Mix Example - WORK AND ENERGY Questions in English

(B) watt is the $SI$ unit of power.
It is defined as the rate of doing work or the rate of energy consumption.
One watt is equal to $1 \, J \, s^{-1}$,which means one joule of work done per second.

(A) In physics,work is defined as the product of the force applied and the displacement caused in the direction of the force. The formula for work is $W = F \times s \times \cos(\theta)$,where $W$ is work,$F$ is force,$s$ is displacement,and $\theta$ is the angle between force and displacement.
Since the rock did not move,the displacement $(s)$ is $0 \, m$.
Therefore,$W = F \times 0 = 0 \, J$.
Thus,the work done by Seema is $0 \, J$.

(B) The formula for kinetic energy $(KE)$ is given by $KE = \frac{1}{2}mv^2$.
Given: Mass $(m)$ = $1\, kg$,Kinetic Energy $(KE)$ = $1\, J$.
Substituting the values into the formula: $1 = \frac{1}{2} \times 1 \times v^2$.
$v^2 = 2$.
$v = \sqrt{2} \approx 1.414\, m/s$.
Therefore,the body will have a kinetic energy of $1\, J$ at a speed of approximately $1.414\, m/s$.

(B) rolling stone is in motion. The energy possessed by an object due to its motion is known as kinetic energy. Therefore,a rolling stone possesses kinetic energy.

(B) running athlete is in motion. The energy possessed by an object due to its motion is known as kinetic energy. Therefore,a running athlete possesses kinetic energy.

(C) The formula for work done is $W = F \cdot s \cdot \cos(\theta)$,where $W$ is work,$F$ is force,$s$ is displacement,and $\theta$ is the angle between force and displacement.
If the displacement $(s)$ of the object is $0$,then $W = F \cdot 0 \cdot \cos(\theta) = 0$.
Therefore,the work done on the object is $0$.

(A) Work done is defined as the product of force and displacement in the direction of force,given by the formula $W = F \times S \times \cos(\theta)$.
In this scenario,the weight lifter is holding the weight stationary on his shoulders.
Since there is no change in the position of the weight,the displacement $(S)$ is $0$.
Therefore,$W = F \times 0 = 0$.
Thus,the work done by the weight lifter is $0 \,J$.

(B) The kinetic energy $(KE)$ of an object is given by the formula $KE = \frac{1}{2}mv^2$,where $m$ is the mass and $v$ is the velocity.
Since both the car and the truck are moving with the same velocity $(v)$,the kinetic energy is directly proportional to the mass $(m)$ of the object $(KE \propto m)$.
Given that the mass of the truck is greater than the mass of the car,the truck will have more kinetic energy.

(N/A) The formula for calculating work done is $W = F \times S \times \cos(\theta)$,where $W$ is work,$F$ is the magnitude of the force applied,$S$ is the displacement,and $\theta$ is the angle between the force and displacement vectors. In the simplest case where force is applied in the direction of displacement,the formula is $W = F \times S$.
The $SI$ unit of work is the joule $(J)$,where $1 \ J = 1 \ N \cdot m$.

(N/A) The kinetic energy $(KE)$ of a body is given by the formula $KE = \frac{1}{2}mv^2$.
As the body is thrown vertically upwards,its velocity $(v)$ decreases due to the gravitational force acting downwards.
At the highest point,the velocity of the body becomes zero $(v = 0)$.
Consequently,the kinetic energy at the highest point becomes $KE = \frac{1}{2}m(0)^2 = 0$.
Therefore,the change in kinetic energy is equal to the initial kinetic energy $(KE_{initial} - 0 = KE_{initial})$,which is entirely converted into potential energy at that point.

(20 W) The power $P$ is defined as the product of force $F$ and velocity $v$ when the force is applied in the direction of motion.
Given:
Force $F = 10\, N$
Velocity $v = 2\, m s^{-1}$
Formula:
$P = F \times v$
Calculation:
$P = 10\, N \times 2\, m s^{-1} = 20\, W$
Therefore,the power of the body is $20\, W$.

(N/A) The commercial unit of electrical energy is kilowatt-hour $(kWh)$.
The $SI$ unit of energy is Joule $(J)$.
Since $1 \text{ kW} = 1000 \text{ W}$ and $1 \text{ hour} = 3600 \text{ seconds}$,
$1 \text{ kWh} = 1000 \text{ J/s} \times 3600 \text{ s} = 3.6 \times 10^{6} \text{ J}$.
Therefore,the relation is $1 \text{ kWh} = 3.6 \times 10^{6} \text{ J}$.

(N/A) In a dry cell,the chemical energy stored in the electrolytes and electrodes is converted into electrical energy through redox reactions.

(C) The work done $(W)$ is defined 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 the displacement.
In the case of the earth moving around the sun,the gravitational force acts towards the center of the sun (centripetal force),while the displacement of the earth is along the tangent to its circular orbit.
Since the force and the displacement are perpendicular to each other,the angle $\theta = 90^\circ$.
Therefore,$W = F \cdot s \cdot \cos(90^\circ) = F \cdot s \cdot 0 = 0$.
Thus,the work done by the earth in moving around the sun is zero.

(B) As the ball moves vertically upwards,its kinetic energy is converted into potential energy due to the work done against gravity.
At the highest point,the velocity of the ball becomes zero,meaning the kinetic energy is zero.
According to the law of conservation of energy,the total mechanical energy remains constant.
Therefore,at the point where velocity is zero,the potential energy reaches its maximum value.

(A) Work is defined as the product of the force applied and the displacement in the direction of the force.
Mathematically,$W = F \times s \times \cos(\theta)$,where $W$ is work,$F$ is force,$s$ is displacement,and $\theta$ is the angle between force and displacement.
In this scenario,the person is standing still,which means the displacement $s = 0 \, m$.
Since the displacement is zero,the work done $W = F \times 0 = 0 \, J$.
Therefore,no work is done by the person.

(A) The expression for kinetic energy is $KE = \frac{1}{2} mv^{2}$.
Since the kinetic energy is directly proportional to the mass $(KE \propto m)$ when the speed $(v)$ is constant,the object with the greater mass will have more kinetic energy.
Given that the mass of the horse $(210\, kg)$ is greater than the mass of the dog $(25\, kg)$,the horse possesses more kinetic energy.

(N/A) Potential energy is the energy possessed by an object by virtue of its position or configuration.
The expression for gravitational potential energy is $PE = mgh$,where $m$ is the mass of the object,$g$ is the acceleration due to gravity,and $h$ is the height above the reference level.
The $SI$ unit of potential energy is the joule $(J)$.

(N/A) Work is defined as the product of the magnitude of the force applied on an object and the displacement of the object in the direction of the force.
The expression for work done $(W)$ is given by:
$W = F \times s$
Where $F$ is the force applied and $s$ is the displacement.
The $SI$ unit of work done is the joule $(J)$,where $1 \ J = 1 \ N \cdot m$.

(A) The work done in lifting an object against gravity is given by the formula: $W = mgh$.
Here,mass $m = 40 \text{ kg}$,height $h = 0.5 \text{ m}$,and acceleration due to gravity $g = 9.8 \text{ m s}^{-2}$.
Substituting these values into the formula:
$W = 40 \times 9.8 \times 0.5$
$W = 40 \times 4.9$
$W = 196 \text{ Joules}$.
Therefore,the total work done is $196 \text{ J}$.

(N/A) The work done $(W)$ by a force is defined 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.
Therefore,the work done depends on the following three factors:
$(i)$ The magnitude of the force $(F)$ applied to the object.
$(ii)$ The magnitude of the displacement $(s)$ produced in the object.
$(iii)$ The angle $(\theta)$ between the direction of the applied force and the direction of the displacement.

(A) Kinetic energy $(K)$ is defined by the relation $K = \frac{1}{2} m v^{2}$.
To relate this to momentum $(p = mv)$,we multiply and divide the right-hand side by $m$:
$K = \frac{1}{2} \frac{m^2 v^2}{m}$
Since $p = mv$,we can substitute $p^2$ for $m^2 v^2$:
$K = \frac{p^2}{2m}$
Rearranging this to solve for momentum,we get $p = \sqrt{2mK}$.

(C) In order to balance the load on his head,the coolie applies a force on it in the upward direction,equal to its weight.
His displacement is along the horizontal direction.
Thus,the angle between the force $F$ and the displacement $S$ is $90^{\circ}$.
Therefore,the work done $W = F S \cos \theta = F S \cos 90^{\circ} = 0$.

(N/A) When we wind our watch,we perform work on the spring inside the watch. As a result,this energy is stored in the spring in the form of elastic potential energy. This stored elastic potential energy is gradually released to power the mechanism of the watch throughout the day.

(C) Work done is defined by the formula $W = F S \cos \theta$,where $F$ is the force,$S$ is the displacement,and $\theta$ is the angle between the force and displacement vectors.
When a body completes one full revolution in a circular path,the final position coincides with the initial position,making the net displacement $S = 0$.
Since the centripetal force acts perpendicular to the direction of motion (tangential velocity) at every point,the work done by the centripetal force is zero.
Therefore,the total work done by the force when a body moves in a circular path is $W = F \times 0 = 0$.

(N/A) The work done $(W)$ in lifting a box of mass $(m)$ through a vertical height $(h)$ against the gravitational force $(F = mg)$ is given by the formula $W = F \times h = mgh$.
In this expression,the work done depends only on the mass of the object $(m)$,the acceleration due to gravity $(g)$,and the vertical displacement $(h)$.
Since the time taken or the speed at which the box is lifted does not appear in this formula,the work done is independent of the rate at which the box is lifted.

(N/A) Yes,it is possible when the force acting on the body is perpendicular to the direction of its motion.
Work done $(W)$ is defined as $W = Fs \cos \theta$,where $\theta$ is the angle between the force $(F)$ and the displacement $(s)$.
If $\theta = 90^{\circ}$,then $\cos 90^{\circ} = 0$,resulting in $W = 0$.
An example of this is the moon revolving around the earth in a circular orbit. The earth exerts a gravitational centripetal force on the moon,which acts towards the center of the orbit,while the moon's velocity is tangential to the orbit. Since the force is always perpendicular to the displacement,the work done by the earth on the moon is zero,even though the moon is in accelerated motion due to the change in the direction of its velocity.

(0) When a body moves along a circular path,the centripetal force acting on it is always directed towards the center of the circle (along the radius).
The displacement of the body at any instant is along the tangent to the circular path at that point.
Since the radius and the tangent at any point on a circle are perpendicular to each other,the angle $\theta$ between the force $(F)$ and the displacement $(S)$ is $90^{\circ}$.
The formula for work done is $W = F S \cos \theta$.
Substituting $\theta = 90^{\circ}$,we get $W = F S \cos 90^{\circ}$.
Since $\cos 90^{\circ} = 0$,the work done $W = 0$.

(NO) Work is defined as the product of force and displacement in the direction of the force $(W = F \cdot d \cdot \cos \theta)$.
In this scenario,the man is applying force to the oars to move the boat relative to the water,so he is doing work against the stream.
However,with respect to the shore,the boat's displacement $(d)$ is $0$ because the man is at rest relative to the shore.
Since $W = F \cdot 0 = 0$,the man does no work with respect to the shore.

(B) If the roads go straight up,the angle of slope (i.e.,$\theta$) would be very large.
According to the laws of physics,the force required to move a vehicle up an incline is proportional to $\sin(\theta)$.
By winding the road gradually,the angle of inclination $\theta$ is significantly reduced.
This reduction in $\theta$ decreases the force required to climb the mountain,making it easier for the engine to pull the vehicle up without overheating or slipping.

(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.

(B) The relationship between momentum $(p)$,mass $(m)$,and kinetic energy $(K)$ is given by the formula $p = \sqrt{2mK}$.
Since the kinetic energy $(K)$ is the same for both bodies,the momentum is directly proportional to the square root of the mass $(p \propto \sqrt{m})$.
Therefore,the body with the larger mass will have a greater momentum.
Thus,the heavy body will have more momentum than the light body.

(A) The relationship between kinetic energy $(K)$,momentum $(p)$,and mass $(m)$ is given by the formula $K = \frac{p^2}{2m}$.
Since the momentum $(p)$ is the same for both bodies,the kinetic energy is inversely proportional to the mass $(K \propto \frac{1}{m})$.
Therefore,the body with the smaller mass (the lighter body) will have a larger denominator,resulting in a greater value for kinetic energy.
Thus,the lighter body has more kinetic energy than the heavier body.

(N/A) Let the initial velocity of the body be $u$ and the final velocity be $v$ after traveling a distance $S$ under the influence of a constant force $F$.
According to the third equation of motion, $v^{2} - u^{2} = 2aS$, where $a$ is the acceleration.
From this, the distance $S$ can be expressed as $S = \frac{v^{2} - u^{2}}{2a}$.
According to Newton's second law of motion, the force applied is $F = ma$.
The work done $W$ by the force is defined as $W = F \times S$.
Substituting the values of $F$ and $S$ into the work equation:
$W = (ma) \times \left( \frac{v^{2} - u^{2}}{2a} \right)$.
Simplifying the expression by canceling $a$:
$W = \frac{1}{2}m(v^{2} - u^{2}) = \frac{1}{2}mv^{2} - \frac{1}{2}mu^{2}$.
Since kinetic energy $K = \frac{1}{2}mv^{2}$, the expression becomes $W = K_{final} - K_{initial} = \Delta K$.
Thus, the work done by the force is equal to the change in the kinetic energy of the body.

(N/A) Case $I$: The force $F$ is acting in the vertically upward direction,while the displacement of the object is in the horizontal direction (from west to east). Since the force and displacement are perpendicular to each other (angle $\theta = 90^\circ$),the work done is $W = F \cdot s \cdot \cos(90^\circ) = 0$.
Case $II$: Here,the force $F$ is applied in the same direction as the displacement. Since the angle between force and displacement is $0^\circ$,the work done is positive $(+ve)$,as $W = F \cdot s \cdot \cos(0^\circ) = F \cdot s$.

(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) Work done is calculated by the formula $W = F \times s$. Given $F = 1\, N$ and $s = 1\, m$,the work done is $1\, N \times 1\, m = 1\, J$.
$(b)$ Yes,it is possible for the work done to be zero even if a force is acting on a body. This occurs when the force is applied at an angle of $90^{\circ}$ to the direction of displacement. Since $W = Fs \cos(\theta)$,if $\theta = 90^{\circ}$,then $\cos(90^{\circ}) = 0$,making the work done zero.
Example: When a satellite moves in a circular orbit around the Earth,the gravitational force acts towards the center of the Earth (perpendicular to the motion),while the displacement is along the tangent to the orbit. Thus,the work done by gravity on the satellite is zero.

(N/A) Potential energy is the energy possessed by a body by virtue of its position or configuration. It is a scalar quantity because it has only magnitude and no direction.
$(b)$ $A$ stretched string of a bow is an example of a body having potential energy due to its change in configuration.

(N/A) Work done is given by the formula $W = F S \cos \theta$,where $\theta$ is the angle between the force vector $F$ and the displacement vector $S$.
Work done is said to be negative when the angle $\theta$ between the force and the displacement is obtuse,i.e.,$90^{\circ} < \theta \leq 180^{\circ}$,because $\cos \theta$ is negative in this range.
$A$ common situation is a body falling vertically downwards through the air. In this case,the force of gravity acts in the direction of motion (downwards),so the work done by gravity is positive. Simultaneously,the air resistance (frictional force) acts in the opposite direction of motion (upwards),so the work done by air resistance is negative.

(N/A) For work to be done on an object,the following two conditions must be satisfied:
$(i)$ $A$ force must be applied to the object.
$(ii)$ The object must undergo a displacement in the direction of the force (or have a component of displacement in the direction of the force,i.e.,the angle $\theta$ between force and displacement must be such that $\cos \theta \neq 0$).

(A) The mass of the box is $m = 10 \, kg$.
The gravitational force (weight) acting on the box is $F = mg = 10 \, kg \times 10 \, m s^{-2} = 100 \, N$,which acts vertically downwards.
The displacement $S = 2 \, m$ is along the horizontal surface.
The angle $\theta$ between the gravitational force (downward) and the displacement (horizontal) is $90^{\circ}$.
The work done $W$ is given by the formula $W = FS \cos \theta$.
Substituting the values: $W = 100 \times 2 \times \cos 90^{\circ}$.
Since $\cos 90^{\circ} = 0$,the work done $W = 100 \times 2 \times 0 = 0 \, J$.
Therefore,the work done by the gravitational force is $0 \, J$ because the force of gravity is perpendicular to the direction of displacement.

(N/A) In a loudspeaker,electrical energy is transformed into sound energy.
$(b)$ In a solar battery,light energy is transformed into electrical energy.

(II, III) In physics,work is said to be done when a force applied on an object causes a displacement in the direction of the force. The formula is $W = F \times s \times \cos(\theta)$.
$(i)$ No work is done because the displacement $(s)$ is $0$,even though a force is applied.
$(ii)$ Work is done because the bullock applies force to the cart,resulting in a displacement of $1 \, km$.
$(iii)$ Work is done because the girl applies force to the trolley,resulting in a displacement of $2 \, m$.
$(iv)$ No work is done because the person is stationary,meaning the displacement $(s)$ is $0$.

(A) Work done by a force is measured as the product of the magnitude of the force applied and the displacement in the direction of the force $(W = F \times S)$.
Given:
Mass of the luggage $(m)$ = $20 \, kg$
Displacement $(h)$ = $1.7 \, m$
Acceleration due to gravity $(g)$ = $10 \, m s^{-2}$
The force applied by the porter to lift the luggage is equal to the weight of the luggage,which is $F = m \times g$.
Therefore,the work done $(W)$ = $m \times g \times h$.
$W = 20 \, kg \times 10 \, m s^{-2} \times 1.7 \, m$.
$W = 340 \, J$.
Thus,the work done by the porter on the luggage is $340 \, J$.

(N/A) Work is defined as the product of the force applied to an object and the displacement produced in the direction of the force. Mathematically,$W = F \times s \times \cos(\theta)$,where $W$ is work,$F$ is force,$s$ is displacement,and $\theta$ is the angle between force and displacement.
Work is measured as the product of the magnitude of force and the displacement in the direction of the force.
Work done by a force is negative when the force acts in the direction opposite to the direction of displacement,i.e.,the angle between force and displacement is $180^{\circ}$ (since $\cos(180^{\circ}) = -1$).

(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) $1 \text{ kWh} = 1000 \text{ W} \times 1 \text{ h} = 1000 \text{ J/s} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ J}$.
$1$ watt is defined as the power of an agent that performs work at the rate of $1$ joule per second.

(N/A) Yes,a body at rest has no momentum,i.e.,$p = 0$. However,it can possess potential energy $(U)$ due to its position or configuration. Therefore,its total energy $E = K + U = U$ is not zero. For example,a stone lying on a roof or a wound-up spring of a clock.
$(b)$ No,a body cannot have momentum without having energy. Momentum $(p = mv)$ implies that the body is in motion $(v \neq 0)$. If a body is in motion,it must possess kinetic energy $(K = \frac{1}{2}mv^2)$. Since kinetic energy is a component of total energy,a body with momentum must possess energy.

(D) $(i)$ The energy possessed by an object due to its position or configuration is called potential energy. The $SI$ unit of potential energy is joule $(J)$.
$(ii)$ Given: mass of the body $(m) = 50 \, kg$,height $(h) = 10 \, m$,and acceleration due to gravity $(g) = 10 \, m s^{-2}$.
The formula for potential energy $(PE)$ is $PE = m \times g \times h$.
Substituting the values: $PE = 50 \, kg \times 10 \, m s^{-2} \times 10 \, m = 5000 \, J$.
Therefore,the potential energy of the body is $5000 \, J$.

(N/A) $(i)$ Power is defined as the rate of doing work or the rate of energy transfer. Its $SI$ unit is the watt $(W)$.
$(ii)$ Given: Mass $(m)$ = $50 \, kg$,Number of steps = $40$,Height of each step = $15 \, cm = 0.15 \, m$,Time $(t)$ = $8 \, s$,Acceleration due to gravity $(g)$ = $10 \, m \, s^{-2}$.
Total height $(h)$ = $40 \times 0.15 \, m = 6 \, m$.
Work done $(W)$ = $mgh = 50 \times 10 \times 6 = 3000 \, J$.
Power $(P)$ = $\frac{W}{t} = \frac{3000}{8} = 375 \, W$.