Mix Examples - WORK AND ENERGY Questions in English

(N/A) When riding a bicycle, the chemical energy stored in the body of the cyclist (muscular energy) is converted into the kinetic energy of the bicycle's pedals and wheels.
As the bicycle moves, this kinetic energy is used to overcome friction between the tires and the road, as well as air resistance, which eventually dissipates as heat energy.
Thus, the transformation chain is: $\text{Chemical Energy} \rightarrow \text{Muscular Energy} \rightarrow \text{Kinetic Energy} \rightarrow \text{Heat Energy (due to friction)}.$

(N/A) The work done by a force is given by the product of the force and the displacement in the direction of the force.
Work done $W = F \times d$
Here,the boy is moving against a constant frictional force $F = 5\, N$ throughout the entire path.
Total distance travelled by the boy is the sum of the straight path,the circular path,and the final straight path.
$1$. Initial straight path distance $= 1.5\, km = 1500\, m$.
$2$. Circular path distance: The boy completes $1.5$ cycles on a circular path of radius $r = 100\, m$. Distance $= 1.5 \times (2 \pi r) = 1.5 \times 2 \times 3.14 \times 100 = 942\, m$.
$3$. Final straight path distance $= 2.0\, km = 2000\, m$.
Total distance $d = 1500\, m + 942\, m + 2000\, m = 4442\, m$.
Since the frictional force acts along the entire path length,the work done against friction is calculated using the total distance covered.
Work done $W = 5\, N \times 4442\, m = 22210\, J$.

(NO) No,an object cannot have momentum if its mechanical energy is zero.
Mechanical energy is the sum of kinetic energy $(KE)$ and potential energy $(PE)$.
If the mechanical energy is zero,then $KE + PE = 0$.
Since kinetic energy $(KE = \frac{1}{2}mv^2)$ cannot be negative,and potential energy is typically defined relative to a reference point,for the total sum to be zero,both $KE$ and $PE$ must be zero.
If $KE = 0$,then $\frac{1}{2}mv^2 = 0$,which implies that the velocity $(v)$ of the object is $0$.
Since momentum $(p)$ is defined as the product of mass and velocity $(p = mv)$,if $v = 0$,then the momentum $(p)$ must also be $0$.

(N/A) According to the third equation of motion: $v^{2} - u^{2} = 2as$
From this, the displacement $s$ can be expressed as: $s = \frac{v^{2} - u^{2}}{2a}$
According to Newton's second law of motion, the force applied is: $F = ma$
Work done $(W)$ by the force $F$ over a distance $s$ is defined as: $W = F \times s$
Substituting the expressions for $F$ and $s$: $W = (ma) \times \left( \frac{v^{2} - u^{2}}{2a} \right)$
Simplifying the expression: $W = \frac{1}{2}m(v^{2} - u^{2}) = \frac{1}{2}mv^{2} - \frac{1}{2}mu^{2}$
Since kinetic energy $(K.E.)$ is given by $\frac{1}{2}mv^{2}$, we get: $W = (K.E.)_{f} - (K.E.)_{i}$
Thus, the work done is equal to the change in kinetic energy.

(10 M/S) $m(A) = m(B) = 1000 \,kg$. Initial velocity of car $(A)$,$v = 36 \,km/h = 36 \times (5/18) \,m/s = 10 \,m/s$.
Frictional force $F = 100 \,N$.
Since the car $(A)$ moves with a uniform speed,the engine must apply a force equal to the opposing frictional force.
Power $P = F \times v = 100 \,N \times 10 \,m/s = 1000 \,W$.
For the collision,we use the law of conservation of linear momentum:
$m_A u_A + m_B u_B = m_A v_A + m_B v_B$
Here,$m_A = 1000 \,kg$,$u_A = 10 \,m/s$,$m_B = 1000 \,kg$,$u_B = 0 \,m/s$,and $v_A = 0 \,m/s$ (as car $(A)$ comes to rest).
$1000 \times 10 + 1000 \times 0 = 1000 \times 0 + 1000 \times v_B$
$10000 = 1000 \times v_B$
$v_B = 10 \,m/s$.

(A) To find the work done on the trolley,we use the Work-Energy Theorem: $W = \Delta KE = KE_f - KE_i$.
Given: Total mass $m = 35 \, kg + 5 \, kg = 40 \, kg$,initial velocity $u = 4 \, m/s$,final velocity $v = 0 \, m/s$.
$W = \frac{1}{2} m v^2 - \frac{1}{2} m u^2 = 0 - \frac{1}{2} \times 40 \times (4)^2 = -20 \times 16 = -320 \, J$.
The work done on the trolley is $-320 \, J$.
$(b)$ Since the girl is sitting on the trolley and not applying any additional force to change the motion relative to the trolley,the work done by the girl is $0 \, J$.

(N/A) The force applied to lift the box is equal to its weight: $F = m \times g = 250 \,kg \times 10 \,m \,s^{-2} = 2500 \,N$.
Work done in lifting the box is $W = F \times s = 2500 \,N \times 1 \,m = 2500 \,J$.
$(b)$ Work done in holding the box is zero because the displacement $s = 0$.
$(c)$ They get tired because,to hold the box,their muscles must continuously exert a force to counteract the gravitational force,which involves internal muscular effort and energy expenditure.

(N/A) $(i)$ Chemical energy
$(ii)$ Heat energy
$(iii)$ Light energy
$(iv)$ Electrical energy
$(v)$ Sound energy
$(vi)$ Solar energy

(D) When a compressed spring is dissolved in an acid,the stored elastic potential energy $(PE)$ is released.
As the chemical bonds of the spring material break and react with the acid,this potential energy is transferred to the surrounding acid molecules.
This energy increases the kinetic energy $(KE)$ of the acid molecules,which is observed as an increase in the temperature of the solution.

(N/A) Power is defined as the rate of doing work. Its $SI$ unit is watt $(W)$.
$(b)$ For a simple pendulum, the interconversion of energy is as shown in the table below:

Position	Potential Energy $(PE)$	Kinetic Energy $(KE)$
Extreme position $1$	Maximum	Zero
Between extreme $1$ and mean	Decreasing	Increasing
Mean position	Zero	Maximum
Between mean and extreme $2$	Increasing	Decreasing
Extreme position $2$	Maximum	Zero

As the pendulum swings, the total mechanical energy $(PE + KE)$ remains constant, demonstrating the law of conservation of energy.

(N/A) The work done is $0 \, J$. In a circular orbit,the force of gravity acts perpendicular to the direction of motion (displacement). Since the angle between force and displacement is $90^{\circ}$,the work done $W = F \cdot s \cdot \cos(90^{\circ}) = 0$.
$(b)$ Given: mass $m = 250 \, g = 0.25 \, kg$,height $h = 2.5 \, m$,acceleration due to gravity $g = 9.8 \, m/s^2$ (or $10 \, m/s^2$).
Since the stone is thrown upwards,the force of gravity acts downwards (opposite to displacement).
Work done $W = -mgh = -(0.25 \, kg) \times (9.8 \, m/s^2) \times (2.5 \, m) = -6.125 \, J$ (or $-6.25 \, J$ if $g = 10 \, m/s^2$ is used).

(N/A) Yes,it is possible in the case of a body moving in a circular path with a constant speed $v$. The body experiences a centripetal acceleration directed towards the center of the circular path. The displacement is always tangential to the circular path,meaning the angle between the force and the displacement is $\theta = 90^{\circ}$. Thus,the work done is:
$W = F S \cos 90^{\circ} = 0$.
$(b)$ We know that the relation between kinetic energy $K$ and linear momentum $p$ is given by $p = \sqrt{2 m K}$.
For two bodies with equal kinetic energies $K$:
$p_{1} = \sqrt{2 m_{1} K}$
$p_{2} = \sqrt{2 m_{2} K}$
Therefore,the ratio of their linear momenta is:
$\frac{p_{1}}{p_{2}} = \sqrt{\frac{2 m_{1} K}{2 m_{2} K}} = \sqrt{\frac{m_{1}}{m_{2}}}$.

(N/A) When the bow is stretched,it stores elastic potential energy. When the arrow is released,this stored potential energy is transformed into the kinetic energy of the arrow,causing it to move forward.
$(b)$ Given:
Mass $(m) = 50 \, kg$
Height $(h) = 100 \, m$
Acceleration due to gravity $(g) = 10 \, m/s^2$
Potential Energy $(PE) = m \times g \times h$
$PE = 50 \times 10 \times 100$
$PE = 50,000 \, J$ or $50 \, kJ$.

(N/A) Power is defined as the rate of doing work or the rate of transfer of energy. Its $SI$ unit is watt $(W)$.
$(b)$ Work done is defined as the product of the force applied on an object and the displacement produced in the direction of the force. Its $SI$ unit is joule $(J)$.
$(c)$ Kinetic energy is the energy possessed by an object due to its motion. Its $SI$ unit is joule $(J)$.

(100 W) Work: The work done by a constant force is defined as the product of the force in the direction of the displacement and the magnitude of the displacement. Its $SI$ unit is joule $(J)$.
Energy: The capacity of a body to do work is termed as energy. Its $SI$ unit is joule $(J)$.
Power: Power is defined as the rate of doing work. Its $SI$ unit is watt $(W)$.
Given: Mass $m = 80 \, kg$,number of steps $= 30$,height of each step $= 12.5 \, cm = 0.125 \, m$,time $t = 30 \, s$,acceleration due to gravity $g = 10 \, m \, s^{-2}$.
Total height $h = 30 \times 0.125 \, m = 3.75 \, m$.
Work done $W = mgh = 80 \times 10 \times 3.75 = 3000 \, J$.
Power $P = W / t = 3000 / 30 = 100 \, W$.

(A) Law of conservation of energy: It states that energy can neither be created nor destroyed; it can only be converted from one form to another. The total energy in an isolated system remains constant.
Example: During the free fall of an object,the potential energy is gradually converted into kinetic energy,but the total mechanical energy remains constant.
$(b)$ $(i)$ Power expended by Girl $A$:
Weight $(mg) = 400 \, N$,Height $(h) = 8 \, m$,Time $(t) = 20 \, s$.
Power $(P) = \frac{\text{Work done}}{\text{Time}} = \frac{mgh}{t} = \frac{400 \times 8}{20} = 160 \, W$.
$(ii)$ Power expended by Girl $B$:
Weight $(mg) = 400 \, N$,Height $(h) = 8 \, m$,Time $(t) = 50 \, s$.
Power $(P) = \frac{mgh}{t} = \frac{400 \times 8}{50} = 64 \, W$.
$(c)$ Power $(P) = 1500 \, W = 1.5 \, kW$,Time $(t) = 10 \, h$.
Energy consumed $= P \times t = 1.5 \, kW \times 10 \, h = 15 \, kWh$.

(A) Work done by a force acting on an object is defined as the product of the magnitude of the force and the displacement of the object in the direction of the force.
The expression for work done is: $W = F \times s$,where $W$ is work,$F$ is force,and $s$ is displacement.
$(b)$ If the displacement of the object is zero,the work done is zero. This is because $W = F \times 0 = 0$.
$(c)$ Kinetic energy is the energy possessed by an object by virtue of its motion. The expression for the kinetic energy $(K)$ of an object of mass $m$ moving with a speed $v$ is $K = \frac{1}{2}mv^2$.

(N/A) No,the statement is incorrect. Work is not done because the gravitational force acting on the Earth (directed towards the Sun) and the displacement (along the tangent to the circular path) are perpendicular to each other,i.e.,$\theta = 90^{\circ}$. Since $W = F S \cos \theta$ and $\cos 90^{\circ} = 0$,the work done $W = 0$.
$(b)$ Since the coolie is applying force to hold the box but there is no displacement of the box $(S = 0)$,the work done by him is zero.
$(c)$ Total mass $m = 70 \, kg + 10 \, kg = 80 \, kg$. Force $F = m \times g = 80 \, kg \times 10 \, m s^{-2} = 800 \, N$. Displacement $S = 100 \, m$. Work done $W = F \times S = 800 \, N \times 100 \, m = 80,000 \, J$.

(C) Average power is defined as the ratio of total work done to the total time taken.
$(b)$ When a football is kicked,it follows a curved path and returns to the ground. The work done by gravity depends only on the vertical displacement. Since the initial and final vertical positions are the same (ground level),the net vertical displacement $h = 0$. Therefore,the work done by gravity $W = mgh = 0$.
$(c)$ Work done by Ram and Shyam is equal because both have the same mass $(40 \, kg)$ and climb the same vertical height $(3 \, m)$.
Power of Ram $= \frac{mgh}{t} = \frac{40 \times 9.8 \times 3}{10} = 117.6 \, W$.
Power of Shyam $= \frac{mgh}{t} = \frac{40 \times 9.8 \times 3}{12} = 98 \, W$.
Since $117.6 \, W > 98 \, W$,the power of Ram is greater than the power of Shyam.

(1:3) Work is defined as the product of the force applied on an object and the displacement of the object in the direction of the force. The magnitude of work depends on two factors: $(1)$ The magnitude of the force applied $(F)$ and $(2)$ The displacement of the object $(s)$ in the direction of the force.
$(b)$ Given: Speed of car $(v_1) = 30 \, m/s$,Speed of truck $(v_2) = 30 \, m/s$. Ratio of masses $(m_1 : m_2) = 1 : 3$.
Since kinetic energy $(E_k) = \frac{1}{2}mv^2$,the ratio of kinetic energies is:
$\frac{E_1}{E_2} = \frac{\frac{1}{2}m_1v_1^2}{\frac{1}{2}m_2v_2^2} = \frac{m_1}{m_2} \times \left(\frac{v_1}{v_2}\right)^2$
Since $v_1 = v_2$,the ratio becomes $\frac{m_1}{m_2} = \frac{1}{3}$.
Therefore,the ratio of kinetic energy is $1:3$.

(N/A) 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. Thus,the total energy of an isolated system remains constant.
$(b)$ The $SI$ unit of energy is the joule $(J)$,and the commercial unit of energy is the kilowatt-hour $(kWh)$.
$1 \, kWh = 1 \, kW \times 1 \, h$
$1 \, kWh = 1000 \, W \times 3600 \, s$
$1 \, kWh = 3,600,000 \, J = 3.6 \times 10^6 \, J$.
$(c)$ Given: Force $(F)$ = $7 \, N$,Displacement $(S)$ = $5 \, m$.
Work done $(W)$ = $F \times S$
$W = 7 \, N \times 5 \, m = 35 \, J$.

(D) $(i)$ When an object is thrown at a certain angle, it rises to a maximum height $h$ and then returns to the ground following a parabolic path. The work done against gravity while rising is $W_{against} = mgh$. The work done by gravity while falling is $W_{by} = mgh$. Since the initial and final points are on the same horizontal line, the net displacement in the vertical direction is $0$. Therefore, the total work done by the force of gravity is $W_{net} = W_{against} + W_{by} = -mgh + mgh = 0 \, J$.
$(ii)$ Given: mass $m = 20 \, kg$, initial velocity $u = 5 \, m s^{-1}$, final velocity $v = 2 \, m s^{-1}$.
According to the work-energy theorem, the work done by a force is equal to the change in kinetic energy:
$W = \Delta KE = \frac{1}{2} m v^2 - \frac{1}{2} m u^2$
$W = \frac{1}{2} \times 20 \times (2)^2 - \frac{1}{2} \times 20 \times (5)^2$
$W = 10 \times 4 - 10 \times 25$
$W = 40 - 250 = -210 \, J$.
The work done by the force is $-210 \, J$.

(A) Given: Force $F = 100 \, N$,horizontal displacement $S_H = 5 \, m$,vertical displacement $S_V = 10 \, m$.
Let $W_1$ be the work done during horizontal movement and $W_2$ be the work done during vertical movement.
For horizontal movement,the force is vertical and displacement is horizontal,so the angle $\theta = 90^{\circ}$.
$W_1 = F \cdot S_H \cdot \cos(90^{\circ}) = 100 \times 5 \times 0 = 0 \, J$.
For vertical movement,the force is vertical and displacement is vertical,so the angle $\theta = 0^{\circ}$.
$W_2 = F \cdot S_V \cdot \cos(0^{\circ}) = 100 \times 10 \times 1 = 1000 \, J$.
Total work done $W = W_1 + W_2 = 0 + 1000 = 1000 \, J$.
$(b)$ Kinetic energy $K$ is given by $K = \frac{1}{2}mv^2$ and momentum $p$ is given by $p = mv$.
From $p = mv$,we get $v = \frac{p}{m}$.
Substituting $v$ in the kinetic energy formula:
$K = \frac{1}{2}m \left(\frac{p}{m}\right)^2 = \frac{1}{2}m \left(\frac{p^2}{m^2}\right) = \frac{p^2}{2m}$.
Thus,the relationship is $K = \frac{p^2}{2m}$.

(N/A) Average power is defined as the average rate of doing work. If a total work $W$ is done in a time $t,$ then average power $P_{av}$ is given by:
$P_{av} = \frac{W}{t}$
Total power consumed by $4$ devices $= 4 \times 500 = 2000\, W = 2\, kW$
Time for which devices run $= 10\, hours$
Energy consumed per day $= 2\, kW \times 10\, h = 20\, kWh$
Energy consumed in the month of November ($30$ days) $= 20\, kWh \times 30 = 600\, kWh$
$(b)$ When a body falls freely under gravity,it possesses kinetic energy due to its motion. When it strikes the ground,its kinetic energy is dissipated into other forms of energy,primarily sound energy and heat energy,due to the impact with the surface.

(A) $(i)$ The kinetic energy $(E)$ of a body is related to its momentum $(p)$ as $E = p^2 / 2m$. Since the momentum is the same for both objects,$E \propto 1/m$.
Therefore,the ratio of the kinetic energy of the light object to the heavy object is $\frac{E_{\text{light}}}{E_{\text{heavy}}} = \frac{m_{\text{heavy}}}{m_{\text{light}}}$.
Since $m_{\text{heavy}} > m_{\text{light}}$,the ratio is greater than $1$. Thus,the lighter object has a larger kinetic energy.
$(ii)$ Given: Initial height $h_1 = 10\, m$. The energy reduces by $40\%$,so the remaining energy is $60\%$ of the initial energy $(E_2 = 0.6 E_1)$.
Using the potential energy formula $E = mgh$,we have $mgh_2 = 0.6 mgh_1$.
$h_2 = 0.6 \times h_1 = 0.6 \times 10\, m = 6\, m$.
The ball will bounce back to a height of $6\, m$.

(B) Mechanical energy is the sum of kinetic energy $(KE)$ and potential energy $(PE)$ of an object. Its two forms are kinetic energy and potential energy. 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. An example is a simple pendulum: at the mean position,the energy is entirely kinetic,while at the extreme positions,it is entirely potential. As the pendulum oscillates,energy continuously converts between kinetic and potential forms.
$(b)$ Given: Mass $m = 1000 \, kg$,initial velocity $u = 72 \, km \, h^{-1} = 72 \times (5/18) \, m \, s^{-1} = 20 \, m \, s^{-1}$,final velocity $v = 0 \, m \, s^{-1}$.
Work done = Change in kinetic energy = $\frac{1}{2} m (v^2 - u^2)$.
Work done = $\frac{1}{2} \times 1000 \times (0^2 - 20^2) = 500 \times (-400) = -200,000 \, J = -2 \times 10^5 \, J$.
The negative sign indicates that work is done against the motion of the car.

(A) Given: $m_1 = m, m_2 = m, v_1 = 2v, v_2 = 3v$. We need to find the ratio $E_1 / E_2$.
We know that kinetic energy $E = \frac{1}{2}mv^2$. Therefore:
$\frac{E_1}{E_2} = \frac{\frac{1}{2}m_1v_1^2}{\frac{1}{2}m_2v_2^2} = \frac{v_1^2}{v_2^2} = \frac{(2v)^2}{(3v)^2} = \frac{4v^2}{9v^2} = \frac{4}{9}$.
$(b)$ Total energy = Kinetic Energy $(KE)$ + Potential Energy $(PE)$.
At height $h$,the total energy is $mgh$ (since $KE = 0$ and $PE = mgh$).
According to the law of conservation of energy,the total energy remains constant at every point during the fall.
At halfway to the ground (height $h/2$),the velocity $v$ is given by $v^2 - u^2 = 2g(h/2)$. Since $u = 0$,$v^2 = gh$.
$KE = \frac{1}{2}mv^2 = \frac{1}{2}mgh$.
$PE = mg(h/2) = \frac{1}{2}mgh$.
Total energy = $KE + PE = \frac{1}{2}mgh + \frac{1}{2}mgh = mgh$.

$(D)$ The two conditions for work to be done are: $(i)$ $A$ force must act on the object, and $(ii)$ The object must undergo displacement in the direction of the force.
$(b)$ The work done by frictional force is negative because the force of friction acts in the direction opposite to the displacement of the object.
$(c)$ Power is defined as the rate of doing work. The relationship is given by $P = W / t$, where $P$ is power, $W$ is work done, and $t$ is time taken.
$(d)$ Given: Work $(W) = 1960 \, J$, Time $(t) = 4 \, \text{minutes} = 4 \times 60 = 240 \, s$.
Using the formula $P = W / t = 1960 / 240 = 8.166... \approx 8.17 \, W$.

(N/A) Kinetic energy $(KE)$ is the work done by a force $F$ to accelerate a body of mass $m$ from rest $(u=0)$ to velocity $v$ over a distance $s$. Using $v^2 - u^2 = 2as$,we get $s = v^2 / (2a)$. Since $F = ma$,the work done $W = F \times s = ma \times (v^2 / 2a) = 1/2 mv^2$. Thus,$KE = 1/2 mv^2$.
$(b)$ Since $KE \propto v^2$,if the velocity becomes $5v$,the new kinetic energy $KE' = 1/2 m(5v)^2 = 25 \times (1/2 mv^2) = 25 \times KE$. The kinetic energy increases by a factor of $25$.
$(c)$ The kinetic energy of a falling body at the ground is equal to its initial potential energy $(PE = mgh)$. For mass $m$ at height $h$,$PE_1 = mgh$. For mass $2m$ at height $2h$,$PE_2 = (2m)g(2h) = 4mgh$. Since $PE_2 > PE_1$,the body with mass $2m$ will have greater kinetic energy.

(N/A) 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.
$(b)$ Given: Mass $m = 1500 \, kg$,Initial velocity $u = 36 \, km/h = 36 \times (5/18) = 10 \, m/s$,Final velocity $v = 72 \, km/h = 72 \times (5/18) = 20 \, m/s$.
Work done $(W)$ = Change in kinetic energy = $\frac{1}{2} m (v^2 - u^2)$.
$W = \frac{1}{2} \times 1500 \times (20^2 - 10^2) = 750 \times (400 - 100) = 750 \times 300 = 225,000 \, J$ or $2.25 \times 10^5 \, J$.
Since the velocity increases,the work done is positive.
$(c)$ An oscillating pendulum has maximum potential energy $(PE)$ at its extreme positions and maximum kinetic energy $(KE)$ at its mean position.

(B) Given: Force $F = 20 \,N$,time $t = 10 \,s$,mass $m = 5 \,kg$,and initial velocity $u = 0 \,m/s$.
First,calculate the acceleration $a$ using Newton's second law: $a = F/m = 20/5 = 4 \,m/s^2$.
Next,calculate the final velocity $v$ after $10 \,s$ using the equation $v = u + at$: $v = 0 + (4 \times 10) = 40 \,m/s$.
Finally,calculate the kinetic energy $KE$ using the formula $KE = 1/2 \times m \times v^2$: $KE = 1/2 \times 5 \times (40)^2 = 1/2 \times 5 \times 1600 = 4000 \,J$.

(B) Given: Mass $m = 4 \, kg$,Force $F = 16 \, N$,Initial velocity $u = 0$,Time $t = 10 \, s$.
Step $1$: Calculate acceleration $a$ using Newton's second law,$F = ma$.
$a = \frac{F}{m} = \frac{16 \, N}{4 \, kg} = 4 \, m/s^2$.
Step $2$: Calculate final velocity $v$ using the equation of motion $v = u + at$.
$v = 0 + (4 \, m/s^2 \times 10 \, s) = 40 \, m/s$.
Step $3$: Calculate kinetic energy $KE$ using the formula $KE = \frac{1}{2}mv^2$.
$KE = \frac{1}{2} \times 4 \, kg \times (40 \, m/s)^2 = 2 \times 1600 = 3200 \, J$.

(A) The boy has to overcome the force of gravity to climb the stairs.
Force of gravity on the boy,$F = m \times g = 40 \, kg \times 9.8 \, m s^{-2} = 392 \, N$.
Total height (distance) covered,$S = 50 \times 10 \, cm = 500 \, cm = 5 \, m$.
$(i)$ Work done by the boy in climbing,$W = F \times S = 392 \, N \times 5 \, m = 1960 \, J$.
$(ii)$ Power developed,$P = \frac{W}{t} = \frac{1960 \, J}{5 \, s} = 392 \, W$.

(C) Given: Mass $m = 10 \, kg$,Height $h = 50 \, cm = 0.5 \, m$,Acceleration due to gravity $g = 10 \, m s^{-2}$.
$(a)$ Potential energy $(PE)$ just before dropping is calculated as $PE = mgh = 10 \times 10 \times 0.5 = 50 \, J$.
$(b)$ According to the law of conservation of energy,the total mechanical energy remains constant. Just before touching the ground,the potential energy is converted into kinetic energy $(KE)$. Therefore,$KE = PE = 50 \, J$.
$(c)$ Using the formula for kinetic energy,$KE = \frac{1}{2}mv^2$,we can find the velocity $v$ as $v = \sqrt{\frac{2 \times KE}{m}} = \sqrt{\frac{2 \times 50}{10}} = \sqrt{10} \approx 3.16 \, m s^{-1}$.

(B) Given: Mass $m = 5\, kg$,Initial velocity $u = 0\, m/s$,Force $F = 20\, N$,Time $t = 10\, s$.
Step $1$: Calculate acceleration using Newton's second law of motion,$F = ma$.
$a = F / m = 20 / 5 = 4\, m/s^2$.
Step $2$: Calculate the final velocity $v$ using the first equation of motion,$v = u + at$.
$v = 0 + (4 \times 10) = 40\, m/s$.
Step $3$: Calculate the kinetic energy $KE$ using the formula $KE = 1/2 mv^2$.
$KE = 1/2 \times 5 \times (40)^2 = 1/2 \times 5 \times 1600 = 5 \times 800 = 4000\, J$.

(A) Work done by each child is calculated using the formula $W = mgh$.
Given $m = 20 \, kg$,$g = 9.8 \, m/s^2$,and $h = 10 \, m$.
Work done $= 20 \times 9.8 \times 10 = 1960 \, J$.
Since $m$,$g$,and $h$ are the same for both children,the work done is equal.
Power is defined as the rate of doing work,$P = W / t$.
For child $A$: $P_A = 1960 / 10 = 196 \, W$.
For child $B$: $P_B = 1960 / 20 = 98 \, W$.
Since $196 \, W > 98 \, W$,child $A$ has more power.

(D) compressed spring stores elastic potential energy due to its deformed state.
When the spring is dissolved in an acid,the chemical bonds holding the spring in its compressed state are broken.
The stored elastic potential energy is released and converted into the kinetic energy of the acid molecules.
This increase in the kinetic energy of the molecules results in a rise in the temperature of the acid solution.
Therefore,the potential energy is ultimately converted into thermal energy.

(FALSE) The statement is False.
In physics,work is defined as the product of force and displacement,and energy is defined as the capacity to do work.
Both work and energy are measured in the same $SI$ unit,which is the Joule $(J)$.