u + at = 50 + (-5)(2) = 40 \ m/s$.
According to the work-energy theorem,the work done $W$ is equal to the change in kinetic energy:
$W = \Delta K = K_f - K_i = \frac{1}{2} m v^2 - \frac{1}{2} m u^2$
$W = \frac{1}{2} \times 10 \ kg \times ((40 \ m/s)^2 - (50 \ m/s)^2)$
$W = 5 \times (1600 - 2500) = 5 \times (-900) = -4500 \ J$.

(D) Given: Mass $m = 6 \,kg$,displacement $x = \frac{t^2}{4} \,m$.
Velocity $v = \frac{dx}{dt} = \frac{d}{dt}(\frac{t^2}{4}) = \frac{2t}{4} = \frac{t}{2} \,m/s$.
At $t = 0 \,s$,velocity $v_i = \frac{0}{2} = 0 \,m/s$.
At $t = 2 \,s$,velocity $v_f = \frac{2}{2} = 1 \,m/s$.
According to the Work-Energy Theorem,the work done $W$ is equal to the change in kinetic energy $\Delta K$.
$W = K_f - K_i = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$.
$W = \frac{1}{2} \times 6 \times (1)^2 - \frac{1}{2} \times 6 \times (0)^2$.
$W = 3 - 0 = 3 \,J$.

(B) Let the radius of the semi-circle be $R$. The horizontal distance covered by bead $Q$ to reach point $B$ is $2R$. Since it moves with a constant horizontal velocity $v$,the time taken is $t_Q = \frac{2R}{v}$.
For bead $P$,the horizontal component of velocity is given as $v_x = v$ at point $S$. As the bead slides down the frictionless semi-circular wire,its speed increases due to gravity. The horizontal component of its velocity $v_x$ will be $v \cos \theta$,where $\theta$ is the angle the velocity vector makes with the horizontal. Since the bead is constrained to move along the semi-circle,its horizontal velocity component $v_x$ will always be greater than or equal to its initial horizontal velocity $v$ as it moves towards $C$ and then towards $B$. Because the average horizontal velocity of $P$ is greater than $v$,and the total horizontal distance $P$ needs to cover is $2R$,the time taken $t_P$ must be less than $t_Q$.