Mix Example - WORK AND ENERGY Questions in English

(A) Let the two objects have masses $m_1$ and $m_2$ such that $m_2 > m_1$. Let their common momentum be $p$.
The kinetic energy $K$ is related to momentum $p$ and mass $m$ by the formula $K = \frac{p^2}{2m}$.
For the first object (lighter),$K_1 = \frac{p^2}{2m_1}$.
For the second object (heavier),$K_2 = \frac{p^2}{2m_2}$.
The ratio of their kinetic energies is $\frac{K_1}{K_2} = \frac{p^2 / 2m_1}{p^2 / 2m_2} = \frac{m_2}{m_1}$.
Since $m_2 > m_1$,it follows that $\frac{m_2}{m_1} > 1$,which implies $K_1 > K_2$.
Therefore,the lighter object has a larger kinetic energy.

(N/A) Potential energy is the energy stored in an object due to its position or configuration. Gravitational potential energy of an object at a height $h$ above the ground is defined as the work done in raising it from the ground to that point against gravity.
For an object of mass $m$ raised through a height $h$, the work done $W$ is given by:
$W = \text{force} \times \text{displacement} = (m \times g) \times h = mgh$
Thus, the potential energy $PE = mgh$.
$(b)$ Given:
Mass $m = 10 \, kg$
Height $h = 6 \, m$
Acceleration due to gravity $g = 9.8 \, m s^{-2}$
Potential energy $PE = mgh = 10 \times 9.8 \times 6 = 588 \, J$.

(112.5 W) Power is defined as the rate of doing work.
Given:
Mass of the boy $(m)$ = $45 \, kg$
Number of steps $(n)$ = $20$
Height of each step = $25 \, cm = 0.25 \, m$
Total height $(h)$ = $20 \times 0.25 \, m = 5 \, m$
Time taken $(t)$ = $20 \, s$
Acceleration due to gravity $(g)$ = $10 \, m s^{-2}$
Work done $(W)$ = Potential energy gained = $mgh$
$W = 45 \times 10 \times 5 = 2250 \, J$
Power $(P)$ = $\frac{\text{Work}}{\text{Time}} = \frac{W}{t}$
$P = \frac{2250}{20} = 112.5 \, W$
Thus,the power of the boy is $112.5 \, W$.

(N/A) The power of a device is said to be $1 \ W$ if it consumes energy at the rate of $1 \ J$ per second.
$(b)$ Given:
Energy $(E) = 1000 \ J$
Time $(t) = 10 \ s$
Power $(P) = E / t$
$P = 1000 \ J / 10 \ s = 100 \ W$
Thus,the power of the lamp is $100 \ W$.

(N/A) No,the total energy of the body is conserved. As a body falls down,its height from the ground decreases,and hence its potential energy decreases. Simultaneously,its velocity and kinetic energy increase. The decrease in potential energy is exactly equal to the gain in kinetic energy. Thus,the sum of potential energy and kinetic energy (mechanical energy) remains constant at any instant of time.
$(b)$
$(i)$ Potential energy is maximum at the initial height $h$,where the object is at rest.
$(ii)$ Kinetic energy is maximum at the ground level,just before the object hits the surface,as the velocity is highest at this point.

(N/A) The work done by the force applied by the boy is positive because the displacement of the ball is in the same direction as the applied force.
$(b)$ The work done by the gravitational force is negative because the gravitational force acts downwards while the displacement of the ball is upwards,meaning the force and displacement are in opposite directions.

(N/A) The force applied to lift the box is equal to its weight,$F = mg = 250 \,kg \times 10 \,m/s^2 = 2500 \,N$. The displacement is $S = 1 \,m$. The work done is $W = F \times S = 2500 \,N \times 1 \,m = 2500 \,J$.
$(b)$ When holding the box,there is no displacement $(S = 0)$. Therefore,the work done is $W = F \times 0 = 0 \,J$.
$(c)$ They get tired because of the continuous muscular effort required to counteract the gravitational force acting on the box,even though no mechanical work is performed.

(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) Power is defined as the rate of doing work or the rate of energy consumption. One watt is the power of an agent that does $1\, J$ of work in $1\, s$.
Given:
Energy consumed $(W)$ = $1000\, J$
Time taken $(t)$ = $10\, s$
Formula:
$P = \frac{W}{t}$
Calculation:
$P = \frac{1000\, J}{10\, s} = 100\, W$
Therefore,the power of the lamp is $100\, W$.

(N/A) flying bird possesses potential energy because of its height above the ground level,which is defined with respect to the Earth's surface.
It also possesses kinetic energy due to the velocity with which it is flying through the air.
The expression for potential energy $(PE)$ is: $PE = mgh$,where $m$ is the mass of the bird,$g$ is the acceleration due to gravity,and $h$ is the height above the ground.
The expression for kinetic energy $(KE)$ is: $KE = \frac{1}{2}mv^2$,where $m$ is the mass of the bird and $v$ is its velocity.

(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) The commercial unit of energy is kilowatt-hour $(kWh)$.
$1 \,kWh = 1 \,kW \times 1 \,h = 1000 \,W \times 3600 \,s = 3.6 \times 10^6 \,J$.
$(b)$ Given: Power $(P) = 60 \,W = 0.06 \,kW$,Time $(t) = 10 \,h$.
Energy consumed $(E) = P \times t = 0.06 \,kW \times 10 \,h = 0.6 \,kWh$.
Since $1 \,kWh = 1 \text{ unit}$,the bulb consumes $0.6 \text{ units}$ of electrical energy in one day.

(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 $(J)$.

(N/A) Power is defined as the rate of doing work or the rate of energy consumption.
The commercial unit of electrical energy is $kWh$ (kilowatt-hour), and the $SI$ unit is Joule $(J)$.
Given:
Power $(P)$ = $1.5 \, kW$
Time $(t)$ = $2 \, \text{hours}$
Electrical energy consumed $(E)$ = $P \times t$
$E = 1.5 \, kW \times 2 \, \text{h} = 3 \, kWh$.
Therefore, the electrical energy consumed in a day is $3 \, \text{units}$ (since $1 \, kWh = 1 \, \text{unit}$).

(2:3) Kinetic energy is the energy possessed by an object due to its motion.
$(b)$ Given the ratio of masses $\frac{m_1}{m_2} = \frac{2}{3}$ and the velocity $v_1 = v_2 = 108 \, km/h$.
The formula for kinetic energy is $KE = \frac{1}{2}mv^2$.
Since the velocity is the same for both,the ratio of kinetic energies is $\frac{KE_1}{KE_2} = \frac{\frac{1}{2}m_1v^2}{\frac{1}{2}m_2v^2} = \frac{m_1}{m_2}$.
Therefore,the ratio of their kinetic energy is $2:3$.

(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 $P = W / t = 3000 / 30 = 100 \, W$.

(N/A) Kinetic energy: The energy possessed by a body by virtue of its motion is called kinetic energy.
Potential energy: The energy possessed by a body by virtue of its position or configuration is called potential energy.
The expression for kinetic energy is $KE = \frac{1}{2} mv^2$.
Given:
Mass $(m)$ = $10 \, kg$
Velocity $(v)$ = $10 \, m s^{-1}$
Calculation:
$KE = \frac{1}{2} \times 10 \times (10)^2$
$KE = 5 \times 100$
$KE = 500 \, J$
The kinetic energy of the stone is $500 \, J$.

(D) $(i)$ Consider a body of mass $m$ initially at rest,i.e.,$u = 0$,on a frictionless surface. Let a constant force $F$ act on the body,producing an acceleration $a$. After a displacement $S$,the body attains a velocity $v$. Using the equation of motion $v^2 - u^2 = 2aS$,we get $a = v^2 / (2S)$.
The work done $W$ by the force is $W = F \times S$. Since $F = ma$,we have $W = (ma) \times S = m \times (v^2 / 2S) \times S = 1/2 mv^2$.
This work done is stored as kinetic energy $(KE)$. Thus,$KE = 1/2 mv^2$. The $SI$ unit of kinetic energy is Joule $(J)$.
$(ii)$ Given: Mass $m = 10 \, kg$,velocity $v = 5 \, m s^{-1}$.
$KE = 1/2 mv^2 = 1/2 \times 10 \times (5)^2 = 5 \times 25 = 125 \, J$.

(N/A) The two forms of energy involved are Kinetic Energy $(KE)$ and Potential Energy $(PE)$.
$(b)$ In the diagram,point $A$ represents the extreme position where $PE$ is maximum and $KE$ is zero. Point $B$ represents the mean position where $KE$ is maximum and $PE$ is zero.
$(c)$ As the pendulum moves from the mean position $(B)$ to the extreme position $(A)$,$KE$ is converted into $PE$,so $KE$ decreases and $PE$ increases. Conversely,as it moves from the extreme position $(A)$ to the mean position $(B)$,$PE$ is converted into $KE$,so $PE$ decreases and $KE$ increases.
$(d)$ The law involved is the Law of Conservation of Energy. It states that energy can neither be created nor destroyed,only transformed from one form to another. In an isolated system,the total energy remains constant.

(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) Given: Mass $(m) = 10 \ kg$,Height $(h) = 10 \ m$,Acceleration due to gravity $(g) = 10 \ m/s^2$.
Potential Energy $(PE) = mgh$,Kinetic Energy $(KE) = \frac{1}{2}mv^2$,Mechanical Energy $(ME) = PE + KE$.

Height $(m)$	$PE$ $(J)$	$KE$ $(J)$	$ME$ $(J)$
$10$	$1000$	$0$	$1000$
$8$	$800$	$200$	$1000$
$5$	$500$	$500$	$1000$
$0$	$0$	$1000$	$1000$

Law of Conservation of Energy: Energy can neither be created nor destroyed; it can only be transformed from one form to another. The total energy in an isolated system remains constant.

(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 $6\, m$.

(N/A) $(i)$ 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 a system,the total energy remains constant if no external non-conservative forces act on it.
$(ii)$ Consider a body of mass $m$ falling freely under gravity from a height $h$ above the ground. Let the body be at point $A$ at height $h$,point $B$ at height $(h-x)$ after falling a distance $x$,and point $C$ just before hitting the ground.
At point $A$ (Height $h$): The body is at rest,so velocity $v = 0$.
Kinetic Energy $(T) = 1/2 mv^2 = 0$.
Potential Energy $(U) = mgh$.
Total Energy $(E) = T + U = 0 + mgh = mgh$ $....(i)$
At point $B$ (Height $h-x$): The body has fallen distance $x$. Using $v^2 - u^2 = 2as$,where $u=0, a=g, s=x$,we get $v^2 = 2gx$.
Kinetic Energy $(T) = 1/2 m(2gx) = mgx$.
Potential Energy $(U) = mg(h-x) = mgh - mgx$.
Total Energy $(E) = T + U = mgx + mgh - mgx = mgh$ $....(ii)$
At point $C$ (Height $0$): The body has fallen distance $h$. Using $v^2 - u^2 = 2as$,where $u=0, a=g, s=h$,we get $V^2 = 2gh$.
Kinetic Energy $(T) = 1/2 mV^2 = 1/2 m(2gh) = mgh$.
Potential Energy $(U) = mg(0) = 0$.
Total Energy $(E) = T + U = mgh + 0 = mgh$ $....(iii)$
From equations $(i), (ii),$ and $(iii)$,it is clear that the total mechanical energy of a freely falling body remains constant at all points.

(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.

(A) Given: Mass $m = 800 \, kg$,height $h = 1500 \, cm = 15 \, m$,time $t = 20 \, s$,and acceleration due to gravity $g = 10 \, m s^{-2}$.
The work done $W$ to lift the water is equal to the potential energy gained: $W = mgh$.
Power $P$ is defined as the rate of doing work: $P = \frac{W}{t} = \frac{mgh}{t}$.
Substituting the values: $P = \frac{800 \times 10 \times 15}{20}$.
$P = \frac{120000}{20} = 6000 \, W$.
Since $1 \, kW = 1000 \, W$,we have $P = 6 \, kW$.

(N/A) Given: Mass of car $m_1 = 750 \,kg$,Velocity $v = 54 \,km \cdot h^{-1} = 54 \times \frac{5}{18} \,m \cdot s^{-1} = 15 \,m \cdot s^{-1}$.
Initial Kinetic Energy $(KE_1)$:
$KE_1 = \frac{1}{2} m_1 v^2 = \frac{1}{2} \times 750 \times (15)^2 = 375 \times 225 = 84375 \,J$.
When a passenger of mass $50 \,kg$ sits in the car,the new total mass $m_2 = 750 \,kg + 50 \,kg = 800 \,kg$.
New Kinetic Energy $(KE_2)$:
$KE_2 = \frac{1}{2} m_2 v^2 = \frac{1}{2} \times 800 \times (15)^2 = 400 \times 225 = 90000 \,J$.

(A) Given: Mass of the man $m = 60 \, kg$,time taken $t = 40 \, s$,number of steps $= 30$,height of each step $= 20 \, cm = 0.2 \, m$.
Total height $h = 30 \times 0.2 \, m = 6 \, m$.
Acceleration due to gravity $g = 10 \, m/s^2$ (taking $g = 10 \, m/s^2$ for simplicity).
Work done $W = mgh = 60 \times 10 \times 6 = 3600 \, J$.
Power $P = \frac{W}{t} = \frac{3600 \, J}{40 \, s} = 90 \, W$.
Thus,the power of the man is $90 \, W$.

(N/A) The work done by the man is equal to the change in the potential energy of the system (man + weight).
The total mass $m$ is the sum of the man's mass and the weight he carries:
$m = 70 \,kg + 10 \,kg = 80 \,kg$
The height $h$ is $100 \,m$.
Taking the acceleration due to gravity $g = 9.8 \,m/s^2$,the work done $W$ is calculated as:
$W = m \times g \times h$
$W = 80 \,kg \times 9.8 \,m/s^2 \times 100 \,m$
$W = 78400 \,J$ (or $7.84 \times 10^4 \,J$).

(B) The work done in moving a body through a certain distance is given by the formula:
$W = F \times S \times \cos \theta$
Assuming the force is applied in the direction of displacement,$\theta = 0^\circ$,so $\cos 0^\circ = 1$.
Given:
Force $(F)$ = $100 \, N$
Displacement $(S)$ = $10 \, m$
Mass $(m)$ = $50 \, kg$ (Note: Mass is not required for this calculation as force is already provided).
Calculation:
$W = F \times S = 100 \, N \times 10 \, m = 1000 \, J$.

(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 $KE = 1/2 \times m \times v^2$: $KE = 1/2 \times 5 \times (40)^2 = 1/2 \times 5 \times 1600 = 4000 \,J$.

(A) The power $P$ expended is defined as the rate of doing work,given by the formula $P = W / t$.
Since the girl climbs at a constant speed,the work done $W$ is equal to the change in potential energy,which is $W = mgh$.
Given values are: mass $m = 40\, kg$,height $h = 6\, m$,and time $t = 15\, s$. Taking acceleration due to gravity $g = 9.8\, m/s^2$.
Substituting these values into the formula:
$P = \frac{mgh}{t} = \frac{40 \times 9.8 \times 6}{15} = \frac{2352}{15} = 156.8\, W$.
Thus,the power expended by the girl is $156.8\, W$.

(D) Let the mass of each body be $m$.
The kinetic energy $(KE)$ of the first body is given by $K_{1} = \frac{1}{2}mv^{2} \quad \dots(i)$.
The kinetic energy $(KE)$ of the second body is given by $K_{2} = \frac{1}{2}m(3v)^{2} = \frac{1}{2}m(9v^{2}) = \frac{9}{2}mv^{2} \quad \dots(ii)$.
To find the ratio of their kinetic energies,we divide equation $(ii)$ by equation $(i)$:
$\frac{K_{2}}{K_{1}} = \frac{\frac{9}{2}mv^{2}}{\frac{1}{2}mv^{2}} = \frac{9}{1}$.
Thus,the ratio of their kinetic energies is $9:1$.

(A) Given: Mass,$m = 2 \, kg$,Velocity,$v = 0.1 \, m s^{-1}$.
The formula for kinetic energy $(KE)$ is given by $KE = \frac{1}{2} m v^2$.
Substituting the given values into the formula:
$KE = \frac{1}{2} \times 2 \times (0.1)^2$
$KE = 1 \times 0.01$
$KE = 0.01 \, J$.
Therefore,the kinetic energy of the body is $0.01 \, J$.

(B) Given: Mass of the body,$m = 100 \, g = 0.1 \, kg$.
Kinetic energy,$KE = 20 \, J$.
We know that the formula for kinetic energy is $KE = \frac{1}{2} m v^2$.
Rearranging the formula to solve for velocity $v$:
$v^2 = \frac{2 \times KE}{m}$.
Substituting the values:
$v^2 = \frac{2 \times 20}{0.1} = \frac{40}{0.1} = 400$.
Taking the square root on both sides:
$v = \sqrt{400} = 20 \, m s^{-1}$.

(N/A) Given: Mass,$m = 10\, kg$,Height,$h = 5\, m$,Acceleration due to gravity,$g = 9.8\, m/s^2$.
$1$. Potential Energy $(PE)$ at the top:
$PE = mgh = 10 \times 9.8 \times 5 = 490\, J$.
$2$. Kinetic Energy $(KE)$ at the ground:
According to the law of conservation of energy,the potential energy at the top is converted into kinetic energy at the ground.
$KE = PE = 490\, J$.
$3$. Speed just before hitting the ground:
Using the equation of motion $v^2 - u^2 = 2gh$,where initial velocity $u = 0\, m/s$:
$v^2 - 0 = 2 \times 9.8 \times 5$
$v^2 = 98$
$v = \sqrt{98} \approx 9.9\, m/s$.

(B) The formula for kinetic energy $(KE)$ is given by $KE = \frac{1}{2}mv^2$,where $m$ is the mass and $v$ is the velocity of the object.
$1$. If we double the mass $(m' = 2m)$,the new kinetic energy becomes $KE' = \frac{1}{2}(2m)v^2 = 2 \times KE$. Thus,the kinetic energy doubles.
$2$. If we double the velocity $(v' = 2v)$,the new kinetic energy becomes $KE'' = \frac{1}{2}m(2v)^2 = \frac{1}{2}m(4v^2) = 4 \times KE$. Thus,the kinetic energy increases four times.
Comparing the two,doubling the velocity has a greater effect on the kinetic energy than doubling the mass.

(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 $KE = \frac{1}{2}mv^2$.
$KE = \frac{1}{2} \times 4 \, kg \times (40 \, m/s)^2 = 2 \times 1600 = 3200 \, J$.

(C) Work done is defined as $W = F S \cos \theta$,where $F$ is the force,$S$ is the displacement,and $\theta$ is the angle between the force and displacement vectors.
$1$. For the horizontal motion: The force applied is vertical ($10 \,N$ upwards) and the displacement is horizontal $(5 \,m)$. The angle $\theta = 90^{\circ}$.
$W_{1} = F S \cos 90^{\circ} = 10 \times 5 \times 0 = 0 \,J$.
$2$. For the vertical motion: The force applied is vertical ($10 \,N$ upwards) and the displacement is vertical ($10 \,m$ upwards). The angle $\theta = 0^{\circ}$.
$W_{2} = F S \cos 0^{\circ} = 10 \times 10 \times 1 = 100 \,J$.
Total work done = $W_{1} + W_{2} = 0 + 100 = 100 \,J$.