Class 9Science · PhysicsFull chapter

Work and Energy

Q: What is the work done by the force of gravity on a satellite moving in a circular orbit around the Earth? Justify your answer.

The work done by gravity on the satellite is zero . The gravitational force on the satellite always acts along the radius, directed towards the centre of the Earth (this is the centripetal force that keeps the satellite in its circular path). The satellite's displacement at every instant is along the tangent to its circular path, which is always perpendicular to the radius. Since the force and the displacement are perpendicular to each other (angle = 90°) at every point of the orbit, the work done, W = F s cosθ, becomes zero because cos 90° = 0.

Q: An object of mass 15 kg is moving with a uniform velocity of 4 m/s. Calculate the kinetic energy possessed by the object.

Given: m = 15 kg, v = 4 m/s Formula: E k = ½ m v 2 Substitution: E k = ½ × 15 × (4) 2 = ½ × 15 × 16 Result: E k = 120 J

Q: A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done any work, in the scientific sense? Justify your answer.

No, in the scientific sense he has done zero work on the bundle, even though he feels tired. Work is done only when a force produces a displacement: W = F × s. While he holds the bundle steadily over his head, the bundle does not move at all, so its displacement, s, is zero. Even though he is exerting a considerable muscular force to support the weight of the bundle, since s = 0, the work done, W = F × 0 = 0. His tiredness is due to biological (muscular) energy being used internally, but this does not count as mechanical work being done on the bundle.

Q: A battery lights a bulb. Describe the energy changes involved in the process, and state the law that governs these changes.

When a battery lights a bulb, the chemical energy stored in the battery is converted into electrical energy , which flows through the circuit and is converted by the bulb's filament into light energy and some heat energy (the filament also gets hot). This sequence of transformations is governed by the law of conservation of energy , which states that energy can neither be created nor destroyed, only converted from one form to another; the total energy released by the battery as chemical energy is exactly equal to the sum of the light energy and heat energy produced by the bulb.

Q: Find the potential energy of an object of mass 10 kg when it is at a height of 6 m above the ground. If the object is allowed to fall, find its kinetic energy when it has fallen halfway down, that is, through a height of 3 m. Take g = 9.8 m s⁻².

Potential energy at 6 m: Given: m = 10 kg, h = 6 m, g = 9.8 m s -2 Formula: E p = m g h Substitution: E p = 10 × 9.8 × 6 Result: E p = 588 J Kinetic energy after falling 3 m: By the law of conservation of energy, the potential energy lost in falling through 3 m is converted entirely into kinetic energy (the object starts from rest, so its initial kinetic energy is zero). Formula: E k = m g × (height fallen) Substitution: E k = 10 × 9.8 × 3 Result: E k = 294 J

Q: State the conditions under which the work done by a force on an object is (i) positive, (ii) negative, and (iii) zero. Give one example of each.

Positive work: done when the force (or its component) acts in the same direction as the displacement of the object. Example: a horse pulling a cart forward, the pulling force and the cart's displacement are in the same direction. Negative work: done when the force acts in a direction opposite to the displacement of the object. Example: friction acting on a block sliding across a rough floor opposes its motion and does negative work on it. Zero work: done when either the displacement of the object is zero, or the force acts perpendicular to the displacement. Example: a person pushing against a wall that does not move (displacement = 0); or the force of gravity on a satellite moving in a circular orbit, which is always perpendicular to its motion.

Q: A body of mass 5 kg is dropped from a height of 20 m above the ground. Calculate its kinetic energy after it has fallen through 8 m. Take g = 10 m s⁻².

Given: m = 5 kg, height fallen = 8 m, g = 10 m s -2 , initial velocity u = 0 (dropped from rest) Formula: by the law of conservation of energy, kinetic energy gained = potential energy lost = m g × (height fallen) Substitution: E k = 5 × 10 × 8 Result: E k = 400 J This can be checked using v 2 = u 2 + 2as: v 2 = 0 + 2 × 10 × 8 = 160, so E k = ½ × 5 × 160 = 400 J, which matches.

Q: Define power. An electric lamp consumes 1000 J of electrical energy in 10 s. Find its power.

Power is defined as the rate of doing work, or the rate at which energy is consumed or transferred: P = W/t. Its SI unit is the watt (W), where 1 W = 1 J/s. Given: W = 1000 J, t = 10 s Formula: P = W ÷ t Substitution: P = 1000 ÷ 10 Result: P = 100 W

Q: State the law of conservation of energy. Using this law, show that the total mechanical energy of a freely falling object (starting from rest) remains constant throughout its fall, neglecting air resistance.

The law of conservation of energy states that energy can neither be created nor destroyed; it can only be transformed from one form into another, and the total energy of an isolated system remains constant. Consider an object of mass m at rest at a height h above the ground, about to fall freely under gravity (air resistance neglected). At the starting point (height h): velocity = 0, so kinetic energy E k = 0; potential energy E p = m g h. Total energy E = 0 + m g h = m g h. After it has fallen a distance x (so it is at height (h − x) above the ground): using v 2 = u 2 + 2gx with u = 0, the velocity is v 2 = 2 g x. So kinetic energy E k = ½ m v 2 = ½ m (2gx) = m g x. Potential energy at this point, E p = m g (h − x). Total energy E = m g x + m g (h − x) = m g h. Just before hitting the ground (x = h): potential energy E p = m g (h − h) = 0, and kinetic energy E k = m g h (all potential energy has converted to kinetic energy). Total energy E = m g h + 0 = m g h. In every case, the sum of kinetic and potential energy is equal to m g h, the same value as at the start. This shows that the total mechanical energy of a freely falling object remains constant throughout its fall — only the form of energy changes, from potential to kinetic, confirming the law of conservation of energy.

The whole chapter in one place — read it, then test yourself. Clear notes, formula sheet, a practice quiz, and worked NCERT solutions & PYQs.

In this chapter

Work Done by a Force Kinetic Energy and the Work-Energy Theorem Potential Energy The Law of Conservation of Energy Power and the Commercial Unit of Energy

Work Done by a Force

Quick answer Work is done on an object only when a force causes a real displacement in the direction of the force; W = F × s, measured in joules.

In everyday language, "work" can mean any tiring activity. In science, however, work has a precise meaning: a force does work on an object only when the force causes a real displacement of the object.

The work done by a constant force acting on an object is defined as the product of the force and the displacement of the object in the direction of the force:

W = F × s

Here F is the force applied (in newton) and s is the displacement of the object along the direction of F (in metre). The SI unit of work is the joule (J). One joule of work is done when a force of 1 N displaces an object through 1 m in the direction of the force, so 1 J = 1 N m.

Worked example: A force of 7 N acts on an object and moves it through 8 m in the direction of the force. What is the work done?

Given: F = 7 N, s = 8 m
Formula: W = F × s
Substitution: W = 7 N × 8 m
Result: W = 56 J

Work can be positive, negative, or zero depending on the angle between the force and the displacement.

Positive work: the force (or its component) acts in the same direction as the displacement, e.g. a horse pulling a cart forward.
Negative work: the force acts opposite to the displacement, e.g. friction acting on a sliding box, or the braking force that slows a moving car.
Zero work: this happens in two common situations — (i) the displacement is zero, e.g. a person pushing hard against a wall that does not move; or (ii) the force acts at right angles to the displacement, e.g. the force of gravity on a satellite moving in a circular orbit does no work because it always acts perpendicular to the satellite's motion, and a coolie carrying a load on his head while walking horizontally does no work against gravity.

Work done by a constant force W = F × s F = force applied (N), s = displacement along F (m)

SI unit of work 1 J = 1 N × 1 m = 1 kg m² s⁻²

Work at angle θ between force and displacement W = F s cosθ W = 0 when θ = 90°

Remember

Work is done only when a force produces displacement of the object on which it acts.
W = F × s; SI unit of work is the joule (J); 1 J = 1 N m.
Work is positive when force and displacement are in the same direction.
Work is negative when force acts opposite to displacement (e.g. friction).
Work is zero when displacement is zero or the force is perpendicular to the displacement.

Kinetic Energy and the Work-Energy Theorem

Quick answer Kinetic energy, E_k = ½mv², is the energy of motion, and the work-energy theorem says the work done by the net force equals the change in kinetic energy.

The capacity of an object to do work is called its energy. Like work, energy is measured in joules (J). A moving object can do work on anything it strikes — a hammer driving a nail, flowing water turning a wheel, a moving cricket ball knocking off the bails. The energy possessed by an object because of its motion is called kinetic energy.

The kinetic energy of an object of mass m moving with velocity v is:

E_k = ½ m v²

Kinetic energy increases very rapidly with speed because it depends on v² — doubling the speed of an object makes its kinetic energy four times as large.

Worked example: Find the kinetic energy of an object of mass 15 kg moving with a uniform velocity of 4 m/s.

Given: m = 15 kg, v = 4 m/s
Formula: E_k = ½ m v²
Substitution: E_k = ½ × 15 × (4)² = ½ × 15 × 16
Result: E_k = 120 J

The work-energy theorem connects work and kinetic energy: the work done by the net force acting on an object is equal to the change produced in its kinetic energy.

W = E_k(final) − E_k(initial) = ½ m v² − ½ m u²

This follows directly from the equation of motion v² = u² + 2as: multiplying both sides by ½m gives ½mv² − ½mu² = (ma) × s = F × s = W, since F = ma. So whenever a force speeds an object up, it does positive work on it and increases its kinetic energy; whenever a retarding force (like friction) slows an object down, it does negative work and decreases its kinetic energy.

Kinetic energy Ek = ½ m v² m in kg, v in m/s, E_k in joule

Work-energy theorem W = ΔEk = ½ m v² − ½ m u²

Equation of motion used in the derivation v² = u² + 2as

Remember

Energy is the capacity to do work; its SI unit is also the joule (J).
Kinetic energy is the energy of motion: E_k = ½mv².
Kinetic energy depends on v², so doubling speed quadruples kinetic energy.
Work-energy theorem: work done by the net force = change in kinetic energy.
A force that speeds an object up does positive work; a retarding force does negative work.

Potential Energy

Quick answer Potential energy is energy stored due to position or configuration; gravitational potential energy of a raised object is E_p = mgh.

An object can possess energy because of its position or configuration, even if it is not moving. This is called potential energy. A stretched bow, a compressed spring, and an object held above the ground all have potential energy because work was done to bring them into that state, and that work gets stored in them.

The potential energy an object gains when it is raised above the ground is called gravitational potential energy. Consider an object of mass m raised through a height h. To lift it at nearly constant velocity, an applied force equal to its weight, mg, must act through the distance h. The work done against gravity is stored in the object as gravitational potential energy:

E_p = m g h

Here m is in kg, g is the acceleration due to gravity (taken as 9.8 m s^-2 unless stated otherwise) and h is the height above a chosen reference level, in metre. Potential energy is always measured relative to some reference level — usually the ground.

Worked example: Find the potential energy of an object of mass 10 kg raised to a height of 6 m above the ground. Take g = 9.8 m s^-2.

Given: m = 10 kg, h = 6 m, g = 9.8 m s^-2
Formula: E_p = m g h
Substitution: E_p = 10 × 9.8 × 6
Result: E_p = 588 J

If the same object is released and allowed to fall freely, this stored potential energy is converted into kinetic energy as it falls, and it becomes entirely kinetic energy (in the absence of air resistance) just before it strikes the ground.

Gravitational potential energy Ep = m g h h = height above the reference (ground) level

Remember

Potential energy is energy stored in an object due to its position or configuration.
Gravitational potential energy: E_p = mgh, measured relative to a chosen reference level.
Raising an object higher, or increasing its mass, increases its potential energy.
Potential energy is converted to kinetic energy when a raised object is allowed to fall.
A stretched spring or bow also stores potential energy due to its changed configuration.

The Law of Conservation of Energy

Quick answer Energy can neither be created nor destroyed, only transformed; for a freely falling body, kinetic plus potential energy stays constant.

Energy can change from one form to another — kinetic to potential, chemical to electrical, electrical to heat and light, and so on — but the total amount of energy in an isolated system never changes. This is the law of conservation of energy: energy can neither be created nor destroyed; it can only be transformed from one form to another, and the total energy before and after the transformation remains exactly the same.

A freely falling object under gravity, with no air resistance, is a simple example of this law in action. Suppose an object of mass m is dropped from height h. At the top, its velocity is zero, so its kinetic energy is zero and its entire energy is potential energy, E_p = mgh. As it falls, it loses height and therefore loses potential energy, but it gains speed and therefore gains kinetic energy. At every instant during the fall, the sum of its kinetic and potential energy remains equal to mgh — the potential energy lost is exactly equal to the kinetic energy gained. Just before the object hits the ground, all of the original potential energy has been converted into kinetic energy.

Worked example: An object of mass 10 kg is dropped from a height of 6 m. Find its kinetic energy after it has fallen through 3 m (halfway down), given g = 9.8 m s^-2.

Given: m = 10 kg, height fallen = 3 m, g = 9.8 m s^-2, initial velocity u = 0
Formula: loss in potential energy = gain in kinetic energy = m g × (height fallen)
Substitution: E_k = 10 × 9.8 × 3
Result: E_k = 294 J (exactly half of the total energy mgh = 588 J available at the start, since the object has fallen half of the total height)

The law of conservation of energy applies to every kind of energy transformation, not just mechanical ones — for example, in an electric bulb, electrical energy is converted into light energy and heat energy, and in a battery-operated toy, chemical energy stored in the battery is converted into electrical energy and then into kinetic energy and sound. In every case, the total energy output equals the total energy input; energy is only ever transformed, never lost.

Total mechanical energy of a freely falling body (no air resistance) E = Ek + Ep = m g h = constant true at every point during the fall

Remember

Law of conservation of energy: energy can neither be created nor destroyed, only transformed from one form to another.
In every energy transformation, the total energy before and after remains constant.
For a freely falling object (no air resistance), the sum of kinetic and potential energy stays constant throughout the fall.
At the highest point of a free fall, energy is entirely potential; just before hitting the ground, it is entirely kinetic.
The law applies to all forms of energy — mechanical, electrical, chemical, light, heat and sound.

Power and the Commercial Unit of Energy

Quick answer Power is the rate of doing work, P = W/t, measured in watts; the kilowatt hour (kWh) is the practical commercial unit of energy.

Two people can do the same amount of work, yet one may finish it faster than the other. To compare how quickly work is done, we use the quantity power. Power is defined as the rate of doing work, that is, the work done in unit time:

P = W / t

The SI unit of power is the watt (W). A power of 1 watt means 1 joule of work is done in 1 second, so 1 W = 1 J/s. Larger amounts of power are often measured in kilowatt (kW), where 1 kW = 1000 W.

If an agent does different amounts of work in different intervals of time, we describe its overall rate of working using average power, found by dividing the total work done (or total energy consumed) by the total time taken.

Worked example: A boy of mass 50 kg runs up a staircase of 45 steps in 9 s. If each step has a height of 15 cm, find his power. Take g = 9.8 m s^-2.

Given: m = 50 kg, number of steps = 45, height of each step = 0.15 m, t = 9 s, g = 9.8 m s^-2
Formula: total height h = number of steps × height of one step; work done W = m g h; power P = W / t
Substitution: h = 45 × 0.15 = 6.75 m, so W = 50 × 9.8 × 6.75 = 3307.5 J, and P = 3307.5 ÷ 9
Result: P = 367.5 W

Because the joule is a very small unit of energy, electricity companies do not use it to bill households. Instead they use a bigger, practical unit called the commercial unit of energy, the kilowatt hour (kWh), commonly called a "unit" of electricity. One kilowatt hour is the energy consumed when a device of power 1 kW (1000 W) is used continuously for 1 hour.

1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J = 3.6 million joules.

Worked example: An electric heater of power 1000 W is used for 3 hours. How many units (kWh) of electrical energy does it consume?

Given: P = 1000 W = 1 kW, t = 3 h
Formula: energy consumed = P × t
Substitution: energy = 1 kW × 3 h
Result: energy = 3 kWh, i.e. 3 units of electricity

Power P = W / t unit: watt (W); 1 W = 1 J/s

Kilowatt 1 kW = 1000 W

Commercial unit of energy (kilowatt hour) 1 kWh = 3.6 × 10⁶ J

Remember

Power is the rate of doing work: P = W/t; SI unit is the watt (W), 1 W = 1 J/s.
1 kilowatt (kW) = 1000 W; average power = total work (or energy) ÷ total time.
The commercial unit of electrical energy is the kilowatt hour (kWh), also called a "unit".
1 kWh = 3.6 × 10^6 J, the energy used by a 1 kW appliance running for 1 hour.
Electricity bills are calculated in kWh because the joule is too small a unit for everyday consumption.

The formula sheet

Every formula in this chapter, in one place — screenshot it before your exam.

W = F × s

Work done by a constant force

1 J = 1 N × 1 m = 1 kg m² s⁻²

SI unit of work

W = F s cosθ

Work at angle θ between force and displacement

Ek = ½ m v²

Kinetic energy

W = ΔEk = ½ m v² − ½ m u²

Work-energy theorem

v² = u² + 2as

Equation of motion used in the derivation

Ep = m g h

Gravitational potential energy

E = Ek + Ep = m g h = constant

Total mechanical energy of a freely falling body (no air resistance)

P = W / t

Power

1 kW = 1000 W

Kilowatt

1 kWh = 3.6 × 10⁶ J

Commercial unit of energy (kilowatt hour)

Test yourself

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0 correct · 0/12 answered

Q1 Work done by a force easy

A force acts on an object but the object does not move at all. What is the work done by the force?

Q2 Work done by a force easy

What is the SI unit of work?

Q3 Work done by a force medium

A force of 20 N moves an object through a distance of 5 m in the direction of the force. What is the work done?

Q4 Work done by a force medium

A coolie carries a load on his head and walks a certain distance on a horizontal platform. What is the work done by him against gravity?

Q5 Work done by a force hard

Which of the following is an example of negative work being done?

Q6 Kinetic energy easy

The kinetic energy of an object of mass m moving with velocity v is given by:

Q7 Kinetic energy medium

A car of mass 1000 kg is moving with a velocity of 20 m/s. What is its kinetic energy?

Q8 Work-energy theorem easy

According to the work-energy theorem, the work done by the net force acting on an object is equal to:

Q9 Potential energy medium

An object of mass 2 kg is raised to a height of 5 m above the ground. Taking g = 10 m s⁻², what is its gravitational potential energy?

Q10 Conservation of energy easy

The law of conservation of energy states that:

Q11 Power easy

The SI unit of power is the:

Q12 Commercial unit of energy medium

One kilowatt hour (1 kWh) of energy is equal to:

NCERT solutions & previous-year questions

Step-by-step model answers — tap a question to reveal the full solution.

NCERT questions 6

1 What is the work done by the force of gravity on a satellite moving in a circular orbit around the Earth? Justify your answer.Work done by a force

The work done by gravity on the satellite is zero.

The gravitational force on the satellite always acts along the radius, directed towards the centre of the Earth (this is the centripetal force that keeps the satellite in its circular path).
The satellite's displacement at every instant is along the tangent to its circular path, which is always perpendicular to the radius.
Since the force and the displacement are perpendicular to each other (angle = 90°) at every point of the orbit, the work done, W = F s cosθ, becomes zero because cos 90° = 0.

2 An object of mass 15 kg is moving with a uniform velocity of 4 m/s. Calculate the kinetic energy possessed by the object.Kinetic energy

Given: m = 15 kg, v = 4 m/s
Formula: E_k = ½ m v²
Substitution: E_k = ½ × 15 × (4)² = ½ × 15 × 16
Result: E_k = 120 J

3 A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done any work, in the scientific sense? Justify your answer.Work done by a force

No, in the scientific sense he has done zero work on the bundle, even though he feels tired.

Work is done only when a force produces a displacement: W = F × s.
While he holds the bundle steadily over his head, the bundle does not move at all, so its displacement, s, is zero.
Even though he is exerting a considerable muscular force to support the weight of the bundle, since s = 0, the work done, W = F × 0 = 0.
His tiredness is due to biological (muscular) energy being used internally, but this does not count as mechanical work being done on the bundle.

4 A battery lights a bulb. Describe the energy changes involved in the process, and state the law that governs these changes.Conservation of energy

When a battery lights a bulb, the chemical energy stored in the battery is converted into electrical energy, which flows through the circuit and is converted by the bulb's filament into light energy and some heat energy (the filament also gets hot).

This sequence of transformations is governed by the law of conservation of energy, which states that energy can neither be created nor destroyed, only converted from one form to another; the total energy released by the battery as chemical energy is exactly equal to the sum of the light energy and heat energy produced by the bulb.

5 Find the potential energy of an object of mass 10 kg when it is at a height of 6 m above the ground. If the object is allowed to fall, find its kinetic energy when it has fallen halfway down, that is, through a height of 3 m. Take g = 9.8 m s⁻².Conservation of energy

Potential energy at 6 m:

Given: m = 10 kg, h = 6 m, g = 9.8 m s^-2
Formula: E_p = m g h
Substitution: E_p = 10 × 9.8 × 6
Result: E_p = 588 J

Kinetic energy after falling 3 m:

By the law of conservation of energy, the potential energy lost in falling through 3 m is converted entirely into kinetic energy (the object starts from rest, so its initial kinetic energy is zero).
Formula: E_k = m g × (height fallen)
Substitution: E_k = 10 × 9.8 × 3
Result: E_k = 294 J

6 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 = 9.8 m s⁻².Power

Given: m = 50 kg, number of steps = 45, height of each step = 0.15 m, t = 9 s, g = 9.8 m s^-2
Formula: total height climbed, h = number of steps × height of each step; work done, W = m g h; power, P = W ÷ t
Substitution: h = 45 × 0.15 = 6.75 m; W = 50 × 9.8 × 6.75 = 3307.5 J; P = 3307.5 ÷ 9
Result: P = 367.5 W

Previous-year board questions 4

Q1 State the conditions under which the work done by a force on an object is (i) positive, (ii) negative, and (iii) zero. Give one example of each. CBSE 2022 3 marks

Positive work: done when the force (or its component) acts in the same direction as the displacement of the object. Example: a horse pulling a cart forward, the pulling force and the cart's displacement are in the same direction.
Negative work: done when the force acts in a direction opposite to the displacement of the object. Example: friction acting on a block sliding across a rough floor opposes its motion and does negative work on it.
Zero work: done when either the displacement of the object is zero, or the force acts perpendicular to the displacement. Example: a person pushing against a wall that does not move (displacement = 0); or the force of gravity on a satellite moving in a circular orbit, which is always perpendicular to its motion.

Q2 A body of mass 5 kg is dropped from a height of 20 m above the ground. Calculate its kinetic energy after it has fallen through 8 m. Take g = 10 m s⁻². CBSE 2020 3 marks

Given: m = 5 kg, height fallen = 8 m, g = 10 m s^-2, initial velocity u = 0 (dropped from rest)
Formula: by the law of conservation of energy, kinetic energy gained = potential energy lost = m g × (height fallen)
Substitution: E_k = 5 × 10 × 8
Result: E_k = 400 J

This can be checked using v² = u² + 2as: v² = 0 + 2 × 10 × 8 = 160, so E_k = ½ × 5 × 160 = 400 J, which matches.

Q3 Define power. An electric lamp consumes 1000 J of electrical energy in 10 s. Find its power. CBSE 2019 2 marks

Power is defined as the rate of doing work, or the rate at which energy is consumed or transferred: P = W/t. Its SI unit is the watt (W), where 1 W = 1 J/s.

Given: W = 1000 J, t = 10 s
Formula: P = W ÷ t
Substitution: P = 1000 ÷ 10
Result: P = 100 W

Q4 State the law of conservation of energy. Using this law, show that the total mechanical energy of a freely falling object (starting from rest) remains constant throughout its fall, neglecting air resistance. CBSE 2023 5 marks

The law of conservation of energy states that energy can neither be created nor destroyed; it can only be transformed from one form into another, and the total energy of an isolated system remains constant.

Consider an object of mass m at rest at a height h above the ground, about to fall freely under gravity (air resistance neglected).

At the starting point (height h): velocity = 0, so kinetic energy E_k = 0; potential energy E_p = m g h. Total energy E = 0 + m g h = m g h.
After it has fallen a distance x (so it is at height (h − x) above the ground): using v² = u² + 2gx with u = 0, the velocity is v² = 2 g x. So kinetic energy E_k = ½ m v² = ½ m (2gx) = m g x. Potential energy at this point, E_p = m g (h − x). Total energy E = m g x + m g (h − x) = m g h.
Just before hitting the ground (x = h): potential energy E_p = m g (h − h) = 0, and kinetic energy E_k = m g h (all potential energy has converted to kinetic energy). Total energy E = m g h + 0 = m g h.

In every case, the sum of kinetic and potential energy is equal to m g h, the same value as at the start. This shows that the total mechanical energy of a freely falling object remains constant throughout its fall — only the form of energy changes, from potential to kinetic, confirming the law of conservation of energy.

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