Work Energy Theorem and Conservation of Mechanical Energy Questions in English

(D) Using the work-energy theorem,the work done by the constant force $F$ to stop the car is equal to the change in kinetic energy: $W = \Delta K$.
Since the final velocity is $0$,the work done is $F \cdot S = \frac{1}{2} m u^2$.
Therefore,$S = \frac{m u^2}{2F}$.
Since $m$ and $F$ are constant,we have $S \propto u^2$.
Given $u_1 = 10 \, m/s$ and $S_1 = 20 \, m$.
For $u_2 = 30 \, m/s$,the ratio is $\frac{S_2}{S_1} = (\frac{u_2}{u_1})^2$.
$\frac{S_2}{20} = (\frac{30}{10})^2 = 3^2 = 9$.
$S_2 = 9 \times 20 = 180 \, m$.

(B) By applying the law of conservation of energy,the potential energy at the top is converted into kinetic energy at the bottom.
Let the potential energy at the bottom be zero. The potential energy at the top is $U = mgR$.
The kinetic energy at the bottom is $K = \frac{1}{2}mv^2$.
According to the law of conservation of energy:
$mgR = \frac{1}{2}mv^2$
$gR = \frac{1}{2}v^2$
$v^2 = 2gR$
$v = \sqrt{2gR}$

(C) According to the law of conservation of energy,the potential energy at the top is converted into kinetic energy at the bottom.
$mgh = \frac{1}{2}mv^2$
Here,$m$ is the mass of the object,$g$ is the acceleration due to gravity,$h$ is the height,and $v$ is the final velocity.
Canceling $m$ from both sides,we get $gh = \frac{1}{2}v^2$,which simplifies to $v = \sqrt{2gh}$.
Since all three objects fall from the same height $h$ and are subject to the same acceleration due to gravity $g$,their final speeds will be independent of their masses.
Therefore,the ratio of their speeds is $v_1 : v_2 : v_3 = \sqrt{2gh} : \sqrt{2gh} : \sqrt{2gh} = 1 : 1 : 1$.

(A) According to the law of conservation of mechanical energy,the potential energy at the top is converted into kinetic energy at the bottom.
Since both objects start from the same height $h$,their initial potential energy is $mgh$.
At the bottom,this energy is converted into kinetic energy: $mgh = \frac{1}{2}mv^2$.
Solving for speed $v$,we get $v = \sqrt{2gh}$.
Since $v$ depends only on $g$ and $h$,both objects will reach the ground with the same speed.
Therefore,the correct option is $A$.

(C) The potential energy lost by the bob as it moves from the $60^\circ$ position to the lowest point is converted into kinetic energy.
Using the principle of conservation of energy: $mgh = \frac{1}{2}mv^2$,where $h$ is the vertical height dropped.
The vertical height $h$ is given by $h = l(1 - \cos \theta)$,where $l = 2\, m$ and $\theta = 60^\circ$.
$h = 2(1 - \cos 60^\circ) = 2(1 - 0.5) = 1\, m$.
Equating energy: $v = \sqrt{2gh} = \sqrt{2 \times 9.8 \times 1} = \sqrt{19.6} \approx 4.43\, m/s$.

(B) According to the law of conservation of mechanical energy,the potential energy at the maximum height is equal to the kinetic energy at the lowest point $A$.
Let $m$ be the mass of the ball,$g$ be the acceleration due to gravity,and $h$ be the height reached.
$PE_{top} = KE_{bottom}$
$mgh = \frac{1}{2}mv^2$
$v = \sqrt{2gh}$
Given $h = 20 \,cm = 0.2 \,m$ and $g = 10 \,m/s^2$.
$v = \sqrt{2 \times 10 \times 0.2} = \sqrt{4} = 2 \,m/s$.

(D) Given: Mass of the bullet $m = 20\, g = 0.02\, kg$.
Initial velocity $u = 250\, m/s$.
Final velocity $v = 0\, m/s$ (as it stops).
Distance penetrated $s = 12\, cm = 0.12\, m$.
Using the work-energy theorem or the third equation of motion $v^2 = u^2 + 2as$:
$0^2 = (250)^2 + 2 \times a \times 0.12$
$a = -\frac{62500}{0.24} = -260416.67\, m/s^2$.
The magnitude of the average force is $F = m|a|$.
$F = 0.02 \times 260416.67 = 5208.33\, N$.
Rounding to the nearest given option,$F \approx 5.2 \times 10^3\, N$.

(D) Given: Mass of the bullet $m = 30 \, g = 30 \times 10^{-3} \, kg$,Initial velocity $u = 120 \, m/s$,Final velocity $v = 0 \, m/s$ (as it stops),Distance $S = 12 \, cm = 0.12 \, m$.
Using the work-energy theorem or the equation of motion $v^2 - u^2 = 2aS$:
$0^2 - (120)^2 = 2 \times a \times 0.12$
$-14400 = 0.24 \times a$
$a = -\frac{14400}{0.24} = -60000 \, m/s^2$.
The resistive force $F = m \times |a| = (30 \times 10^{-3} \, kg) \times (60000 \, m/s^2) = 1800 \, N$.

(A) According to the Work-Energy Theorem,the work done $W$ is equal to the change in kinetic energy: $W = \Delta K = \frac{1}{2} m(v_f^2 - v_i^2)$.
Given $x = 3t - 4t^2 + t^3$,the velocity $v$ is the derivative of position with respect to time: $v = \frac{dx}{dt} = 3 - 8t + 3t^2$.
At $t = 0 \,s$,$v_i = 3 - 8(0) + 3(0)^2 = 3 \,m/s$.
At $t = 4 \,s$,$v_f = 3 - 8(4) + 3(4)^2 = 3 - 32 + 48 = 19 \,m/s$.
The mass $m = 3 \,g = 0.003 \,kg$.
Substituting the values: $W = \frac{1}{2} \times 0.003 \times (19^2 - 3^2) = 0.0015 \times (361 - 9) = 0.0015 \times 352 = 0.528 \,J$.
Since $1 \,J = 1000 \,mJ$,the work done is $0.528 \times 1000 = 528 \,mJ$.

(B) According to the Work-Energy Theorem,the work done by the net force on an object is equal to the change in its kinetic energy.
Work done $W = F_x \cdot s$,where $F_x$ is the horizontal component of the force $(F \cos \theta)$ and $s$ is the displacement.
Given: $s = 0.4\,m$,$W = 1\,J$.
$W = F_x \cdot s$
$1\,J = F_x \cdot 0.4\,m$
$F_x = \frac{1}{0.4} = 2.5\,N$.
Therefore,the horizontal component of the force is $2.5\,N$.

(C) The work done by a force $F$ on a particle is given by $W = \int \vec{F} \cdot d\vec{r} = \int \vec{F} \cdot \vec{v} dt$.
Since the force is always perpendicular to the velocity,$\vec{F} \cdot \vec{v} = 0$.
Therefore,the work done $W = 0$.
According to the work-energy theorem,the change in kinetic energy $\Delta K = W$.
Since $W = 0$,the change in kinetic energy $\Delta K = 0$,which implies that the kinetic energy of the particle remains constant.

(B) The initial kinetic energy $(K_i)$ is $490\, J$.
At the height $h$ where the kinetic energy becomes half of its original value,the kinetic energy $(K_f)$ is $\frac{490}{2} = 245\, J$.
According to the law of conservation of energy,the loss in kinetic energy is equal to the gain in potential energy $(U)$.
Loss in $K.E. = K_i - K_f = 490 - 245 = 245\, J$.
Gain in potential energy $U = mgh$.
Equating the two: $mgh = 245$.
Substituting the given values: $2 \times 9.8 \times h = 245$.
$19.6 \times h = 245$.
$h = \frac{245}{19.6} = 12.5\, m$.

(C) According to the law of conservation of energy,when a body is projected vertically upwards with an initial velocity $u$,it will return to the point of projection with the same speed $v = u$,assuming air resistance is negligible.
Given mass $m = 2 \, kg$ and initial velocity $u = 2 \, m/s$.
The velocity just before striking the ground will be $v = 2 \, m/s$.
The kinetic energy is given by the formula $K.E. = \frac{1}{2} m v^2$.
Substituting the values: $K.E. = \frac{1}{2} \times 2 \times (2)^2 = 1 \times 4 = 4 \, J$.

(B) Let the thickness of each plank be $s$. If the initial speed of the bullet is $u_1 = 100 \ m/s$,it stops after covering a total distance of $2s$. Using the equation of motion $v^2 = u^2 - 2as$,where $v=0$ is the final velocity and $a$ is the constant retardation:
$0 = u_1^2 - 2a(2s) \implies 2a = \frac{u_1^2}{2s}$.
Since the retardation $a$ is constant,the total distance $S$ covered by the bullet is proportional to the square of the initial speed $(S \propto u^2)$.
If the speed is doubled,$u_2 = 2u_1$. The new distance covered $S_2$ will be:
$S_2 = \left(\frac{u_2}{u_1}\right)^2 S_1 = (2)^2 (2s) = 4 \times 2s = 8s$.
Since each plank has thickness $s$,the number of planks penetrated is $8s / s = 8$.

(C) Since the surface is smooth,mechanical energy is conserved throughout the motion.
Let the initial height be $h_1 = 100 \, m$ and the final height be $h_2 = 20 \, m$.
The initial velocity $u = 0 \, m/s$.
According to the law of conservation of energy:
$PE_{initial} + KE_{initial} = PE_{final} + KE_{final}$
$mgh_1 + 0 = mgh_2 + \frac{1}{2}mv^2$
$mg(h_1 - h_2) = \frac{1}{2}mv^2$
$v = \sqrt{2g(h_1 - h_2)}$
Substituting the values $(g = 10 \, m/s^2)$:
$v = \sqrt{2 \times 10 \times (100 - 20)}$
$v = \sqrt{20 \times 80}$
$v = \sqrt{1600}$
$v = 40 \, m/s$.

(C) According to the work-energy theorem,the work done by the retarding force $(F)$ is equal to the change in kinetic energy of the vehicle.
$W = \Delta K.E.$
Since the vehicles are brought to rest,the work done $W = F \times s$,where $s$ is the stopping distance.
Therefore,$F \times s = K.E.$
$s = \frac{K.E.}{F}$
Since both the lorry and the car have the same initial kinetic energy $(K.E.)$ and are subjected to the same retarding force $(F)$,the stopping distance $(s)$ must be the same for both vehicles.

(A) 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, $W = \Delta K = \frac{1}{2}m(v_f^2 - v_i^2)$.
For the work done to be positive, the final kinetic energy must be greater than the initial kinetic energy, which implies that the final velocity $v_f$ must be greater than the initial velocity $v_i$.
Looking at the velocity-time graph:
$1$. In the region $AB$, the velocity increases with time, so $v_f > v_i$. Thus, the work done is positive.
$2$. In the region $BC$, the velocity is constant, so $v_f = v_i$. Thus, the work done is zero.
$3$. In the region $CD$, the velocity decreases, so $v_f < v_i$. Thus, the work done is negative.
$4$. In the region $DE$, the velocity is constant, so $v_f = v_i$. Thus, the work done is zero.
Therefore, the work done by the force on the particle is positive from $A$ to $B$.

(D) According to the work-energy theorem,the work done $(W)$ on a particle is equal to the change in its kinetic energy.
Since the particle is initially at rest,its initial kinetic energy is $0$.
Therefore,$W = K_f - K_i = \frac{1}{2}mv^2 - 0 = \frac{1}{2}mv^2$.
Here,$m$ is the mass of the particle,which is constant.
Thus,$W \propto v^2$.
This relationship represents a parabola opening along the $W$-axis.
Therefore,the graph between $W$ and $v$ will be parabolic in nature,as shown in option $(D)$.

(C) 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.
$W = \Delta K$
Since the body starts from rest,the initial kinetic energy is $0$. Thus,$K = W$.
For a body moving with a constant acceleration $a$ under a constant force $F$,the work done $W$ over a distance $x$ is given by:
$W = F \cdot x$
Since $F = m \cdot a$ is constant,we have:
$K = (m \cdot a) \cdot x$
This implies that $K \propto x$.
Therefore,the graph of kinetic energy $K$ versus distance $x$ is a straight line passing through the origin.

(A) According to the work-energy theorem,the work done by friction is equal to the change in kinetic energy of the block.
This work done by friction is dissipated as thermal energy.
Given:
Mass $m = 100\, g = 0.1\, kg$
Initial velocity $v_1 = 10\, m/s$
Final velocity $v_2 = 5\, m/s$
Thermal energy $\Delta Q = \Delta K = \frac{1}{2}m(v_1^2 - v_2^2)$
$\Delta Q = \frac{1}{2} \times 0.1 \times (10^2 - 5^2)$
$\Delta Q = 0.05 \times (100 - 25)$
$\Delta Q = 0.05 \times 75 = 3.75\, J$.

(A) The mechanical energy $(ME)$ of a body is the sum of its potential energy $(PE)$ and kinetic energy $(KE)$.
According to the law of conservation of mechanical energy,in the absence of non-conservative forces like air resistance,the total mechanical energy of a system remains conserved.
As the body falls,its potential energy decreases while its kinetic energy increases by the same amount.
Therefore,the total mechanical energy remains constant at every instant during the fall.

(D) The law of conservation of energy states that energy can neither be created nor destroyed,but it can be transformed from one form to another.
This implies that for an isolated system,the total energy remains constant.
Therefore,the sum of all forms of energy (kinetic,potential,thermal,etc.) in an isolated system is conserved.
Option $D$ correctly describes this fundamental principle.

(A) According to the work-energy theorem,the work done by the retarding force is equal to the change in kinetic energy.
$W = F \times d = \Delta K$
Since both vehicles have the same initial kinetic energy $(K)$ and come to rest,the change in kinetic energy is the same for both $(\Delta K = K)$.
Given that the retarding force $(F)$ applied is also the same for both vehicles,we have:
$d = \frac{K}{F}$
Since both $K$ and $F$ are identical for the car and the truck,the distance $(d)$ traveled by both before coming to rest will be the same.

(D) The speed of the stone at the moment it hits the ground can be determined using the principle of conservation of mechanical energy.
According to the law of conservation of energy,the total mechanical energy at the top of the cliff must equal the total mechanical energy at the ground level.
Let $m$ be the mass of the stone,$g$ be the acceleration due to gravity,$h$ be the height of the cliff,and $v$ be the initial speed.
Initial energy at the top: $E_i = \frac{1}{2}mv^2 + mgh$.
Final energy at the ground: $E_f = \frac{1}{2}mv_f^2$,where $v_f$ is the final speed.
Since $E_i = E_f$,we have $\frac{1}{2}mv_f^2 = \frac{1}{2}mv^2 + mgh$.
Simplifying this,$v_f = \sqrt{v^2 + 2gh}$.
As seen from the final expression,the final speed $v_f$ depends only on the initial speed $v$,the height $h$,and the acceleration due to gravity $g$.
It is independent of the angle or direction of projection. Therefore,the speed at which the stone hits the ground is the same regardless of the direction in which it is thrown.

(B) According to the law of conservation of mechanical energy,the potential energy at the top is converted into kinetic energy at the bottom.
Initial potential energy = $mgr$.
Final kinetic energy = $\frac{1}{2}mv^2$.
Equating the two: $mgr = \frac{1}{2}mv^2$.
Solving for $v$: $v^2 = 2rg$.
Therefore,$v = \sqrt {2rg}$.

(B) The law of conservation of energy states that energy can neither be created nor destroyed; it can only be transformed from one form to another.
For example,when an object falls from a certain height,its potential energy is converted into kinetic energy,and at every instant,the total mechanical energy of the system remains constant.

(A) The work-energy theorem states that the work done by the net force acting on a particle is equal to the change in its kinetic energy $(W_{net} = \Delta K)$.
This theorem is derived from Newton's second law of motion and is applicable to all types of forces,whether they are internal,external,conservative,or non-conservative.
Therefore,the work-energy theorem is valid in the presence of all types of forces.

(C) Let the velocity imparted to the lower mass be $v_0$. Since the rod is rigid,the angular velocity $\omega$ of the rod is given by $\omega = \frac{v_0}{L}$.
The velocity of the mass at the midpoint is $v_{mid} = \omega \cdot \frac{L}{2} = \frac{v_0}{2}$.
By the principle of conservation of mechanical energy,the loss in kinetic energy equals the gain in potential energy as the rod moves from the vertical to the horizontal position.
Initial kinetic energy $K_i = \frac{1}{2}mv_0^2 + \frac{1}{2}m\left(\frac{v_0}{2}\right)^2 = \frac{1}{2}mv_0^2 + \frac{1}{8}mv_0^2 = \frac{5}{8}mv_0^2$.
Final kinetic energy $K_f = 0$ (at the horizontal position).
Gain in potential energy $U_f - U_i = mg\left(\frac{L}{2}\right) + mgL = \frac{3}{2}mgL$.
Equating the loss in kinetic energy to the gain in potential energy:
$\frac{5}{8}mv_0^2 = \frac{3}{2}mgL$.
$v_0^2 = \frac{3}{2} \cdot \frac{8}{5} gL = \frac{12}{5}gL$.
$v_0 = \sqrt{\frac{12Lg}{5}}$.

(D) According to the work-energy theorem,the work done by all forces is equal to the change in kinetic energy of the body.
Here,the work done against air resistance $(W_{air})$ is equal to the loss in kinetic energy.
Mass of the bullet,$m = 10 \; g = 0.01 \; kg$.
Initial velocity,$u = 1000 \; m/s$.
Final velocity,$v = 500 \; m/s$.
Work done against air resistance $W = \Delta K = K_i - K_f = \frac{1}{2} m (u^2 - v^2)$.
$W = \frac{1}{2} \times 0.01 \times [(1000)^2 - (500)^2]$.
$W = 0.005 \times [1,000,000 - 250,000]$.
$W = 0.005 \times 750,000$.
$W = 3750 \; J$.

(B) Let the chain be initially hanging symmetrically on both sides of the pin,with length $l$ on each side. The center of mass of each half is at a distance $l/2$ below the pin.
The initial potential energy of the system is $U_i = 2 \times [(\frac{m}{2})g(-\frac{l}{2})] = -\frac{mgl}{2}$.
When the chain just leaves the pin,the entire length $2l$ is hanging vertically. The center of mass of the chain is at a distance $l$ below the pin.
The final potential energy of the system is $U_f = mg(-l) = -mgl$.
By the law of conservation of energy,the loss in potential energy equals the gain in kinetic energy:
$|U_f - U_i| = K_f - K_i$
$|-\frac{mgl}{2} - (-mgl)| = \frac{1}{2}mv^2 - 0$
$\frac{mgl}{2} = \frac{1}{2}mv^2$
$v^2 = gl$
$v = \sqrt{gl}$

(D) According to the Work-Energy Theorem,the work done by the resistive force is equal to the change in kinetic energy of the bullet.
$W = \Delta K.E. = \frac{1}{2}m(v^2 - u^2)$
Here,$m = 10 \ g = 0.01 \ kg$,$u = 800 \ m/s$,$v = 100 \ m/s$,and displacement $s = 1 \ m$.
$F \cdot s = \frac{1}{2} \cdot 0.01 \cdot (100^2 - 800^2)$
$F \cdot 1 = 0.005 \cdot (10000 - 640000)$
$F = 0.005 \cdot (-630000)$
$F = -3150 \ N$
The negative sign indicates that the force is resistive. Therefore,the magnitude of the average resistive force is $3150 \ N$.

(B) According to the work-energy theorem,the total work done by all forces (gravity and spring force) on the ball is equal to the change in its kinetic energy.
The ball starts from rest at height $h$ above the spring and comes to rest momentarily after compressing the spring by distance $d$. Thus,the initial and final kinetic energy of the ball is $0$.
Change in kinetic energy $\Delta K = K_f - K_i = 0 - 0 = 0$.
Work done by gravity $W_g = mg(h + d)$.
Work done by the spring force $W_s = -\frac{1}{2}kd^2$.
According to the work-energy theorem,$W_{total} = W_g + W_s = \Delta K$.
$mg(h + d) - \frac{1}{2}kd^2 = 0$.

(D) Given the relation: $t = \sqrt{x} + 3 \implies \sqrt{x} = t - 3 \implies x = (t - 3)^2$.
Velocity $v = \frac{dx}{dt} = \frac{d}{dt}(t - 3)^2 = 2(t - 3)$.
At $t = 0 \text{ s}$, $v_i = 2(0 - 3) = -6 \text{ m/s}$.
At $t = 6 \text{ s}$, $v_f = 2(6 - 3) = 6 \text{ m/s}$.
According to the Work-Energy Theorem, the work done $W$ is equal to the change in kinetic energy $\Delta K$.
$W = \Delta K = \frac{1}{2}m(v_f^2 - v_i^2)$.
Substituting the values: $W = \frac{1}{2}m(6^2 - (-6)^2) = \frac{1}{2}m(36 - 36) = 0 \text{ J}$.

(B) According to the law of conservation of energy,the potential energy at point $A$ is converted into kinetic energy at point $B$.
Given:
Mass $m = 50 \ gm = 50 \times 10^{-3} \ kg = 0.05 \ kg$
Length $l = h = 1.5 \ m$
Acceleration due to gravity $g = 10 \ m/s^2$
Kinetic energy at $B$ = Potential energy at $A$
$K.E. = mgh$
$K.E. = 0.05 \times 10 \times 1.5$
$K.E. = 0.5 \times 1.5 = 0.75 \ J$
$K.E. = 7.5 \times 10^{-1} \ J$

(C) Let $F$ be the average resistive force exerted by the wood. Using the work-energy theorem,the total work done on the knife by all forces equals the change in its kinetic energy.
The forces acting on the knife during its penetration into the wood are gravity ($mg$ downwards) and the resistive force ($F$ upwards).
The total distance moved by the knife from the point of release to the final stop is $(h + d)$.
Work done by gravity = $mg(h + d)$
Work done by resistive force = $-Fd$
According to the work-energy theorem:
$W_{\text{total}} = \Delta K$
$mg(h + d) - Fd = 0 - 0$
$mg(h + d) = Fd$
$F = \frac{mg(h + d)}{d}$
$F = mg(1 + \frac{h}{d})$

(D) Given that the body starts from rest (implied by uniform acceleration starting from $t=0$),the acceleration $a$ is given by $a = \frac{v - 0}{t_1} = \frac{v}{t_1}$.
At any time $t$,the velocity $v_t$ of the body is $v_t = u + at = 0 + \left( \frac{v}{t_1} \right) t = \frac{v}{t_1} t$.
According to the Work-Energy Theorem,the work done $W$ is equal to the change in kinetic energy:
$W = \Delta K = \frac{1}{2} m v_t^2 - 0$.
Substituting the value of $v_t$:
$W = \frac{1}{2} m \left( \frac{v}{t_1} t \right)^2 = \frac{1}{2} m \frac{v^2}{t_1^2} t^2$.

(A) Given: Mass $m = 1 \ kg$, displacement $x = \frac{t^3}{3}$.
Velocity $v = \frac{dx}{dt} = \frac{d}{dt}(\frac{t^3}{3}) = t^2$.
At $t = 0 \ s$, initial velocity $u = (0)^2 = 0 \ m/s$.
At $t = 1 \ s$, final velocity $v = (1)^2 = 1 \ m/s$.
According to the Work-Energy Theorem, the work done $W$ is equal to the change in kinetic energy:
$W = \Delta K = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$.
$W = \frac{1}{2} \times 1 \times (1)^2 - \frac{1}{2} \times 1 \times (0)^2$.
$W = 0.5 - 0 = 0.5 \ J$.

(B) According to the Work-Energy Theorem,the total work done by all forces is equal to the change in kinetic energy: $W = \Delta KE = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$.
Given the velocity function $v = a\sqrt{x}$.
At $x = 0$,the initial velocity $v_i = a\sqrt{0} = 0$.
At $x = d$,the final velocity $v_f = a\sqrt{d}$.
Substituting these values into the work-energy equation:
$W = \frac{1}{2} m (a\sqrt{d})^2 - \frac{1}{2} m (0)^2$.
$W = \frac{1}{2} m a^2 d - 0$.
$W = \frac{1}{2} m a^2 d$.

(D) According to the work-energy theorem,the work done (energy required) is equal to the change in kinetic energy.
For the first case,the energy required to accelerate from $v_1 = 10 \ m/s$ to $v_2 = 20 \ m/s$ is:
$E_1 = \frac{1}{2} m (v_2^2 - v_1^2) = \frac{1}{2} m (20^2 - 10^2) = \frac{1}{2} m (400 - 100) = \frac{1}{2} m (300)$.
For the second case,the energy required to accelerate from rest $(v_0 = 0 \ m/s)$ to $v_1 = 10 \ m/s$ is:
$E_2 = \frac{1}{2} m (v_1^2 - v_0^2) = \frac{1}{2} m (10^2 - 0^2) = \frac{1}{2} m (100)$.
Taking the ratio:
$\frac{E_1}{E_2} = \frac{\frac{1}{2} m (300)}{\frac{1}{2} m (100)} = 3$.
Therefore,the energy required is $3$ times.

(D) The initial kinetic energy of the object is $K_i = \frac{1}{2} m v^2 = \frac{1}{2} m (4)^2 = 8m \ J$.
Let the height be $h$ where the kinetic energy becomes half of its initial value,i.e.,$K_f = \frac{1}{2} K_i = 4m \ J$.
According to the law of conservation of mechanical energy,the loss in kinetic energy is equal to the gain in potential energy: $\Delta K = \Delta U$.
$K_i - K_f = mgh$.
$8m - 4m = mgh$.
$4m = mgh$.
Using $g = 10 \ m/s^2$,we get $4 = 10h$.
Therefore,$h = 0.4 \ m$.

(B) Given: Mass $m = 8 \ kg$,Position $x = \frac{1}{2} t^2$.
Velocity $v = \frac{dx}{dt} = \frac{d}{dt}(\frac{1}{2} t^2) = t \ m/s$.
At $t = 0 \ s$,velocity $v_i = 0 \ m/s$.
At $t = 2 \ s$,velocity $v_f = 2 \ m/s$.
According to the Work-Energy Theorem,the work done $W$ is equal to the change in kinetic energy $\Delta K$.
$W = \Delta K = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2$.
$W = \frac{1}{2} \times 8 \times (2)^2 - \frac{1}{2} \times 8 \times (0)^2$.
$W = 4 \times 4 - 0 = 16 \ J$.

(B) The force of friction is given by $f = \mu mg = 0.2 \times 5 \times 10 = 10 \ N$.
The net force acting on the body is $F_{net} = F - f = 25 \ N - 10 \ N = 15 \ N$.
According to the work-energy theorem,the kinetic energy gained by the body is equal to the work done by the net force.
Kinetic Energy = $F_{net} \times d = 15 \ N \times 10 \ m = 150 \ J$.

(A) According to the work-energy theorem,the total work done by all forces is equal to the change in kinetic energy: $W = \Delta KE = \frac{1}{2} m v^2$.
Given $v = k\sqrt{s}$,we have $v^2 = k^2 s$.
Differentiating with respect to time $t$: $2v \frac{dv}{dt} = k^2 \frac{ds}{dt}$.
Since $\frac{ds}{dt} = v$,we get $2v \frac{dv}{dt} = k^2 v$,which simplifies to $\frac{dv}{dt} = \frac{k^2}{2}$.
Integrating with respect to time $t$ (starting from rest at $t=0$): $v = \int_0^t \frac{k^2}{2} dt = \frac{k^2}{2} t$.
Now,substituting $v$ into the work equation: $W = \frac{1}{2} m \left( \frac{k^2}{2} t \right)^2 = \frac{1}{2} m \left( \frac{k^4}{4} t^2 \right) = \frac{1}{8} m k^4 t^2$.

(A) According to the law of conservation of mechanical energy,the total mechanical energy remains constant in the absence of non-conservative forces like friction.
Initial mechanical energy at height $h_1 = 100 \ m$ is $E_i = mgh_1 + 0 = 20 \times 10 \times 100 = 20,000 \ J$.
Final mechanical energy at height $h_f = 20 \ m$ is $E_f = mgh_f + \frac{1}{2}mv^2 = 20 \times 10 \times 20 + \frac{1}{2} \times 20 \times v^2 = 4,000 + 10v^2$.
Equating initial and final energy: $20,000 = 4,000 + 10v^2$.
$10v^2 = 16,000$.
$v^2 = 1,600$.
$v = 40 \ m/s$.

(B) According to the work-energy theorem,the work done by the net force on an object is equal to the change in its kinetic energy.
Let the initial velocity be $v_0$. The initial kinetic energy is $K_i = \frac{1}{2}mv_0^2$.
When the velocity becomes half $(v = v_0/2)$,the kinetic energy is $K_f = \frac{1}{2}m(v_0/2)^2 = \frac{1}{8}mv_0^2$.
The work done by the resistive force $F$ over a distance $d = 6 \ cm$ is:
$W = K_f - K_i = \frac{1}{8}mv_0^2 - \frac{1}{2}mv_0^2 = -\frac{3}{8}mv_0^2$.
Since $W = F \cdot d$,we have $F \cdot 6 = -\frac{3}{8}mv_0^2$,which gives $F = -\frac{1}{16}mv_0^2$.
Now,let the bullet travel an additional distance $d_1$ before coming to rest. The initial kinetic energy for this stage is $\frac{1}{8}mv_0^2$ and the final kinetic energy is $0$.
Applying the work-energy theorem again:
$F \cdot d_1 = 0 - \frac{1}{8}mv_0^2$.
Substituting $F = -\frac{1}{16}mv_0^2$:
$(-\frac{1}{16}mv_0^2) \cdot d_1 = -\frac{1}{8}mv_0^2$.
Solving for $d_1$,we get $d_1 = \frac{1/8}{1/16} = 2 \ cm$.

(D) According to the work-energy theorem,the work done on a particle is equal to the change in its kinetic energy.
$W = \Delta K = \frac{1}{2} m (v_2^2 - v_1^2)$
Initial velocity vector: $\vec{v}_1 = 4 \hat{i} + 16 \hat{k} \; m s^{-1}$.
Initial speed squared: $v_1^2 = |\vec{v}_1|^2 = 4^2 + 16^2 = 16 + 256 = 272 \; m^2 s^{-2}$.
Final velocity vector: $\vec{v}_2 = 8 \hat{i} + 20 \hat{j} \; m s^{-1}$.
Final speed squared: $v_2^2 = |\vec{v}_2|^2 = 8^2 + 20^2 = 64 + 400 = 464 \; m^2 s^{-2}$.
Given mass $m = 0.01 \; kg$.
Substituting the values into the work-energy equation:
$W = \frac{1}{2} \times 0.01 \times (464 - 272)$
$W = 0.005 \times 192$
$W = 0.96 \; J$.

(C) Initial kinetic energy $K_i = 490 \, J$.
At the required height $h$,the kinetic energy $K_f$ becomes half of the initial value,so $K_f = \frac{490}{2} = 245 \, J$.
According to the law of conservation of energy,the loss in kinetic energy is equal to the gain in potential energy:
$K_i - K_f = mgh$.
$490 - 245 = 2 \times 9.8 \times h$.
$245 = 19.6 \times h$.
$h = \frac{245}{19.6} = 12.5 \, m$.

(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.
$W = \Delta K$
$F \cdot s = \frac{1}{2}mv^2$
Since $F$ and $v$ are constant,we have $s \propto m$.
Let the initial mass be $m_1 = m$ and the final mass be $m_2 = m + 0.5m = 1.5m$.
Let the initial distance be $s_1 = s$ and the final distance be $s_2$.
Using the proportionality $s_2 / s_1 = m_2 / m_1$,we get:
$s_2 / s = 1.5m / m$
$s_2 = 1.5s$.
Therefore,the car will come to rest at a distance of $1.5s$.

(C) According to the law of conservation of mechanical energy,the decrease in potential energy is equal to the increase in kinetic energy.
Let $m$ be the mass of the object.
The decrease in potential energy is given as $U$.
The kinetic energy gained by the object is $K = \frac{1}{2}mv^2$.
Equating the two: $U = \frac{1}{2}mv^2$.
Solving for mass $m$: $m = \frac{2U}{v^2}$.

(A) 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 $B$.
$PE_A + KE_A = PE_B + KE_B$
Since the bob starts from rest at point $A$,$KE_A = 0$.
$mgh_1 = mgh_2 + \frac{1}{2}mv^2$
$gh_1 = gh_2 + \frac{1}{2}v^2$
$v^2 = 2g(h_1 - h_2)$
Given $g = 9.8 \, m/s^2$,$h_1 = 5 \, m$,and $h_2 = 2 \, m$:
$v^2 = 2 \times 9.8 \times (5 - 2)$
$v^2 = 2 \times 9.8 \times 3$
$v^2 = 58.8$
$v = \sqrt{58.8} \approx 7.67 \, m/s$
Rounding to the nearest provided option,the velocity is $7.6 \, m/s$.