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Chapter 10 Work And Energy
Having explored the concepts of motion, forces, and gravitation, we now turn to three interconnected concepts: work, energy, and power. These ideas are fundamental to understanding how the world around us functions, from biological processes to the operation of machines and natural phenomena.
In our daily lives, both living beings and machines require energy to perform activities. Animals need energy to move, find food, and survive. Humans use energy for everything from basic life processes to strenuous physical and mental activities. Machines, like engines and vehicles, require fuel or electricity to operate. This necessity for energy drives many processes in nature and technology.
Work
The term 'work' has a specific meaning in science that differs from its everyday use. While studying hard or pushing against an immovable object might be considered hard work in common language, the scientific definition is more precise.
Work Done In Everyday Life vs. Scientific Conception
In everyday life, work often refers to any physical or mental effort or labour, especially if it's purposeful or leads to a result. Examples include studying for an exam, performing household chores, or having a conversation.
However, in science, the definition of work is much stricter. For **work to be done** on an object in the scientific sense, two conditions must be satisfied:
- A **force** must act on the object.
- The object must undergo **displacement** (change in position).
**Crucially, the displacement must have a component in the direction of the force.**
Let's look at activities often considered 'work' in everyday life and see if they qualify scientifically:
- Pushing a huge rock that does not move: You exert force and get tired, but there is **no displacement** of the rock. Scientifically, **no work is done** on the rock.
- Standing still with a heavy load on your head: You exert force to support the load and feel tired, but there is **no displacement** of the load (relative to the ground). Scientifically, **no work is done** on the load.
- Reading or thinking: These involve mental effort and consume energy, but there is no force causing displacement of an object being worked upon in the physical sense. Scientifically, **no work is done**.
Activities that *do* involve scientific work:
- Pushing a pebble that moves: You exert a force, and the pebble is displaced. Work is done.
- A girl pulling a trolley that moves: The girl exerts a force, and the trolley is displaced. Work is done.
- Lifting a book: You apply an upward force, and the book is displaced upwards. Work is done.
- A bullock pulling a cart that moves: The bullock exerts a force, and the cart is displaced. Work is done.
- Climbing stairs or a tree: Your muscles exert an upward force, and your body is displaced upwards. Work is done against gravity.
Thus, the scientific definition of work is very specific and requires both a force and a displacement in the direction of the force.
Work Done By A Constant Force
When a **constant force (F)** acts on an object and causes a **displacement (s)** in the **same direction as the force**, the work done (W) is defined as the product of the force and the displacement.
$$ \text{Work done} (W) = \text{Force} (F) \times \text{Displacement} (s) $$
$$ W = Fs $$
Work has only magnitude; it is a scalar quantity. Its direction is not specified.
The SI unit of work is the **joule (J)**. One joule of work is done when a force of **1 Newton (N)** displaces an object by **1 metre (m)** along the direction of the force.
$$ 1 \text{ Joule (J)} = 1 \text{ Newton (N)} \times 1 \text{ metre (m)} $$
$$ 1 \text{ J} = 1 \text{ N m} $$
Example 10.1. A force of 5 N is acting on an object. The object is displaced through 2 m in the direction of the force (Fig. 10.2). 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.
Given 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. 10.3). 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.
Given 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 in this case is 56 J.
What if the force and displacement are not in the same direction?
- If the force is acting in the **opposite direction** to the displacement (angle between force and displacement is 180°), the work done by the force is considered **negative**. For example, when a moving object slows down due to friction, the friction force is opposite to the direction of motion, so the work done by friction is negative. If a force $F$ causes a displacement $s$ in the opposite direction, $W = F \times (-s) = -Fs$.
- If the force is acting **perpendicular** to the direction of displacement (angle between force and displacement is 90°), the work done by that force is **zero**. For example, the force of gravity on a satellite moving in a circular orbit around the Earth (assuming constant speed) is always directed towards the Earth's center, while the displacement at any instant is tangential to the orbit (perpendicular to the force). Thus, the work done by gravity on the satellite is zero. Similarly, when carrying a load horizontally, the force you exert is upwards (against gravity), while your displacement is horizontal. The work done by your lifting force is zero.
In general, when a constant force $F$ acts on an object and it is displaced by $s$, the work done is $W = Fs \cos \theta$, where $\theta$ is the angle between the direction of the force and the direction of displacement. If $\theta = 0^\circ$ (same direction), $W = Fs \cos 0^\circ = Fs$ (positive work). If $\theta = 180^\circ$ (opposite direction), $W = Fs \cos 180^\circ = -Fs$ (negative work). If $\theta = 90^\circ$ (perpendicular), $W = Fs \cos 90^\circ = 0$ (zero work).
Example 10.2. 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 raised), $s = 1.5$ m.
The porter applies an upward force to lift the luggage. The minimum force required is equal to the weight of the luggage, $F = mg$. We take $g \approx 9.8$ m/s² or 10 m/s² for simplicity as in the example solution provided (using $g=10$ m/s² as in the textbook solution derivation).
$F = m \times g = 15 \text{ kg} \times 10 \text{ m/s}^2 = 150 \text{ N}$.
The displacement is upwards, in the same direction as the force applied by the porter.
Work done by the porter, $W = F \times s = 150 \text{ N} \times 1.5 \text{ m} = 225 \text{ J}$.
If using $g=9.8$ m/s²:
$F = 15 \text{ kg} \times 9.8 \text{ m/s}^2 = 147 \text{ N}$.
$W = 147 \text{ N} \times 1.5 \text{ m} = 220.5 \text{ J}$.
Following the text's calculation using $g=10$ m/s²: Work done is 225 J.
Energy
Life and all activities in the universe require **energy**. Energy is the capacity to do work. When an object has energy, it has the capability to exert a force and cause displacement (do work) on another object.
When work is done on an object, energy is transferred to it. The object that does the work loses energy, and the object on which work is done gains energy. An object possessing energy can, in turn, do work on other objects by transferring its energy.
The amount of energy possessed by an object 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 J is the energy needed to do 1 joule of work. A larger unit, **kilojoule (kJ)**, is often used, where 1 kJ = 1000 J.
Forms Of Energy
Energy exists in many different forms in nature. Some common forms of energy include:
- **Mechanical energy:** The sum of kinetic energy and potential energy.
- **Heat energy:** Associated with the random motion of atoms and molecules.
- **Chemical energy:** Stored in the bonds of chemical compounds, released during chemical reactions (e.g., burning fuel, digestion of food).
- **Electrical energy:** Associated with the flow of electric charge.
- **Light energy:** A form of electromagnetic radiation visible to the eye.
- **Nuclear energy:** Stored in the nucleus of atoms, released during nuclear reactions.
Most of these energy forms on Earth ultimately originate from the Sun (solar energy), which drives photosynthesis, weather patterns, and the formation of fossil fuels. However, some sources, like nuclear energy and geothermal energy (from the Earth's interior), are not derived from the Sun.
Kinetic Energy
**Kinetic energy** is the energy possessed by an object **due to its motion**. Any moving object has kinetic energy. The faster an object moves, the more kinetic energy it has and the greater its capacity to do work (or cause an impact).
Examples of objects with kinetic energy: a falling object, a speeding car, a rolling stone, wind (moving air), flowing water, a moving bullet.
The amount of kinetic energy a moving object possesses is defined as the amount of work done on it to bring it from rest to its current velocity.
Potential Energy
**Potential energy** is the energy stored in an object due to its **position or configuration (shape)**. This energy has the potential to be converted into other forms of energy, like kinetic energy, and do work.
Examples of potential energy:
- A stretched rubber band or a compressed spring: Storing energy due to a change in shape (elastic potential energy). When released, this energy is converted into kinetic energy.
- A toy car wound up with a key: Energy is stored in the wound spring (elastic potential energy), which is released to make the car move.
- An object raised to a height: Storing energy due to its position in a gravitational field (gravitational potential energy). When released, this energy is converted into kinetic energy.
In these cases, work is done on the object (stretching, winding, lifting), and this work is stored as potential energy within the object. This stored potential energy can then be used to do work or cause motion.
Potential Energy Of An Object At A Height
When an object is lifted to a certain height above the ground, work is done against the force of gravity. This work is stored in the object as **gravitational potential energy (EP)**. The gravitational potential energy of an object at a point above the ground is defined as the work done in raising it from the ground to that point against gravity.
Consider an object of mass $m$ lifted vertically upwards by a height $h$. The minimum force required to lift the object at a constant velocity is equal to its weight, $F = mg$.
The work done in lifting the object is $W = \text{Force} \times \text{Displacement} = (mg) \times h = mgh$.
This work done is stored as the gravitational potential energy of the object at height $h$.
$$ E_p = mgh $$
The reference level for zero potential energy is arbitrary and can be chosen for convenience (e.g., ground level). The potential energy value depends on this chosen zero level. However, the change in potential energy between two points is independent of the chosen zero level. The work done by gravity on an object moving between two points depends only on the vertical distance between the points, not the path taken.
Example 10.5. 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 of object, $m = 10$ kg.
Height above the ground, $h = 6$ m.
Acceleration due to gravity, $g = 9.8$ m/s².
Potential energy, $E_p = mgh = (10 \text{ kg}) \times (9.8 \text{ m/s}^2) \times (6 \text{ m})$.
$E_p = 98 \times 6 \text{ J} = 588 \text{ J}$.
The potential energy possessed by the object is 588 J.
Example 10.6. 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 of object, $m = 12$ kg.
Potential energy, $E_p = 480$ J.
Acceleration due to gravity, $g = 10$ m/s².
We use the formula for potential energy: $E_p = mgh$.
$480 \text{ J} = (12 \text{ kg}) \times (10 \text{ m/s}^2) \times h$
$480 \text{ J} = 120 \text{ N} \times h$
$h = \frac{480 \text{ J}}{120 \text{ N}} = \frac{480 \text{ N m}}{120 \text{ N}} = 4 \text{ m}$.
The object is at a height of 4 m above the ground.
Are Various Energy Forms Interconvertible?
Yes, energy can be converted from one form to another. Nature exhibits numerous energy transformations:
- Solar energy (light) is converted into chemical energy (food) by green plants through photosynthesis.
- Wind energy (kinetic energy of air) is converted into electrical energy by wind turbines.
- Chemical energy in fuels is converted into heat and light energy when burned, and into mechanical energy in engines.
- Electrical energy is converted into light energy (bulb), heat energy (heater), mechanical energy (motor), sound energy (speaker), etc.
- In the water cycle, solar energy causes evaporation (converting water into vapour with potential energy), which then releases heat upon condensation and creates kinetic energy as water falls (rain/rivers) and flows, which can be converted to electrical energy in hydropower plants.
Energy conversion is a fundamental process that underpins most natural phenomena and technological applications.
Law Of Conservation Of Energy
A crucial principle governing energy transformations is the **Law of Conservation of Energy**. This law states that **energy can neither be created nor destroyed; it can only be transformed from one form to another**. In any process where energy is transformed, the total amount of energy in the system remains constant – the total energy before the transformation is equal to the total energy after the transformation.
This law holds true in all situations and for all types of energy transformations.
Consider an object of mass $m$ falling freely from a height $h$ above the ground (neglecting air resistance):
- At the starting height $h$: Velocity $v=0$. Potential energy $E_p = mgh$. Kinetic energy $E_k = \frac{1}{2}mv^2 = 0$. Total mechanical energy $E_{total} = E_p + E_k = mgh + 0 = mgh$.
- As the object falls: Its height $h$ decreases, so $E_p$ decreases. Its velocity $v$ increases, so $E_k$ increases. The decrease in potential energy is converted into an equal increase in kinetic energy.
- Just before hitting the ground ($h \approx 0$): Potential energy $E_p \approx mg(0) = 0$. Kinetic energy $E_k$ is maximum, as the velocity $v$ is maximum. The total mechanical energy $E_{total} = E_p + E_k = 0 + \frac{1}{2}mv_{max}^2$.
According to the Law of Conservation of Energy, the total mechanical energy remains constant throughout the free fall:
$$ \text{At any point}, \quad E_p + E_k = mgh + \frac{1}{2}mv^2 = \text{Constant} $$
This constant total energy is equal to the initial potential energy $mgh$ (at the start) and the final kinetic energy $\frac{1}{2}mv_{max}^2$ (just before impact).
$$ mgh_{initial} + \frac{1}{2}mv_{initial}^2 = mgh_{any\ point} + \frac{1}{2}mv_{any\ point}^2 = mgh_{final} + \frac{1}{2}mv_{final}^2 $$
The sum of the kinetic energy and potential energy of an object is called its **total mechanical energy**. During transformations like free fall, mechanical energy is conserved if no non-conservative forces (like air resistance or friction) are acting.
If non-conservative forces are present, mechanical energy is not conserved; some mechanical energy is transformed into other forms, like heat and sound energy.
Question 1. What is the kinetic energy of an object?
Answer:
Kinetic energy is the energy possessed by an object due to its motion. It is the energy that a moving object has by virtue of its movement.
Question 2. Write an expression for the kinetic energy of an object.
Answer:
The kinetic energy ($E_k$) of an object of mass $m$ moving with a velocity $v$ is given by the expression:
$$ E_k = \frac{1}{2}mv^2 $$
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:
Given: Initial kinetic energy $E_k = 25$ J when initial velocity $v = 5$ m/s.
Using the formula $E_k = \frac{1}{2}mv^2$, we have:
$25 \text{ J} = \frac{1}{2}m(5 \text{ m/s})^2 = \frac{1}{2}m(25 \text{ m}^2/\text{s}^2)$.
$25 \text{ J} = 12.5m \text{ m}^2/\text{s}^2$.
From this, we can find the mass $m$: $m = \frac{25}{12.5} \text{ kg} = 2 \text{ kg}$.
Now, we calculate the kinetic energy for different velocities:
When velocity is doubled, $v' = 2 \times v = 2 \times 5 \text{ m/s} = 10 \text{ m/s}$.
New kinetic energy, $E_k' = \frac{1}{2}m(v')^2 = \frac{1}{2}(2 \text{ kg})(10 \text{ m/s})^2 = \frac{1}{2}(2)(100) \text{ J} = 100 \text{ J}$.
Alternatively, since $E_k \propto v^2$, doubling the velocity makes the kinetic energy $2^2 = 4$ times the original. $E_k' = 4 \times E_k = 4 \times 25 \text{ J} = 100 \text{ J}$.
When velocity is increased three times, $v'' = 3 \times v = 3 \times 5 \text{ m/s} = 15 \text{ m/s}$.
New kinetic energy, $E_k'' = \frac{1}{2}m(v'')^2 = \frac{1}{2}(2 \text{ kg})(15 \text{ m/s})^2 = \frac{1}{2}(2)(225) \text{ J} = 225 \text{ J}$.
Alternatively, since $E_k \propto v^2$, increasing the velocity three times makes the kinetic energy $3^2 = 9$ times the original. $E_k'' = 9 \times E_k = 9 \times 25 \text{ J} = 225 \text{ J}$.
When its velocity is doubled, the kinetic energy will be 100 J. When its velocity is increased three times, the kinetic energy will be 225 J.
Rate Of Doing Work (Power)
Different agents (people, machines) can do the same amount of work but may take different amounts of time to do it. Also, they may transfer or consume energy at different rates. The concept that describes how fast work is done or energy is transferred is called **power**.
**Power** is defined as the **rate of doing work** or the **rate of transfer of energy**.
$$ \text{Power} (P) = \frac{\text{Work done} (W)}{\text{Time taken} (t)} \quad \text{or} \quad P = \frac{\text{Energy transferred} (E)}{\text{Time taken} (t)} $$
$$ P = \frac{W}{t} $$
The SI unit of power is the **watt (W)**, named after James Watt. One watt is the power of an agent that does **1 joule of work in 1 second**. Equivalently, 1 watt is the rate of energy consumption of 1 joule per second.
$$ 1 \text{ Watt (W)} = \frac{1 \text{ Joule (J)}}{1 \text{ second (s)}} \quad \text{or} \quad 1 \text{ W} = 1 \text{ J/s} $$
Larger rates of power are often expressed in kilowatts (kW), where 1 kilowatt = 1000 watts (1 kW = 1000 W = 1000 J/s).
The power of an agent can vary over time. In such cases, we often use the concept of **average power**, which is the total work done (or total energy consumed) divided by the total time taken.
Example 10.7. 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:
Work done by each girl is equal to the force exerted (their weight) multiplied by the vertical distance climbed.
Force (Weight) of each girl, $F = 400$ N.
Height climbed, $s = 8$ m.
Work done by each girl, $W = F \times s = 400 \text{ N} \times 8 \text{ m} = 3200 \text{ J}$.
Now, calculate the power expended by each girl using $P = W/t$:
For Girl A:
Time taken, $t_A = 20$ s.
Power of Girl A, $P_A = \frac{W}{t_A} = \frac{3200 \text{ J}}{20 \text{ s}} = 160 \text{ W}$.
For Girl B:
Time taken, $t_B = 50$ s.
Power of Girl B, $P_B = \frac{W}{t_B} = \frac{3200 \text{ J}}{50 \text{ s}} = 64 \text{ W}$.
Power expended by girl A is 160 W, and power expended by girl B is 64 W. Girl A is more powerful because she did the same amount of work in less time.
Example 10.8. 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 = 0.15 m.
Total height climbed, $h = \text{Number of steps} \times \text{Height per step} = 45 \times 0.15 \text{ m} = 6.75 \text{ m}$.
Time taken, $t = 9$ s.
Acceleration due to gravity, $g = 10$ m/s².
The work done by the boy against gravity is equal to the change in his potential energy, which is the force exerted (his weight) multiplied by the total height climbed.
Force (Weight of the boy), $F = mg = (50 \text{ kg}) \times (10 \text{ m/s}^2) = 500 \text{ N}$.
Work done, $W = F \times h = 500 \text{ N} \times 6.75 \text{ m} = 3375 \text{ J}$.
Now, calculate the power using $P = W/t$:
$P = \frac{3375 \text{ J}}{9 \text{ s}} = 375 \text{ W}$.
The power of the boy is 375 W.
Question 1. What is power?
Answer:
Power is the rate at which work is done or the rate at which energy is transferred or consumed. It measures how quickly energy is used or transformed.
Question 2. Define 1 watt of power.
Answer:
One watt (1 W) is defined as the power of an agent that does 1 joule of work in 1 second. Equivalently, it is the rate of energy transfer or consumption of 1 joule per second.
Question 3. A lamp consumes 1000 J of electrical energy in 10 s. What is its power?
Answer:
Given:
Energy consumed, $E = 1000$ J.
Time taken, $t = 10$ s.
Power, $P = \frac{\text{Energy consumed}}{\text{Time taken}} = \frac{1000 \text{ J}}{10 \text{ s}} = 100 \text{ W}$.
The power of the lamp is 100 W.
Question 4. Define average power.
Answer:
Average power is defined as the total work done (or total energy consumed) divided by the total time taken for the work to be done or energy to be consumed. It represents the overall rate of doing work or transferring energy over a period of time, even if the instantaneous power varies.
$$ \text{Average Power} = \frac{\text{Total Work Done}}{\text{Total Time Taken}} $$Commercial Unit of Energy
While the joule (J) is the SI unit of energy, it is a very small unit for expressing the large amounts of energy used in households, industries, etc. The commercial unit of energy is the **kilowatt-hour (kWh)**.
1 kilowatt-hour (kWh) is the energy consumed when an agent uses power at a rate of 1 kilowatt (kW) for 1 hour (h).
Let's convert kWh to joules:
$1 \text{ kWh} = 1 \text{ kilowatt} \times 1 \text{ hour}$
$1 \text{ kWh} = (1000 \text{ watts}) \times (3600 \text{ seconds})$
$1 \text{ kWh} = (1000 \text{ J/s}) \times (3600 \text{ s})$
$1 \text{ kWh} = 3,600,000 \text{ J} = 3.6 \times 10^6 \text{ J}$.
The energy consumed by households and industries is typically measured in kilowatt-hours, often referred to as 'units' in electricity bills.
Question 9. A certain household has consumed 250 units of energy during a month. How much energy is this in joules?
Answer:
Given energy consumed = 250 units.
1 unit of energy is equal to 1 kilowatt-hour (kWh).
We know that 1 kWh = $3.6 \times 10^6$ J.
So, 250 units = 250 kWh.
Energy in joules = $250 \times (3.6 \times 10^6) \text{ J}$
Energy in joules = $250 \times 3,600,000 \text{ J}$
Energy in joules = $900,000,000 \text{ J} = 9 \times 10^8 \text{ J}$.
The household has consumed $9 \times 10^8$ joules of energy.
Question 14. An electric heater is rated 1500 W. How much energy does it use in 10 hours?
Answer:
Given:
Power rating of the heater, $P = 1500$ W.
Time of use, $t = 10$ hours.
We know that Energy = Power $\times$ Time.
It's often convenient to calculate energy in kWh for electrical appliances used over hours.
Power in kW = $1500 \text{ W} / 1000 \text{ W/kW} = 1.5 \text{ kW}$.
Energy used (in kWh) = $P (\text{in kW}) \times t (\text{in hours}) = 1.5 \text{ kW} \times 10 \text{ h} = 15 \text{ kWh}$.
To convert this to joules: 1 kWh = $3.6 \times 10^6$ J.
Energy used (in J) = $15 \text{ kWh} \times (3.6 \times 10^6 \text{ J/kWh})$
Energy used (in J) = $54 \times 10^6 \text{ J} = 5.4 \times 10^7 \text{ J}$.
The electric heater uses 15 kWh or $5.4 \times 10^7$ J of energy in 10 hours.
Intext Questions
Page No. 115
Question 1. A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force (Fig. 10.3). 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. 116
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. 119
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. 123
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.
Answer:
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.
Answer:
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?
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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?
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Question 17. Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?
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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 joules consumed in 10 hours by four devices of power 500 W each.
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Question 21. A freely falling object eventually stops on reaching the ground. What happenes to its kinetic energy?
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