Textbook - WORK AND ENERGY Questions in English

(A) When a force $F$ acts on an object to displace it through a distance $S$ in its direction,the work done $W$ on the body by the force is given by:
Work done = Force $\times$ Displacement
$W = F \times S$
Given:
Force $(F)$ = $7\, N$
Displacement $(S)$ = $8\, m$
Therefore,work done $(W)$ = $7\, N \times 8\, m$
$W = 56\, N\cdot m$
Since $1\, N\cdot m = 1\, J$,the work done is $56\, J$.

(C) Work is said to be done when the following two conditions are satisfied:
$(i)$ $A$ force acts on the body.
$(ii)$ The body undergoes displacement due to the applied force,and the displacement occurs in the direction of the force.

(N/A) When a force $F$ displaces a body through a distance $s$ in the direction of the applied force,the work done $W$ on the body is defined as the product of the force and the displacement.
Work done = Force $\times$ Displacement
$W = F \times s$

(N/A) $1\, J$ is defined as the amount of work done when a force of $1\, N$ displaces an object through a distance of $1\, m$ in the direction of the applied force.
Mathematically,$W = F \times s$,where $W = 1\, J$,$F = 1\, N$,and $s = 1\, m$.

(A) Work done by the bullocks is calculated using the formula:
Work done = Force $\times$ Displacement
$W = F \times d$
Given:
Applied force,$F = 140 \, N$
Displacement,$d = 15 \, m$
Substituting the values into the formula:
$W = 140 \, N \times 15 \, m = 2100 \, J$
Therefore,the total work done in ploughing the length of the field is $2100 \, J$.

(B) Kinetic energy is the energy possessed by a body by virtue of its motion.
Every moving object possesses kinetic energy.
$A$ body uses kinetic energy to perform work.
For example,the kinetic energy of a hammer is used to drive a nail into a log of wood,and the kinetic energy of air is used to run windmills.

(N/A) If a body of mass $m$ is moving with a velocity $v$,then its kinetic energy $E_k$ is given by the expression:
$E_k = \frac{1}{2}mv^2$
Its $SI$ unit is Joule $(J)$.

(D) The expression for kinetic energy is $E_k = 1/2 mv^2$.
Given that the initial kinetic energy $E_k = 25\, J$ at a velocity $v = 5\, m\, s^{-1}$.
$(i)$ When the velocity is doubled,the new velocity becomes $v' = 2v = 10\, m\, s^{-1}$.
Since $E_k \propto v^2$,the new kinetic energy $E_k' = 1/2 m(2v)^2 = 4 \times (1/2 mv^2) = 4 \times 25\, J = 100\, J$.
$(ii)$ When the velocity is increased three times,the new velocity becomes $v'' = 3v = 15\, m\, s^{-1}$.
Similarly,the new kinetic energy $E_k'' = 1/2 m(3v)^2 = 9 \times (1/2 mv^2) = 9 \times 25\, J = 225\, J$.

(B) Power is defined as the rate of doing work or the rate of transfer of energy.
If $W$ is the amount of work done in time $t$,then power $P$ is given by the expression:
$P = \frac{W}{t} = \frac{\text{Energy}}{\text{Time}}$
The $SI$ unit of power is the watt $(W)$,which is defined as $1 \text{ joule per second } (J/s)$.

(N/A) body is said to have a power of $1$ watt if it does work at the rate of $1$ joule per second.
Mathematically,it is expressed as:
$1\,W = 1\,J/1\,s$
This means that $1$ watt is the power consumed when $1$ joule of work is done in $1$ second.

(C) Power is defined as the rate of doing work or the rate of energy consumption.
The formula for power $(P)$ is given by:
$P = \frac{W}{t}$
Where:
$W$ (Work done or Energy consumed) $= 1000\, J$
$t$ (Time taken) $= 10\, s$
Substituting the values into the formula:
$P = \frac{1000\, J}{10\, s} = 100\, J/s$
Since $1\, J/s = 1\, W$ (Watt):
$P = 100\, W$
Therefore,the power of the lamp is $100\, W$.

(N/A) Average power is defined as the total work done by an object divided by the total time taken to perform that work.
Since a body may perform different amounts of work in different time intervals,calculating the average power provides a better representation of the rate of work done over a specific duration.
Mathematically,it is expressed as:
$\text{Average Power} = \frac{\text{Total work done}}{\text{Total time taken}}$

(N/A) $1$. Suma is swimming in a pond: Suma is doing work as she is able to move herself by applying force with the movement of her arms and legs in the water.
$2$. $A$ donkey is carrying a load on its back: The donkey is not doing any work (in the sense of physics) as the weight he is carrying (the direction of force) and the displacement are perpendicular to each other $(W = Fs \cos 90^{\circ} = 0)$.
$3$. $A$ windmill is lifting water from a well: The windmill is lifting water from a well and doing work against gravity.
$4$. $A$ green plant is carrying out photosynthesis: No force and displacement are present here,so the work done is zero.
$5$. An engine is pulling a train: During the pulling of a train,the engine does work against the friction present between the railway track and the wheels.
$6$. Food grains are getting dried in the sun: During the drying of grains,there is no force applied,nor is there any displacement. So,no work is done.
$7$. $A$ sailboat is moving due to wind energy: Work is done by the wind as it moves the sailboat in the direction of the force (force of blowing air).

(A) The work done by a force is given by the formula $W = F \cdot s \cdot \cos(\theta)$, where $F$ is the force, $s$ is the displacement, and $\theta$ is the angle between the force and displacement vectors.
In this case, the force of gravity acts vertically downwards $(F = mg)$.
The displacement $s$ is the straight-line distance between the initial and final points.
Since the initial and final points lie on the same horizontal line, the vertical displacement is $0$.
Because the gravitational force acts only in the vertical direction, the work done by gravity is $W = F \times (\text{vertical displacement}) = mg \times 0 = 0 \ J$.

(N/A) The process involves the following energy transformations:
$1$. Inside the battery,chemical energy is converted into electrical energy.
$2$. As the current flows through the bulb,this electrical energy is converted into light energy (which illuminates the bulb) and heat energy (which warms the bulb).

(D) 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 of the object.
Work done $(W)$ $= \Delta KE = KE_{final} - KE_{initial}$
$W = \frac{1}{2} m v^2 - \frac{1}{2} m u^2$
$W = \frac{1}{2} m (v^2 - u^2)$
Substituting the values:
$W = \frac{1}{2} \times 20 \times (2^2 - 5^2)$
$W = 10 \times (4 - 25)$
$W = 10 \times (-21)$
$W = -210 \, J$
Since the question asks for the magnitude of work done by the force (which is a retarding force), the absolute value is $210 \, J$.

(N/A) The work done by the gravitational force acting on the body is zero.
This is because the gravitational force acts vertically downward,while the displacement of the object is horizontal.
Since the angle $\theta$ between the force and the displacement is $90^\circ$,the work done $W$ is calculated as:
$W = F \cdot s \cdot \cos(90^\circ)$
Since $\cos(90^\circ) = 0$,the work done $W = 0$.

(N/A) The potential energy of a freely falling object decreases as it falls,while its kinetic energy increases progressively due to the increase in its velocity.
According to the law of conservation of energy,the total mechanical energy of an object,which is the sum of its kinetic energy and potential energy,remains constant at all points during its fall.
Therefore,the decrease in potential energy is exactly compensated by an equal increase in kinetic energy,meaning the law of conservation of energy is not violated.

(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) In physics,work is defined as the product of force and displacement $(W = F \times s)$. Since the rock does not move,the displacement $(s)$ is $0$,meaning no mechanical work is done on the rock.
However,energy is still being expended by your body. When you push the rock,your muscles contract and perform internal work. This energy is primarily dissipated as heat into the environment and the contact surface between your hands and the rock. $A$ negligible amount may be stored as elastic potential energy due to the slight deformation of the rock's surface.

(A) We know that $1$ unit of electrical energy is equal to $1 \text{ kilowatt-hour (kWh)}$.
$1 \text{ kWh} = 1 \text{ kW} \times 1 \text{ h} = 1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ J}$.
Given that the household consumed $250$ units.
Therefore,total energy in joules $= 250 \times 3.6 \times 10^6 \text{ J}$.
$= 900 \times 10^6 \text{ J}$.
$= 9 \times 10^8 \text{ J}$.
Hence,the energy consumed is $9 \times 10^8 \text{ J}$.

(B) We know that,potential energy $(PE) = mgh$.
Given:
Mass $(m) = 40\, kg$
Acceleration due to gravity $(g) = 9.8\, m/s^2$
Height $(h) = 5\, m$
Potential energy $= 40 \times 9.8 \times 5 = 1960\, J$.
According to the law of conservation of energy,the total mechanical energy remains constant.
When the object is half-way down,it has fallen a distance of $2.5\, m$. At this point,half of the initial potential energy is converted into kinetic energy.
Therefore,kinetic energy $= 1/2 \times 1960\, J = 980\, J$.

(C) When a satellite moves around the Earth in a circular orbit,the gravitational force acts towards the center of the Earth (centripetal force).
The displacement of the satellite at any instant is along the tangent to the circular path.
Since the radius (direction of force) is always perpendicular to the tangent (direction of displacement),the angle $\theta$ between the force and displacement is $90^{\circ}$.
The formula for work done is $W = F \cdot s \cdot \cos(\theta)$.
Substituting $\theta = 90^{\circ}$,we get $W = F \cdot s \cdot \cos(90^{\circ}) = F \cdot s \cdot 0 = 0$.
Therefore,the work done by the force of gravity on a satellite moving around the Earth is zero.

(A) Yes,it is true. There may be displacement in the absence of force.
We know that,$F = ma$.
In the absence of force,$F = 0$,then $ma = 0$.
Since $m \neq 0$,it follows that $a = 0$.
If $a = 0$,the object is either at rest or in a state of uniform motion in a straight line.
In the case where the object is already moving in a straight line,it will continue to move with a constant velocity,which results in displacement over time.
Therefore,in the absence of force,there may be displacement of an object.

(N/A) In scientific terms,the person has done no work.
Work is defined as the product of the force applied and the displacement in the direction of the force $(W = F \times s)$.
Since the bundle of hay is held stationary over the head,the displacement $(s)$ is $0$.
Therefore,$W = F \times 0 = 0$.
The person feels tired because his muscular energy is being consumed and converted into thermal energy to maintain the posture,but this does not constitute mechanical work.

(B) We know that,$\text{Energy} = \text{Power} \times \text{time}$.
Given:
$\text{Power} = 1500 \, W$
$\text{Time} = 10 \, \text{hours} = 10 \times 60 \times 60 \, \text{seconds} = 36000 \, \text{seconds}$.
Therefore,
$\text{Energy} = 1500 \, W \times 36000 \, \text{s} = 54,000,000 \, J$.
In scientific notation,this is $5.4 \times 10^7 \, J$.

(N/A) When a pendulum bob is displaced to one side (point $B$ or $C$) and released,it oscillates about its mean position $(A)$.
At the extreme positions ($B$ and $C$),the bob is at its maximum height,so its potential energy $(P.E.)$ is maximum and kinetic energy $(K.E.)$ is zero. As it moves towards the mean position $(A)$,$P.E.$ decreases and $K.E.$ increases. At the mean position $(A)$,the bob is at its lowest point,so $P.E.$ is zero and $K.E.$ is maximum. Throughout the motion,the total mechanical energy $(P.E. + K.E.)$ remains constant,illustrating the law of conservation of energy.
The bob eventually comes to rest because of air resistance,which opposes its motion and does work against it.
Its energy is eventually dissipated into the surroundings as heat energy due to friction with air molecules.
No,this is not a violation of the law of conservation of energy. The energy is not lost; it is merely transformed into other forms (heat and sound) in the surroundings,keeping the total energy of the system and its surroundings constant.

(C) The object is in motion,so its initial kinetic energy is $K_i = 1/2 mv^2$.
When the object comes to rest,its final velocity becomes $0$,so its final kinetic energy is $K_f = 0$.
According to the work-energy theorem,the work done on an object is equal to the change in its kinetic energy.
Work done $W = K_f - K_i$.
$W = 0 - 1/2 mv^2 = -1/2 mv^2$.
The negative sign indicates that the work is done against the direction of motion to bring the object to rest. The magnitude of the work required is $1/2 mv^2$.

(A) Mass of the car $m = 1500 \, kg$.
Velocity of the car $v = 60 \, km/h$.
Convert velocity to $m/s$: $v = \frac{60 \times 1000}{3600} = \frac{50}{3} \, m/s$.
The kinetic energy $(KE)$ of the car is given by the formula $KE = \frac{1}{2} mv^2$.
$KE = \frac{1}{2} \times 1500 \times (\frac{50}{3})^2 \, J$.
$KE = 750 \times \frac{2500}{9} \, J$.
$KE = 750 \times 277.777... \, J = 208333.3 \, J$.
According to the work-energy theorem, the work required to stop the car is equal to the change in its kinetic energy.
Work done $= KE_{initial} - KE_{final} = 208333.3 \, J - 0 \, J = 208333.3 \, J$.
Thus, the work required to stop the car is $208333.3 \, J$.

(N/A) The work done by a force is given by the formula $W = F \cdot s \cdot \cos(\theta)$,where $\theta$ is the angle between the force and the displacement.
$1$. In the first case,the force $F$ is acting vertically upwards while the displacement is horizontal (west to east). The angle between them is $90^{\circ}$. Since $\cos(90^{\circ}) = 0$,the work done is zero.
$2$. In the second case,the force $F$ is acting in the same direction as the displacement (west to east). The angle between them is $0^{\circ}$. Since $\cos(0^{\circ}) = 1$,the work done is positive.
$3$. In the third case,the force $F$ is acting in the opposite direction to the displacement (east to west). The angle between them is $180^{\circ}$. Since $\cos(180^{\circ}) = -1$,the work done is negative.

(A) Yes,$I$ agree with Soni.
According to Newton's second law of motion,the acceleration $(a)$ of an object is directly proportional to the net force $(F_{net})$ acting on it,given by the formula $F_{net} = ma$,where $m$ is the mass of the object.
If several forces act on an object such that their vector sum (resultant force) is zero,then the net force $F_{net} = 0$.
Since $F_{net} = ma$,if $F_{net} = 0$ and $m \neq 0$,then $a = 0$.
Therefore,an object can have zero acceleration even when multiple forces are acting on it,provided they balance each other out.

(A) The power of four devices $= 4 \times 500\, W = 2000\, W$.
Time $= 10$ hours.
Energy consumed $= \text{Power} \times \text{Time}$.
Energy $= 2000\, W \times 10\, h = 20000\, Wh$.
Since $1\, kWh = 1000\, Wh$, we divide by $1000$.
Energy $= 20000 / 1000 = 20\, kWh$.

(B) When a freely falling body eventually stops on reaching the ground,its kinetic energy is transformed into other forms of energy.
$1$. Heat energy: The collision between the object and the ground generates heat,warming both surfaces.
$2$. Sound energy: The impact produces a sound.
$3$. Potential energy: Some energy may be used in the deformation or change of shape of the object upon impact.