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Chapter 11: Work And Energy
Building upon our understanding of motion, its causes (force), and gravitation, we now introduce concepts that are fundamental to understanding many natural phenomena: work, energy, and power. These concepts help us describe and quantify interactions and changes in the physical world.
All living beings require energy to perform life processes and other activities like moving, thinking, or engaging in physical labour. Machines also need energy to operate, often obtained from fuel sources like petrol or diesel. Energy is essential for both life and technology.
Work
In everyday language, 'work' often refers to any kind of effort, whether physical or mental, that is useful or leads to exhaustion. However, the scientific definition of work is much more specific and often differs from our common understanding.
Not Much ‘Work’ In Spite Of Working Hard!
Consider someone studying hard for exams, spending a lot of time and mental effort. In everyday terms, they are doing 'hard work'. However, in science, if this activity does not involve physical displacement caused by a force, very little or no 'work' is done in the scientific sense.
Similarly, if you push a huge rock with great effort, get tired, but the rock does not move, according to science, no work is done on the rock because there is no displacement. Standing with a heavy load on your head also involves no scientific work being done on the load, as its position relative to the ground doesn't change, even though you might be exerting force to support it and feeling tired.
Conversely, activities like climbing stairs or a tree, which involve physical effort and movement against gravity over a distance, are considered to involve a significant amount of scientific work.
The key distinction in science is that 'work' has a precise definition based on specific physical conditions.
Scientific Conception Of Work
From a scientific perspective, for work to be done, two essential conditions must be met:
- A force must act on an object.
- The object must undergo displacement (change in position).
If either of these conditions is not fulfilled (no force acts or no displacement occurs), then no work is done in the scientific sense, regardless of the effort expended or exhaustion felt.
Examples of Scientific Work:
- Pushing a pebble that moves.
- A girl pulling a trolley that moves.
- Lifting a book to a certain height (force is applied, and the book moves upwards).
- A bullock pulling a cart that moves.
Work Done By A Constant Force
When a constant force acts on an object, and the object is displaced in the direction of the force, the work done is defined quantitatively.
Let a constant force $F$ act on an object, and let the object be displaced by a distance $s$ in the direction of the force.
The work done ($W$) by the force is defined as the product of the magnitude of the force and the distance displaced in the direction of the force.
$$ \text{Work done} = \text{Force} \times \text{Displacement} $$ $$ W = F \times s $$Work is a scalar quantity; it has only magnitude and no direction.
Units of Work:
- The SI unit of force is Newton (N).
- The SI unit of displacement is metre (m).
So, the SI unit of work is Newton metre (N m). This unit is also called the joule (J), in honour of James Prescott Joule.
Definition of 1 Joule (J): 1 J is the amount of work done on an object when a force of 1 Newton (N) displaces it by 1 metre (m) along the line of action of the force.
If the force is zero ($F=0$) or the displacement is zero ($s=0$), the work done ($W = F \times s$) is zero. This aligns with the scientific conditions for work.
When the force and displacement are in the same direction, the work done by the force is considered positive. Examples: a baby pulling a toy car horizontally, gravity doing work on a falling object (displacement downwards, force downwards).
When the force acts in the direction opposite to the displacement, the work done by the force is considered negative. For example, if an object is moving to the right, and a retarding force (like friction or applied brake force) acts to the left, the displacement is to the right (positive), but the force is to the left (negative direction). The work done by this retarding force is $W = F \times (-s)$ or $(-F) \times s = -Fs$. Gravity does negative work on an object being lifted upwards (displacement upwards, force downwards).
Summary of Work Direction:
- Work done is positive if force and displacement are in the same direction.
- Work done is negative if force and displacement are in opposite directions.
- Work done is zero if force or displacement is zero, or if the force is perpendicular to the displacement.
Example 11.1. A force of 5 N is acting on an object. The object is displaced through 2 m in the direction of the force (Fig. 11.2). If the force acts on the object all through the displacement, then work done is 5 N × 2 m =10 N m or 10 J.
Answer:
Given: Force, $F = 5$ N. Displacement, $s = 2$ m. The force acts in the direction of displacement.
Work done, $W = F \times s = 5 \text{ N} \times 2 \text{ m} = 10 \text{ N m} = 10 \text{ J}$.
The work done is 10 J.
Question 1. A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force (Fig. 11.3). Let us take it that the force acts on the object through the displacement. What is the work done in this case?
Answer:
Given: Force, $F = 7$ N. Displacement, $s = 8$ m. The force acts in the direction of displacement.
Work done, $W = F \times s = 7 \text{ N} \times 8 \text{ m} = 56 \text{ N m} = 56 \text{ J}$.
The work done is 56 J.
Example 11.2. A porter lifts a luggage of 15 kg from the ground and puts it on his head 1.5 m above the ground. Calculate the work done by him on the luggage.
Answer:
Given: Mass of luggage, $m = 15$ kg. Displacement (height lifted), $s = 1.5$ m.
To lift the luggage, the porter applies a force equal to the weight of the luggage, acting upwards. Weight = $m \times g$. Using $g = 9.8$ m s⁻² (or approximately 10 m s⁻² for simplified calculations as in the text example, let's use 9.8 m s⁻² unless specified otherwise for specific problems).
Force applied by porter, $F = m \times g = 15 \text{ kg} \times 9.8 \text{ m s}^{-2} = 147 \text{ N}$.
The force is applied upwards, and the displacement is upwards. So, force and displacement are in the same direction.
Work done by the porter on the luggage, $W = F \times s = 147 \text{ N} \times 1.5 \text{ m} = 220.5 \text{ J}$.
The work done by the porter on the luggage is 220.5 J.
(Note: If we use $g=10$ m s⁻² as in the text's example solution, $F = 15 \times 10 = 150$ N, and $W = 150 \times 1.5 = 225$ J).
Energy
Life and all activities in the universe depend on energy. Energy is the capacity to do work. When an object has energy, it can exert a force and cause displacement in another object, thereby doing work on it.
When an object does work, it loses energy. The object on which work is done gains energy.
The amount of energy an object possesses is measured by its capacity to do work. Therefore, the unit of energy is the same as the unit of work, which is the joule (J).
1 Joule (J) is the amount of energy required to do 1 joule of work.
A larger unit of energy is the kilojoule (kJ), where 1 kJ = 1000 J.
Various phenomena in nature and technological processes involve the acquisition, transfer, and transformation of energy.
Forms Of Energy
Energy exists in many forms. Some common forms of energy include:
- Mechanical Energy: The energy possessed by an object due to its state of motion or position. It is the sum of Kinetic Energy and Potential Energy.
- Heat Energy: Energy associated with the random motion of atoms and molecules.
- Chemical Energy: Energy stored in the bonds of chemical compounds.
- Electrical Energy: Energy associated with the flow of electric charge.
- Light Energy: Energy transmitted through electromagnetic waves (like visible light).
- Nuclear Energy: Energy stored within the nucleus of an atom.
Kinetic Energy
Objects in motion possess energy that allows them to do work. This energy is called kinetic energy.
Observations illustrating Kinetic Energy:
- A fast-moving cricket ball hitting stationary wickets causes them to be thrown away.
- A raised hammer falling on a nail drives the nail into wood.
- A moving trolley hitting a block displaces it (doing work).
An object moving faster has more kinetic energy than an identical object moving slower. The kinetic energy of an object increases with its speed.
Definition: Kinetic energy is the energy possessed by an object due to its motion. It is equal to the amount of work done on the object to make it acquire that velocity from rest.
Expression for Kinetic Energy:
Consider an object of mass $m$ starting from rest ($u=0$). A force $F$ acts on it, displacing it by a distance $s$ and giving it a velocity $v$.
The work done is $W = F \times s$.
From Newton's Second Law, $F = ma$.
From the equations of motion for uniform acceleration, $v^2 - u^2 = 2as$. Since $u=0$, $v^2 = 2as$, so $s = \frac{v^2}{2a}$.
Substituting $F=ma$ and $s=\frac{v^2}{2a}$ into the work done formula $W=Fs$:
$W = (ma) \times \left(\frac{v^2}{2a}\right)$
$W = m \times \frac{v^2}{2} = \frac{1}{2}mv^2$.
Since the work done on the object from rest is equal to its kinetic energy ($E_k$), the kinetic energy of an object of mass $m$ moving with velocity $v$ is given by:
$$ E_k = \frac{1}{2}mv^2 $$If the object's velocity changes from an initial velocity $u$ to a final velocity $v$, the work done on it (or the change in kinetic energy) is $W = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 = \frac{1}{2}m(v^2 - u^2)$. This is the work-energy theorem for constant force.
Example 11.3. An object of mass 15 kg is moving with a uniform velocity of 4 m s⁻¹. What is the kinetic energy possessed by the object?
Answer:
Given: Mass, $m = 15$ kg. Velocity, $v = 4$ m s⁻¹.
Kinetic energy, $E_k = \frac{1}{2}mv^2$
$E_k = \frac{1}{2} \times 15 \text{ kg} \times (4 \text{ m s}^{-1})^2$
$E_k = \frac{1}{2} \times 15 \text{ kg} \times 16 \text{ m}^2\text{s}^{-2}$
$E_k = \frac{1}{2} \times 240 \text{ kg m}^2\text{s}^{-2} = 120 \text{ J}$.
The kinetic energy of the object is 120 J.
Example 11.4. What is the work to be done to increase the velocity of a car from 30 km h⁻¹ to 60 km h⁻¹ if the mass of the car is 1500 kg?
Answer:
Given: Mass, $m = 1500$ kg. Initial velocity, $u = 30 \text{ km h}^{-1}$. Final velocity, $v = 60 \text{ km h}^{-1}$.
Convert velocities to SI units (m s⁻¹):
- $u = 30 \text{ km h}^{-1} = 30 \times \frac{1000}{3600} \text{ m s}^{-1} = \frac{300}{36} \text{ m s}^{-1} = \frac{25}{3} \text{ m s}^{-1}$.
- $v = 60 \text{ km h}^{-1} = 60 \times \frac{1000}{3600} \text{ m s}^{-1} = \frac{600}{36} \text{ m s}^{-1} = \frac{50}{3} \text{ m s}^{-1}$.
The work done to increase the velocity is equal to the change in kinetic energy ($\Delta E_k$).
Initial kinetic energy, $E_{ki} = \frac{1}{2}mu^2 = \frac{1}{2} \times 1500 \text{ kg} \times \left(\frac{25}{3} \text{ m s}^{-1}\right)^2 = 750 \times \frac{625}{9} \text{ J} = \frac{468750}{9} \text{ J} = \frac{156250}{3} \text{ J}$.
Final kinetic energy, $E_{kf} = \frac{1}{2}mv^2 = \frac{1}{2} \times 1500 \text{ kg} \times \left(\frac{50}{3} \text{ m s}^{-1}\right)^2 = 750 \times \frac{2500}{9} \text{ J} = \frac{1875000}{9} \text{ J} = \frac{625000}{3} \text{ J}$.
Work done, $W = E_{kf} - E_{ki} = \frac{625000}{3} \text{ J} - \frac{156250}{3} \text{ J} = \frac{625000 - 156250}{3} \text{ J} = \frac{468750}{3} \text{ J} = 156250 \text{ J}$.
The work to be done is 156250 J.
Potential Energy
Energy can also be possessed by an object due to its position or configuration (shape). This stored energy, which has the potential to do work, is called potential energy.
Observations illustrating Potential Energy:
- A stretched rubber band has energy stored in it (potential energy) because work was done to stretch it. When released, it can do work (e.g., spring back).
- A stretched or compressed slinky has potential energy.
- Winding the key of a toy car stores potential energy in the spring inside. This stored energy is later used to move the car.
- An object lifted to a height gains energy. When released, this energy can be used to do work (e.g., the object falls and can displace something). The higher the object is lifted, the more energy it gains.
Potential energy is the energy transferred to an object and stored within it, not causing an immediate change in its velocity or speed, but giving it the capability to do work later due to its altered position or configuration.
Example: A stretched bow has potential energy due to its change in shape. When the arrow is released, this potential energy is converted into the kinetic energy of the arrow.
Potential Energy Of An Object At A Height
When an object is raised to a certain height above the ground, work is done on it against the force of gravity. This work done is stored in the object as gravitational potential energy.
Definition: The gravitational potential energy of an object at a point above the ground is the work done in raising it from the ground to that point against gravity.
Expression for Gravitational Potential Energy:
Consider an object of mass $m$ lifted vertically upwards by a height $h$ from the ground. The minimum force required to lift the object against gravity is equal to its weight, $F = mg$.
The work done ($W$) in lifting the object to height $h$ is $W = \text{Force} \times \text{Displacement}$ (since force and displacement are in the same upward direction).
$W = (mg) \times h = mgh$.
This work done is stored as the gravitational potential energy ($E_p$) of the object at height $h$.
$$ E_p = mgh $$The reference level for potential energy (where $h=0$ and $E_p=0$) can be chosen arbitrarily (e.g., ground level, tabletop). The potential energy at a point is measured relative to this chosen zero level.
Importantly, the work done by gravity (and thus the change in gravitational potential energy) depends only on the change in vertical height between the initial and final positions, not on the actual path taken to move the object.
Example 11.5. Find the energy possessed by an object of mass 10 kg when it is at a height of 6 m above the ground. Given, g = 9.8 m s⁻².
Answer:
Given: Mass, $m = 10$ kg. Height, $h = 6$ m. Acceleration due to gravity, $g = 9.8$ m s⁻².
Potential energy, $E_p = mgh$
$E_p = 10 \text{ kg} \times 9.8 \text{ m s}^{-2} \times 6 \text{ m}$
$E_p = 98 \text{ N} \times 6 \text{ m} = 588 \text{ J}$.
The potential energy of the object at a height of 6 m is 588 J.
Example 11.6. An object of mass 12 kg is at a certain height above the ground. If the potential energy of the object is 480 J, find the height at which the object is with respect to the ground. Given, g = 10 m s⁻².
Answer:
Given: Mass, $m = 12$ kg. Potential energy, $E_p = 480$ J. Acceleration due to gravity, $g = 10$ m s⁻².
Using the formula $E_p = mgh$, we can find the height $h$:
$480 \text{ J} = 12 \text{ kg} \times 10 \text{ m s}^{-2} \times h$
$480 \text{ J} = 120 \text{ N} \times h$
$h = \frac{480 \text{ J}}{120 \text{ N}} = \frac{480 \text{ N m}}{120 \text{ N}} = 4 \text{ m}$.
The object is at a height of 4 m above the ground.
Are Various Energy Forms Interconvertible?
Energy can be transformed from one form to another. Nature provides countless examples of energy conversions (e.g., plants convert light energy to chemical energy through photosynthesis, wind converts kinetic energy of air to mechanical energy in windmills, heat energy from the Sun drives the water cycle).
Many human activities and devices also involve energy conversions (e.g., electric bulb converts electrical energy to light and heat, a car engine converts chemical energy of fuel to heat and mechanical energy).
Law Of Conservation Of Energy
During energy transformations, what happens to the total amount of energy? The Law of Conservation of Energy states that energy cannot be created or destroyed; it can only be converted from one form to another. The total energy of an isolated system remains constant over time, regardless of the transformations occurring within it.
Statement: Energy can only be converted from one form to another; it can neither be created nor destroyed. The total energy before and after the transformation remains the same.
Consider an object of mass $m$ falling freely from a height $h$.
- At the initial height $h$, its velocity is $u=0$. Potential energy $E_p = mgh$. Kinetic energy $E_k = \frac{1}{2}mu^2 = 0$. Total mechanical energy = $E_p + E_k = mgh$.
- As it falls, its height decreases, so $E_p$ decreases. Its velocity increases, so $E_k$ increases.
- At any point during the fall, if its velocity is $v$ and height is $h'$, its potential energy is $mgh'$ and kinetic energy is $\frac{1}{2}mv^2$. The total mechanical energy at this point is $mgh' + \frac{1}{2}mv^2$.
- Just before hitting the ground, height $h$ is almost 0, potential energy is almost 0. Its velocity $v$ is maximum, so kinetic energy $\frac{1}{2}mv^2$ is maximum.
According to the Law of Conservation of Energy, the total mechanical energy remains constant throughout the free fall (assuming no air resistance or other dissipative forces):
$$ E_p + E_k = \text{constant} $$ $$ mgh + \frac{1}{2}mv^2 = \text{constant} $$The decrease in potential energy is exactly equal to the increase in kinetic energy. Energy is continuously transformed from potential energy to kinetic energy during the fall.
The sum of kinetic energy and potential energy is called the total mechanical energy.
In real-world scenarios, some energy is lost to dissipative forces like air resistance and friction, often converting mechanical energy into heat or sound. However, the total energy of the *entire system* (including the air, ground, etc.) remains conserved.
Rate Of Doing Work
We can do work or transfer energy, but we can do it at different rates. The rate at which work is done or energy is transferred is called power.
A person or machine with more power can do the same amount of work in less time, or do more work in the same amount of time, compared to a less powerful one.
Definition: Power is the rate of doing work or the rate of transfer of energy.
If an agent does a work $W$ in time $t$, its power $P$ is:
$$ \text{Power} = \frac{\text{Work}}{\text{Time}} \quad \text{or} \quad P = \frac{W}{t} $$Similarly, if energy $E$ is consumed or transferred in time $t$, Power $P = E/t$.
Units of Power:
- The SI unit of work/energy is Joule (J).
- The SI unit of time is second (s).
So, the SI unit of power is Joule per second (J/s). This unit is called the watt (W), in honour of James Watt.
Definition of 1 Watt (W): 1 Watt is the power of an agent that does work at the rate of 1 joule per second. 1 W = 1 J/s.
Larger units of power are often used, such as the kilowatt (kW), where 1 kW = 1000 W.
The power of an agent can vary over time. For such cases, the concept of average power is used, which is the total work done (or energy consumed) divided by the total time taken.
Example 11.7. Two girls, each of weight 400 N climb up a rope through a height of 8 m. We name one of the girls A and the other B. Girl A takes 20 s while B takes 50 s to accomplish this task. What is the power expended by each girl?
Answer:
The work done by each girl is against the force of gravity, which is equal to their weight. The displacement is the height climbed.
Weight of each girl (Force), $F = 400$ N. Height (Displacement), $s = 8$ m.
Work done by each girl, $W = F \times s = 400 \text{ N} \times 8 \text{ m} = 3200 \text{ J}$.
(i) Power expended by girl A:
Time taken by A, $t_A = 20$ s.
Power of A, $P_A = \frac{\text{Work done}}{\text{Time taken}} = \frac{3200 \text{ J}}{20 \text{ s}} = 160 \text{ J/s} = 160 \text{ W}$.
(ii) Power expended by girl B:
Time taken by B, $t_B = 50$ s.
Power of B, $P_B = \frac{\text{Work done}}{\text{Time taken}} = \frac{3200 \text{ J}}{50 \text{ s}} = 64 \text{ J/s} = 64 \text{ W}$.
Girl A expends 160 W of power, and girl B expends 64 W of power. Girl A is more powerful as she did the same work in less time.
Example 11.8. A boy of mass 50 kg runs up a staircase of 45 steps in 9 s. If the height of each step is 15 cm, find his power. Take g = 10 m s⁻².
Answer:
Given: Mass of the boy, $m = 50$ kg. Number of steps = 45. Height of each step = 15 cm. Time taken = 9 s. $g = 10$ m s⁻².
Total height climbed = Number of steps $\times$ Height of each step.
Total height, $h = 45 \times 15 \text{ cm} = 675 \text{ cm}$.
Convert total height to metres: $h = 675 \text{ cm} = 675/100 \text{ m} = 6.75 \text{ m}$.
The work done by the boy against gravity is equal to the increase in his potential energy.
Work done, $W = mgh = 50 \text{ kg} \times 10 \text{ m s}^{-2} \times 6.75 \text{ m}$
$W = 500 \text{ N} \times 6.75 \text{ m} = 3375 \text{ J}$.
Power, $P = \frac{\text{Work done}}{\text{Time taken}} = \frac{3375 \text{ J}}{9 \text{ s}} = 375 \text{ J/s} = 375 \text{ W}$.
The boy's power is 375 W.
Commercial Unit Of Energy
The joule (J) is a small unit of energy for everyday use, especially when dealing with large amounts of energy consumption in households and industries. A larger unit, the kilowatt-hour (kW h), is commonly used.
Definition of 1 Kilowatt-hour (kW h): 1 kilowatt-hour is the energy consumed by a device operating at a power of 1 kilowatt (kW) for a duration of 1 hour (h).
We can convert kilowatt-hour to joules:
1 kW h = 1 kW $\times$ 1 h
1 kW = 1000 W = 1000 J/s
1 h = 60 minutes $\times$ 60 seconds/minute = 3600 s
1 kW h = $(1000 \text{ J/s}) \times (3600 \text{ s})$
1 kW h = $3,600,000 \text{ J}$
1 kW h = 3.6 $\times 10^6$ J
Electrical energy consumed in households, industries, and commercial establishments is typically measured in 'units', where 1 unit = 1 kilowatt-hour.
Example 11.9. An electric bulb of 60 W is used for 6 h per day. Calculate the ‘units’ of energy consumed in one day by the bulb.
Answer:
Given: Power of the bulb, $P = 60$ W. Time used per day, $t = 6$ h.
Energy consumed = Power $\times$ Time
To calculate energy in kW h, convert power from W to kW:
$P = 60 \text{ W} = 60/1000 \text{ kW} = 0.06 \text{ kW}$.
Energy consumed = $0.06 \text{ kW} \times 6 \text{ h} = 0.36 \text{ kW h}$.
Since 1 unit = 1 kW h, the energy consumed is 0.36 units.
Intext Questions
Page No. 148
Question 1. A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force (Fig. 11.3). Let us take it that the force acts on the object through the displacement. What is the work done in this case?
Answer:
Page No. 149
Question 1. When do we say that work is done?
Answer:
Question 2. Write an expression for the work done when a force is acting on an object in the direction of its displacement.
Answer:
Question 3. Define 1 J of work.
Answer:
Question 4. A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in ploughing the length of the field?
Answer:
Page No. 152
Question 1. What is the kinetic energy of an object?
Answer:
Question 2. Write an expression for the kinetic energy of an object.
Answer:
Question 3. The kinetic energy of an object of mass, m moving with a velocity of 5 $m s^{-1}$ is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?
Answer:
Page No. 156
Question 1. What is power?
Answer:
Question 2. Define 1 watt of power.
Answer:
Question 3. A lamp consumes 1000 J of electrical energy in 10 s. What is its power?
Answer:
Question 4. Define average power.
Answer:
Exercises
Question 1. Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term ‘work’.
- Suma is swimming in a pond.
- A donkey is carrying a load on its back.
- A wind-mill is lifting water from a well.
- A green plant is carrying out photosynthesis.
- An engine is pulling a train.
- Food grains are getting dried in the sun.
- A sailboat is moving due to wind energy.
Answer:
Question 2. An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?
Answer:
Question 3. A battery lights a bulb. Describe the energy changes involved in the process.
Answer:
Question 4. Certain force acting on a 20 kg mass changes its velocity from 5 m $s^{-1}$ to 2 m $s^{-1}$. Calculate the work done by the force.
Answer:
Question 5. A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer.
Answer:
Question 6. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
Answer:
Question 7. What are the various energy transformations that occur when you are riding a bicycle?
Answer:
Question 8. Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?
Answer:
Question 9. A certain household has consumed 250 units of energy during a month. How much energy is this in joules?
Answer:
Question 10. An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down.
Answer:
Question 11. What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.
Answer:
Question 12. Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.
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Question 13. A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.
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Question 14. An electric heater is rated 1500 W. How much energy does it use in 10 hours?
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Question 15. Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?
Answer:
Question 16. An object of mass, m is moving with a constant velocity, v. How much work should be done on the object in order to bring the object to rest?
Answer:
Question 17. Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?
Answer:
Question 18. In each of the following a force, F is acting on an object of mass, m. The direction of displacement is from west to east shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.
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Question 19. Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?
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Question 20. Find the energy in kW h consumed in 10 hours by four devices of power 500 W each.
Answer:
Question 21. A freely falling object eventually stops on reaching the ground. What happenes to its kinetic energy?
Answer: