Chapter Notes

Work and Energy

25 min read

Work

In previous chapters, we explored how to describe the motion of objects, the causes of motion, and gravitation. Now, we introduce another important concept: work. Work is closely related to energy and power.

All living beings need food to survive. The basic activities they perform are called life processes. The energy for these processes comes from food. We also need energy for activities like playing, singing, reading, writing, thinking, jumping, cycling, and running. More strenuous activities require more energy.

Animals also use energy for activities like jumping, running, fighting, escaping enemies, finding food, or finding shelter. We also use animals to lift weights, carry loads, pull carts, or plow fields. All these activities require energy.

Machines also require energy to work. Some engines need fuels like petrol and diesel. Both living beings and machines need energy to function.

What is Work?

The term "work" has different meanings in everyday life and in science.

Not Much 'Work' in Spite of WORKING HARD!

Example
Kamali is preparing for examinations. She spends a lot of time studying, reading books, drawing diagrams, organizing her thoughts, collecting question papers, attending classes, discussing problems with her friends, and performing experiments. She expends a lot of energy on these activities. In common language, she is 'working hard'. But, according to the scientific definition of work, she may be doing very little 'work'.
Example
You are pushing a huge rock with all your might. The rock doesn't move, even though you get completely exhausted. In science, you haven't done any work on the rock because there is no displacement.
Example
You stand still for a few minutes holding a heavy load on your head. You get tired and use a lot of energy. However, in science, you are not doing work on the load.
Example
You climb up the steps of a staircase to reach the second floor of a building to see the view. This activity involves a lot of work according to the scientific definition.

In daily life, we consider any useful physical or mental effort as work. However, activities like playing, talking with friends, humming, watching a movie, or attending a function are not always considered work.

Scientific Conception of Work

In science, work has a specific meaning. Consider these situations:

Example
You push a pebble on the ground, and it moves. You applied a force on the pebble, and it was displaced. In this case, work is done.
Example
A girl pulls a trolley, and the trolley moves. The girl exerted a force on the trolley, and it moved. Therefore, work is done.
Example
You lift a book. You apply a force, and the book moves upwards. A force was applied, and the book moved. Therefore, work is done.

These examples show that two conditions must be met for work to be done:

  • A force must act on an object.
  • The object must be displaced.

If either of these conditions is not met, then no work is done.

Example
A bullock is pulling a cart, and the cart moves. A force is applied on the cart, and the cart moves. Work is done in this situation.

Work Done by a CONSTANT FORCE

To understand how work is defined in science, consider a constant force acting in the direction of the displacement.

Let a constant force FF act on an object. The object is displaced through a distance ss in the direction of the force. Let WW be the work done. We define work as the product of the force and displacement.

Work done=force×displacement\text{Work done} = \text{force} \times \text{displacement}

W=FsW = Fs

Work done by a force acting on an object is equal to the magnitude of the force multiplied by the distance moved in the direction of the force. Work has only magnitude and no direction.

If F=1 NF = 1 \text{ N} and s=1 ms = 1 \text{ m}, then the work done is 1 N m1 \text{ N m}. The unit of work is newton metre (N m), also known as joule (J).

Joule The amount of work done on an object when a force of 1 N1 \text{ N} displaces it by 1 m1 \text{ m} along the line of action of the force.

Note
If the force on the object is zero, or if the displacement of the object is zero, then the work done is zero.
Example
Example 10.1 A force of 5 N5 \text{ N} is acting on an object. The object is displaced through 2 m2 \text{ m} in the direction of the force. If the force acts on the object all through the displacement, then work done is 5 N×2 m=10 N m5 \text{ N} \times 2 \text{ m} = 10 \text{ N m} or 10 J10 \text{ J}.

Consider a baby pulling a toy car parallel to the ground. The baby exerts a force in the direction of the car's displacement. In this case, the work done is equal to the product of the force and displacement, and it is considered positive.

Now, consider an object moving with a uniform velocity. A retarding force, FF, is applied in the opposite direction. The angle between the force and displacement is 180∘180^\circ. The object stops after a displacement ss. In this case, the work done by the force FF is taken as negative and is denoted by a minus sign. The work done is F×(−s)F \times (-s) or (−F×s)(-F \times s).

The work done by a force can be either positive or negative.

Example
Lift an object up. Work is done by the force you exert on the object, and the object moves upwards. The force you exert is in the direction of displacement. However, gravity is also acting on the object. Your force is doing positive work, while gravity is doing negative work.

Work done is negative when the force acts opposite to the direction of displacement and positive when the force is in the direction of displacement.

Example
Example 10.2 A porter lifts luggage of 15 kg15 \text{ kg} from the ground and puts it on his head 1.5 m1.5 \text{ m} above the ground. Calculate the work done by him on the luggage.

Given

  • Mass of luggage, m=15 kgm = 15 \text{ kg}
  • Displacement, s=1.5 ms = 1.5 \text{ m}

To Find

Work done, WW

Formula

W=F×s=mg×sW = F \times s = mg \times s

Solution

Substitute the given values into the formula

W=15 kg×10 m s−2×1.5 m=225 kg m2s−2W = 15 \text{ kg} \times 10 \text{ m s}^{-2} \times 1.5 \text{ m} = 225 \text{ kg m}^2 \text{s}^{-2}

W=225 N m=225 JW = 225 \text{ N m} = 225 \text{ J}

Final Answer Work done is 225 J225 \text{ J}.

Energy

Life is impossible without energy. The demand for energy is always increasing. The Sun is the biggest natural source of energy. Many of our energy sources are derived from the Sun. We can also get energy from the nuclei of atoms, the interior of the earth, and the tides.

Energy The word energy is often used in our daily life, but in science, it has a definite and precise meaning.

Example
When a fast-moving cricket ball hits a stationary wicket, the wicket is thrown away.
Example
An object raised to a certain height can do work. A raised hammer falling on a nail drives the nail into a piece of wood.
Example
Children wind up a toy car, and when placed on the floor, it starts moving.
Example
When we press a balloon filled with air, its shape changes. If we press it gently, it returns to its original shape when the force is removed. However, if we press it hard, it can explode.

In all these examples, the objects can do work.

Energy An object with the ability to do work is said to possess energy. The object doing the work loses energy, and the object on which the work is done gains energy.

An object that possesses energy can exert a force on another object. When this happens, energy is transferred. The second object may move as it receives energy and do some work. Therefore, any object that possesses energy can do work.

The energy possessed by an object is measured by its capacity to do work. The unit of energy is the same as that of work, which is the joule (J). 1 J1 \text{ J} is the energy required to do 11 joule of work. A larger unit of energy is the kilojoule (kJ), where 1 kJ=1000 J1 \text{ kJ} = 1000 \text{ J}.

Forms of Energy

Energy exists in many different forms:

  • Mechanical energy (potential energy + kinetic energy)
  • Heat energy
  • Chemical energy
  • Electrical energy
  • Light energy

Kinetic Energy

Example
Take a heavy ball and drop it on a thick bed of sand from a height of 25 cm25 \text{ cm}. The ball creates a depression. Repeat this from heights of 50 cm50 \text{ cm}, 1 m1 \text{ m} and 1.5 m1.5 \text{ m}. The depths of the depressions will vary, with the deepest being from 1.5 m1.5 \text{ m}. The ball makes a deeper dent because it has more energy when dropped from a greater height.
Example
Set up a trolley with a wooden block in front of it. Place a mass on the pan so the trolley moves and hits the block. The block gets displaced, meaning the trolley did work on the block, transferring energy. This energy comes from the moving trolley.

A moving object can do work. A faster-moving object can do more work than a slower one.

Example
A moving bullet, blowing wind, a rotating wheel, or a speeding stone can do work.

Kinetic Energy Objects in motion possess energy called kinetic energy.

Example
A falling coconut, a speeding car, a rolling stone, a flying aircraft, flowing water, blowing wind, or a running athlete possess kinetic energy.

Kinetic energy is the energy possessed by an object due to its motion. The kinetic energy of an object increases with its speed.

The kinetic energy of a body moving with a certain velocity is equal to the work done on it to make it acquire that velocity.

Consider an object of mass mm moving with a uniform velocity uu. It is displaced through a distance ss when a constant force FF acts on it in the direction of its displacement. The work done, WW, is FsFs. The work done on the object causes a change in its velocity from uu to vv. Let aa be the acceleration produced.

From the equations of motion, we have:

v2−u2=2asv^2 - u^2 = 2as

s=v2−u22as = \frac{v^2 - u^2}{2a}

From Newton's Second Law of Motion, F=maF = ma. Thus, the work done by the force FF is:

W=ma×(v2−u22a)W = ma \times \left(\frac{v^2 - u^2}{2a}\right)

W=12m(v2−u2)W = \frac{1}{2}m(v^2 - u^2)

If the object starts from rest, u=0u = 0, then:

W=12mv2W = \frac{1}{2}mv^2

The work done is equal to the change in the kinetic energy of an object.

Kinetic Energy The kinetic energy (EkE_k) possessed by an object of mass mm moving with a uniform velocity vv is:

Ek=12mv2E_k = \frac{1}{2}mv^2

Example
Example 10.3 An object of mass 15 kg15 \text{ kg} is moving with a uniform velocity of 4 m s−14 \text{ m s}^{-1}. What is the kinetic energy possessed by the object?

Given

  • Mass of the object, m=15 kgm = 15 \text{ kg}
  • Velocity of the object, v=4 m s−1v = 4 \text{ m s}^{-1}

To Find

Kinetic energy, EkE_k

Formula

Ek=12mv2E_k = \frac{1}{2}mv^2

Solution

Substitute the given values into the formula

Ek=12×15 kg×(4 m s−1)2=120 JE_k = \frac{1}{2} \times 15 \text{ kg} \times (4 \text{ m s}^{-1})^2 = 120 \text{ J}

Final Answer The kinetic energy of the object is 120 J120 \text{ J}.

Example
Example 10.4 What is the work to be done to increase the velocity of a car from 30 km h−130 \text{ km h}^{-1} to 60 km h−160 \text{ km h}^{-1} if the mass of the car is 1500 kg1500 \text{ kg}?

Given

  • Mass of the car, m=1500 kgm = 1500 \text{ kg}
  • Initial velocity of the car, u=30 km h−1u = 30 \text{ km h}^{-1}
  • Final velocity of the car, v=60 km h−1v = 60 \text{ km h}^{-1}

To Find

Work done, WW

Formula

Eki=12mu2E_{ki} = \frac{1}{2}mu^2

Ekf=12mv2E_{kf} = \frac{1}{2}mv^2

W=Ekf−EkiW = E_{kf} - E_{ki}

Solution

First, convert the velocities to m s−1\text{m s}^{-1}:

u=30×100060×60 m s−1=253 m s−1u = \frac{30 \times 1000}{60 \times 60} \text{ m s}^{-1} = \frac{25}{3} \text{ m s}^{-1}

v=60×100060×60 m s−1=503 m s−1v = \frac{60 \times 1000}{60 \times 60} \text{ m s}^{-1} = \frac{50}{3} \text{ m s}^{-1}

Next, calculate the initial kinetic energy:

Eki=12×1500 kg×(253 m s−1)2=1562503 JE_{ki} = \frac{1}{2} \times 1500 \text{ kg} \times \left(\frac{25}{3} \text{ m s}^{-1}\right)^2 = \frac{156250}{3} \text{ J}

Then, calculate the final kinetic energy:

Ekf=12×1500 kg×(503 m s−1)2=6250003 JE_{kf} = \frac{1}{2} \times 1500 \text{ kg} \times \left(\frac{50}{3} \text{ m s}^{-1}\right)^2 = \frac{625000}{3} \text{ J}

Finally, calculate the work done:

W=Ekf−Eki=6250003 J−1562503 J=156250 JW = E_{kf} - E_{ki} = \frac{625000}{3} \text{ J} - \frac{156250}{3} \text{ J} = 156250 \text{ J}

Final Answer The work done is 156250 J156250 \text{ J}.

Potential Energy

Example
Take a rubber band. Hold it at one end and pull it from the other. The band stretches. Release the band, and it returns to its original length. The band acquired energy in its stretched position. This is because you did work on the rubber band to stretch it.
Example
Take a slinky. Have a friend hold one end while you hold the other. Move away from your friend, stretching the slinky. When you release it, it snaps back. The slinky acquired energy when stretched because you did work on it. The slinky also acquires energy when it is compressed.
Example
Wind up a toy car using its key. Place the car on the ground, and it moves. It acquired energy from your winding action. The more windings, the more energy it acquires.
Example
Lift an object to a certain height. The object can now do work because it will fall when released. This implies that it has acquired some energy. If raised to a greater height, it can do more work and, therefore, possesses more energy. The energy comes from the work you did to lift it.

In these situations, energy is stored due to the work done on the object.

Potential Energy The energy transferred to an object is stored as potential energy if it is not used to cause a change in the velocity or speed of the object.

Example
When you stretch a rubber band, the energy transferred to the band is its potential energy. When you wind the key of a toy car, the energy transferred to the spring inside is stored as potential energy.

Potential Energy The potential energy possessed by an object is the energy present in it by virtue of its position or configuration.

Example
Take a bamboo stick and make a bow. Place an arrow on it with one end supported by the stretched string. Stretch the string and release the arrow. The arrow flies off the bow. The potential energy stored in the bow, due to the change of shape, is used as kinetic energy to throw the arrow.

Potential Energy of an Object AT A HEIGHT

An object increases its energy when raised through a height. This is because work is done on it against gravity.

Gravitational Potential Energy The energy present in such an object is the gravitational potential energy. 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 mm. Let it be raised through a height hh from the ground. A force is required to do this. The minimum force required to raise the object is equal to the weight of the object, mgmg. The object gains energy equal to the work done on it. Let the work done on the object against gravity be WW. That is,

work done, W=force×displacement\text{work done, } W = \text{force} \times \text{displacement}

W=mg×hW = mg \times h

W=mghW = mgh

Since the work done on the object is equal to mghmgh, an energy equal to mghmgh units is gained by the object. This is the potential energy (EpE_p) of the object.

Ep=mghE_p = mgh

The potential energy of an object at a height depends on the ground level or the zero level you choose. An object in a given position can have a certain potential energy with respect to one level and a different value with respect to another level.

The work done by gravity depends on the difference in vertical heights of the initial and final positions of the object and not on the path along which the object is moved.

Example
A block is raised from position A to B by taking two different paths. Let the height AB=hAB = h. In both situations, the work done on the object is mghmgh.
Example
Example 10.5 Find the energy possessed by an object of mass 10 kg10 \text{ kg} when it is at a height of 6 m6 \text{ m} above the ground. Given, g=9.8 m s−2g = 9.8 \text{ m s}^{-2}.

Given

  • Mass of the object, m=10 kgm = 10 \text{ kg}
  • Displacement (height), h=6 mh = 6 \text{ m}
  • Acceleration due to gravity, g=9.8 m s−2g = 9.8 \text{ m s}^{-2}

To Find

Potential energy, EpE_p

Formula

Ep=mghE_p = mgh

Solution

Substitute the given values into the formula

Ep=10 kg×9.8 m s−2×6 m=588 JE_p = 10 \text{ kg} \times 9.8 \text{ m s}^{-2} \times 6 \text{ m} = 588 \text{ J}

Final Answer The potential energy is 588 J588 \text{ J}.

Example
Example 10.6 An object of mass 12 kg12 \text{ kg} is at a certain height above the ground. If the potential energy of the object is 480 J480 \text{ J}, find the height at which the object is with respect to the ground. Given, g=10 m s−2g = 10 \text{ m s}^{-2}.

Given

  • Mass of the object, m=12 kgm = 12 \text{ kg}
  • Potential energy, Ep=480 JE_p = 480 \text{ J}
  • Acceleration due to gravity, g=10 m s−2g = 10 \text{ m s}^{-2}

To Find

Height, hh

Formula

Ep=mghE_p = mgh

Solution

Substitute the given values into the formula

480 J=12 kg×10 m s−2×h480 \text{ J} = 12 \text{ kg} \times 10 \text{ m s}^{-2} \times h

h=480 J120 kg m s−2=4 mh = \frac{480 \text{ J}}{120 \text{ kg m s}^{-2}} = 4 \text{ m}

Final Answer The object is at the height of 4 m4 \text{ m}.

Are Various Energy Forms INTERCONVERTIBLE?

Can we convert energy from one form to another? In nature, we find many examples of energy conversion.

Example
Green plants produce food through photosynthesis, using energy from the Sun.
Example
Air moves from place to place due to differences in temperature and pressure, which are forms of energy.
Example
Fuels like coal and petroleum are formed from the remains of plants and animals over millions of years, storing chemical energy.
Example
The water cycle is sustained by energy conversions involving solar energy, evaporation, condensation, and precipitation.

Many human activities and gadgets involve the conversion of energy from one form to another.

Example
A light bulb converts electrical energy into light and heat energy.
Example
A car engine converts chemical energy from fuel into mechanical energy to move the car.

Law of Conservation of Energy

The form of energy can be changed from one form to another. But what happens to the total energy of a system during or after the process?

Law of Conservation of Energy Whenever energy gets transformed, the total energy remains unchanged. 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.

The law of conservation of energy is valid in all situations and for all kinds of transformations.

Example
Let an object of mass mm fall freely from a height hh. At the start, the potential energy is mghmgh, and the kinetic energy is zero because its velocity is zero. The total energy is mghmgh. As it falls, its potential energy changes into kinetic energy. If vv is the velocity of the object at a given instant, the kinetic energy would be 12mv2\frac{1}{2}mv^2. As the fall continues, potential energy decreases, and kinetic energy increases. When the object is about to reach the ground, h=0h=0, and vv is the highest. Therefore, kinetic energy is largest, and potential energy is least. However, the sum of potential energy and kinetic energy remains the same at all points.

mgh+12mv2=constantmgh + \frac{1}{2}mv^2 = \text{constant}

Mechanical Energy The sum of kinetic energy and potential energy of an object.

During the free fall of an object, the decrease in potential energy at any point is equal to the increase in kinetic energy.

Example
An object of mass 20 kg20 \text{ kg} is dropped from a height of 4 m4 \text{ m}.
Height (m)Potential Energy (Ep=mghE_p = mgh) (J)Kinetic Energy (Ek=12mv2E_k = \frac{1}{2}mv^2) (J)Ep+EkE_p + E_k (J)
420×10×4=80020 \times 10 \times 4 = 8000800
320×10×3=60020 \times 10 \times 3 = 600200800
220×10×2=40020 \times 10 \times 2 = 400400800
120×10×1=20020 \times 10 \times 1 = 200600800
Just above ground0800800

(Using g=10 m s−2g = 10 \text{ m s}^{-2} for calculations)

Rate of Doing Work

Do we all work at the same rate? Do machines consume or transfer energy at the same rate? Agents that transfer energy do work at different rates.

Example
Two children, A and B, weigh the same. Both climb a rope separately and reach a height of 8 m8 \text{ m}. A takes 15 s15 \text{ s}, while B takes 20 s20 \text{ s}. The work done by each is the same, but A took less time than B. A did more work in a given time.

A stronger person may do certain work in less time. A more powerful vehicle completes a journey in a shorter time than a less powerful one.

Power Power measures the speed of work done, that is, how fast or slow work is done. Power is defined as the rate of doing work or the rate of transfer of energy.

If an agent does work WW in time tt, then power is given by:

Power=worktime\text{Power} = \frac{\text{work}}{\text{time}}

P=WtP = \frac{W}{t}

The unit of power is watt (W), named in honor of James Watt (1736-1819). 11 watt is the power of an agent that does work at the rate of 11 joule per second. Power is 1 W1 \text{ W} when the rate of consumption of energy is 1 J s−11 \text{ J s}^{-1}.

1 watt=1joulesecond1 \text{ watt} = 1 \frac{\text{joule}}{\text{second}}

1 W=1 J s−11 \text{ W} = 1 \text{ J s}^{-1}

Larger rates of energy transfer are expressed in kilowatts (kW).

1 kilowatt=1000 watts1 \text{ kilowatt} = 1000 \text{ watts}

1 kW=1000 W1 \text{ kW} = 1000 \text{ W}

1 kW=1000 J s−11 \text{ kW} = 1000 \text{ J s}^{-1}

The power of an agent may vary with time, meaning the agent may do work at different rates at different times.

Average Power We obtain average power by dividing the total energy consumed by the total time taken.

Example
Example 10.7 Two girls, each of weight 400 N400 \text{ N}, climb up a rope through a height of 8 m8 \text{ m}. One girl is A, and the other is B. Girl A takes 20 s20 \text{ s}, while B takes 50 s50 \text{ s}. What is the power expended by each girl?

Given

(For Girl A)

  • Weight of the girl, mg=400 Nmg = 400 \text{ N}
  • Displacement (height), h=8 mh = 8 \text{ m}
  • Time taken, t=20 st = 20 \text{ s}

(For Girl B)

  • Weight of the girl, mg=400 Nmg = 400 \text{ N}
  • Displacement (height), h=8 mh = 8 \text{ m}
  • Time taken, t=50 st = 50 \text{ s}

To Find

(i) Power expended by girl A, PAP_A (ii) Power expended by girl B, PBP_B

Formula

P=mghtP = \frac{mgh}{t}

Solution

(i) Power expended by girl A

PA=400 N×8 m20 s=160 WP_A = \frac{400 \text{ N} \times 8 \text{ m}}{20 \text{ s}} = 160 \text{ W}

Answer for part (i) = 160 W160 \text{ W}

(ii) Power expended by girl B

PB=400 N×8 m50 s=64 WP_B = \frac{400 \text{ N} \times 8 \text{ m}}{50 \text{ s}} = 64 \text{ W}

Answer for part (ii) = 64 W64 \text{ W}

Final Answer Power expended by girl A is 160 W160 \text{ W}. Power expended by girl B is 64 W64 \text{ W}.

Example
Example 10.8 A boy of mass 50 kg50 \text{ kg} runs up a staircase of 45 steps in 9 s9 \text{ s}. If the height of each step is 15 cm15 \text{ cm}, find his power. Take g=10 m s−2g = 10 \text{ m s}^{-2}.

Given

  • Mass of the boy, m=50 kgm = 50 \text{ kg}
  • Number of steps, 4545
  • Height of each step, 15 cm=0.15 m15 \text{ cm} = 0.15 \text{ m}
  • Time taken, t=9 st = 9 \text{ s}
  • Acceleration due to gravity, g=10 m s−2g = 10 \text{ m s}^{-2}

To Find

Power, PP

Formula

P=mghtP = \frac{mgh}{t}

Solution

First, find the total height of the staircase:

h=45×0.15 m=6.75 mh = 45 \times 0.15 \text{ m} = 6.75 \text{ m}

Then, substitute the given values into the formula:

P=50 kg×10 m s−2×6.75 m9 s=375 WP = \frac{50 \text{ kg} \times 10 \text{ m s}^{-2} \times 6.75 \text{ m}}{9 \text{ s}} = 375 \text{ W}

Final Answer Power is 375 W375 \text{ W}.

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