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Work, Energy, and Simple Machines

NCERT Class 9 · Science Based on NCERT Class 9 Science textbook · Free CBSE study kit

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Chapter Notes

Overview
This chapter introduces the scientific definitions of work, energy, and power, moving beyond everyday usage to precise physical quantities. It explains how work done by a force changes an object's energy, leading to the work-energy theorem and the concepts of kinetic and potential energy. The chapter also covers the conservation of mechanical energy and introduces simple machines like pulleys, inclined planes, and levers that make tasks easier by changing the magnitude or direction of force. The one big idea is that energy is the capacity to do work, and it can be transferred or transformed but not created or destroyed, with mechanical energy conserved in the absence of non-conservative forces.
Key Concepts (21)
Work
Work done on an object by a constant force is the product of the force applied and the displacement in the direction of the force.
e.g. Lifting a 5 kg wheat bag vertically by 1 m requires work equal to force (mg) times displacement (1 m).
Joule
One joule of work is done on an object when a constant force of 1 newton displaces it by 1 metre in the direction of the force.
e.g. If you push a box with 1 N force and it moves 1 m, you have done 1 J of work.
Zero Work
Work done is zero if the force acting on the object is zero, or if there is no displacement of the object, or if the force acts perpendicular to the displacement.
e.g. Pushing a rigid wall does no work because the wall does not displace, even though you feel tired.
Positive Work
Work done by a force on an object is positive when the displacement is in the same direction as the applied force.
e.g. Pushing a wheelchair forward — the force and displacement are in the same direction.
Negative Work
Work done by a force on an object is negative when the displacement is in the direction opposite to that of the force.
e.g. A goalkeeper stopping a ball applies force opposite to the ball's motion, doing negative work on the ball.
Energy
Energy is the capacity to do work.
e.g. A moving cricket ball has energy to knock over wickets; a raised flowerpot has energy to damage objects below.
Work-Energy Theorem
Work done on an object equals the change in its energy.
e.g. When a fielder throws a ball, the work done on the ball increases its kinetic energy.
Kinetic Energy
The energy possessed by an object due to its motion is called kinetic energy.
e.g. A moving bicycle or a rolling ball possesses kinetic energy.
Potential Energy
The energy stored by an object as a result of its deformation or in a system of objects due to their relative positions is called potential energy.
e.g. A stretched rubber band or a compressed spring stores potential energy due to deformation.
Gravitational Potential Energy
The potential energy of an object due to its position relative to the Earth, given by mgh near the Earth's surface.
e.g. A ball raised to a height of 2 m has more gravitational potential energy than when at 1 m.
Mechanical Energy
The sum of the kinetic energy and the potential energy of an object is called its mechanical energy.
e.g. A falling object has constant mechanical energy if no external forces (like friction) act on it.
Conservation of Mechanical Energy
The mechanical energy of an object remains constant if no other external forces act on it.
e.g. A pendulum bob reaches the same height on each swing if friction is neglected, showing constant mechanical energy.
Power
Power is the rate at which work is done.
e.g. Running up stairs in 1 minute requires more power than walking up in 5 minutes, even though the same work is done.
Watt
One watt is equal to 1 joule of work done per second.
e.g. A 100 W bulb uses 100 J of electrical energy every second.
Simple Machine
A device that helps perform tasks by changing the magnitude or direction of the force that needs to be applied.
e.g. A pulley, an inclined plane, and a lever are simple machines.
Effort
The force applied to a simple machine.
e.g. The force you pull down on a rope in a pulley system is the effort.
Load
The force that needs to be overcome by a simple machine.
e.g. The weight of the object being lifted is the load.
Mechanical Advantage
The ratio of the load to the effort.
e.g. If a machine lifts a 100 N load with a 20 N effort, its mechanical advantage is 5.
Fixed Pulley
A pulley fixed at a point that changes the direction of the effort but does not reduce its magnitude; mechanical advantage is 1.
e.g. Raising a flag by pulling the rope downward uses a fixed pulley.
Movable Pulley
A pulley that moves with the load, often used in a system to achieve a mechanical advantage greater than 1.
e.g. A crane uses a system of pulleys to lift heavy loads with less effort.
Inclined Plane
A sloping surface that allows a load to be raised with a smaller force over a longer distance.
e.g. Pushing a heavy box up a ramp to a platform requires less force than lifting it vertically.
Formulas & Equations (8)
W = F × s
Reads as: Work equals force multiplied by displacement in the direction of the force.
Variables: W = work done (J), F = constant force applied (N), s = displacement in the direction of force (m)
Use when: Use when a constant force acts on an object and causes displacement in the same direction.
⚠ Students forget that displacement must be in the direction of the force; using total displacement instead of component leads to wrong answers.
1 J = 1 N × 1 m = 1 kg m² s⁻²
Reads as: One joule equals one newton times one metre, which equals one kilogram metre squared per second squared.
Variables: J = joule, N = newton, kg = kilogram, m = metre, s = second
Use when: Use to convert units or understand the definition of a joule.
⚠ Students confuse joule with newton; remember joule is a unit of work/energy, newton is a unit of force.
W = (1/2) m (v² - u²)
Reads as: Work done equals half times mass times the difference of final velocity squared and initial velocity squared.
Variables: W = work done (J), m = mass (kg), v = final velocity (m/s), u = initial velocity (m/s)
Use when: Use to find work done when force is constant and velocities are known, derived from kinematic equations.
⚠ Students forget to square the velocities or misplace the factor of 1/2.
K = (1/2) m v²
Reads as: Kinetic energy equals half times mass times velocity squared.
Variables: K = kinetic energy (J), m = mass (kg), v = velocity (m/s)
Use when: Use to calculate kinetic energy of a moving object.
⚠ Students forget to square the velocity or use the wrong units (e.g., km/h instead of m/s).
U = m g h
Reads as: Potential energy equals mass times acceleration due to gravity times height.
Variables: U = gravitational potential energy (J), m = mass (kg), g = acceleration due to gravity (m/s²), h = height above reference point (m)
Use when: Use to calculate gravitational potential energy near Earth's surface.
⚠ Students forget that g is approximately 10 m/s² (or 9.8 m/s²) and use wrong value; also forget that height is measured from a reference point.
P = W / t
Reads as: Power equals work done divided by time taken.
Variables: P = average power (W), W = work done (J), t = time taken (s)
Use when: Use to calculate power when work and time are known.
⚠ Students confuse power with work; power is rate of doing work, not total work.
1 W = 1 J s⁻¹
Reads as: One watt equals one joule per second.
Variables: W = watt, J = joule, s = second
Use when: Use to convert units or understand the definition of a watt.
⚠ Students think watt is a unit of energy; it is a unit of power.
Mechanical Advantage = Load / Effort
Reads as: Mechanical advantage equals load divided by effort.
Variables: Load = force to be overcome (N), Effort = force applied (N)
Use when: Use to calculate how much a machine multiplies force.
⚠ Students invert the ratio, writing effort/load instead of load/effort.
How It Works
Calculating Work Done by a Constant Force
  1. Identify the constant force acting on the object.
  2. Determine the displacement of the object in the direction of the force.
  3. Multiply the force by the displacement: W = F × s.
  4. If force and displacement are in opposite directions, work is negative; if perpendicular, work is zero.
💡 Displacement must be measured in the direction of the force; if the force is at an angle, only the component in the direction of displacement does work.
Deriving Kinetic Energy Using Work-Energy Theorem
  1. Start with an object of mass m, initial velocity u, final velocity v, under constant force F causing displacement s.
  2. Use kinematic equation: v² = u² + 2as, so s = (v² - u²) / (2a).
  3. Work done W = F × s = (ma) × [(v² - u²) / (2a)] = (1/2) m (v² - u²).
  4. If u = 0, then W = (1/2) m v², which equals the kinetic energy K gained.
💡 This derivation assumes constant force and acceleration; the work-energy theorem holds even for non-constant forces.
Conservation of Mechanical Energy During Free Fall
  1. At the top (height h, velocity 0): PE = mgh, KE = 0, total ME = mgh.
  2. At an intermediate point (height h', velocity v): PE = mgh', KE = (1/2)mv², total ME = mgh' + (1/2)mv².
  3. Using v² = 2g(h - h'), KE = mg(h - h'), so total ME = mgh' + mg(h - h') = mgh.
  4. Just before hitting ground (height 0, velocity v): PE = 0, KE = (1/2)mv² = mgh, total ME = mgh.
💡 Mechanical energy is conserved only if no external forces like friction or air resistance act on the object.
Using an Inclined Plane to Reduce Effort
  1. Place a smooth plank against a raised platform to form an inclined plane.
  2. Attach a spring balance to the object (e.g., a cart) and pull it slowly up the plank.
  3. Observe that the force required (effort) is less than the weight of the object.
  4. As the plank becomes less steep (longer length for same height), the required force decreases further, but the distance over which force is applied increases.
💡 The total work done (force × distance) remains the same as lifting vertically; the inclined plane only reduces the force at the cost of increased distance.
Using a Fixed Pulley
  1. Attach a pulley to a rigid support at the top.
  2. Pass a rope through the groove of the pulley.
  3. Attach the load to one end of the rope.
  4. Apply effort by pulling the other end of the rope downward.
  5. The load moves upward, but the magnitude of effort required equals the load (mechanical advantage = 1).
💡 A fixed pulley only changes the direction of force, making it easier to pull downward rather than lift upward; it does not reduce the force magnitude.
Exam Traps — Don't Make These Mistakes
⚠ Confusing Work with Effort or Fatigue
✗ A person pushing a wall gets tired, so work is done on the wall.
→ In science, work is defined only when there is displacement in the direction of force. Pushing a wall with no displacement means zero work, even though the person feels tired due to internal energy use in muscles.
✓ No work is done on the wall because displacement is zero. The person feels tired because their muscles use internal energy, but scientifically, work done on the wall is zero.
⚠ Ignoring Direction in Work Calculation
✗ Work done = force × displacement, regardless of direction.
→ Work depends on the component of displacement in the direction of the force. If force and displacement are perpendicular, work is zero. If opposite, work is negative.
✓ Work done = force × displacement in the direction of the force. If the force is perpendicular to displacement, work is zero. If opposite, work is negative.
⚠ Misapplying Kinetic Energy Formula with Velocity Units
✗ A car moving at 72 km/h has kinetic energy (1/2) × m × (72)².
→ Velocity must be in m/s for the formula to yield joules. 72 km/h = 20 m/s. Using km/h directly gives a wrong numerical value.
✓ First convert velocity to m/s: 72 km/h = 20 m/s. Then kinetic energy = (1/2) × m × (20)².
⚠ Confusing Potential Energy with Kinetic Energy in Pendulum
✗ At the highest point of a pendulum swing, the bob has maximum kinetic energy.
→ At the highest point, the bob momentarily stops, so velocity is zero and kinetic energy is zero. Potential energy is maximum at that point.
✓ At the highest point, the bob has maximum potential energy and zero kinetic energy. At the lowest point, it has maximum kinetic energy and zero potential energy.
⚠ Forgetting the Factor of 1/2 in Kinetic Energy
✗ Kinetic energy = m v².
→ The correct formula is (1/2) m v². Omitting the 1/2 factor doubles the calculated value.
✓ Kinetic energy = (1/2) × mass × (velocity)².
⚠ Misinterpreting Mechanical Advantage of Fixed Pulley
✗ A fixed pulley has mechanical advantage greater than 1 because it makes lifting easier.
→ A fixed pulley only changes the direction of force, not its magnitude. The effort equals the load, so mechanical advantage is exactly 1.
✓ A fixed pulley has a mechanical advantage of 1 because the effort required equals the load. It only provides convenience by changing the direction of force.
Compare & Contrast
Source of energy
Kinetic Energy
Potential Energy
Kinetic energy is due to the motion of an object, while potential energy is due to its position or deformation.
Direction of force relative to displacement
Positive Work
Negative Work
Positive work occurs when force and displacement are in the same direction (energy is added to the object), while negative work occurs when they are opposite (energy is removed from the object).
Mechanical advantage
Fixed Pulley
Movable Pulley
A fixed pulley has a mechanical advantage of 1 (only changes direction), while a movable pulley can have a mechanical advantage greater than 1 (reduces the effort needed).
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 Very Short Answer (1 mark), 1 Short Answer (3 marks), 1 Long Answer (5 marks) — total approximately 11 marks.
Most asked: Definition and calculation of work (positive, negative, zero) · Kinetic energy formula and numerical problems · Potential energy (gravitational) formula and numerical problems · Work-energy theorem · Conservation of mechanical energy in free fall and pendulum · Power calculations · Mechanical advantage of simple machines (pulley, inclined plane)
Don't skip: Numerical problems combining work, kinetic energy, and potential energy (e.g., free fall, inclined plane, escape ramp); Derivation of kinetic energy using work-energy theorem; Conservation of mechanical energy in a pendulum or falling object; Calculating power from work and time
One-Line Revision (read aloud before the exam)
  1. Work is done only when a force causes displacement in its direction; if displacement is zero or perpendicular, work is zero.
  2. One joule is the work done when a force of 1 newton displaces an object by 1 metre in the direction of the force.
  3. Positive work increases an object's energy; negative work decreases it.
  4. Energy is the capacity to do work, and its SI unit is the joule.
  5. The work-energy theorem states that work done on an object equals the change in its energy.
  6. Kinetic energy of an object of mass m moving with velocity v is (1/2) m v².
  7. Potential energy of an object at height h near Earth's surface is m g h.
  8. Mechanical energy is the sum of kinetic and potential energy.
  9. In the absence of external forces like friction, mechanical energy is conserved.
  10. Power is the rate of doing work, measured in watts (1 W = 1 J/s).
  11. A fixed pulley has a mechanical advantage of 1 and only changes the direction of force.
  12. An inclined plane reduces the force needed to lift a load but increases the distance over which the force is applied.
  13. Mechanical advantage is the ratio of load to effort.
  14. When a goalkeeper stops a ball, she does negative work on the ball because force and displacement are opposite.
  15. A pendulum bob has maximum potential energy at the highest points and maximum kinetic energy at the lowest point.

MCQs — 10 Questions with Answers

Q1. A boy applies a force of 10 N to move a toy car through a distance of 2 m in the direction of the force. How much work is done?

  • A. 5 J
  • B. 20 J ✓
  • C. 12 J
  • D. 0.2 J

Answer: B — Work = Force × Displacement = 10 N × 2 m = 20 J.

Q2. Which of the following is NOT correct about work done?

  • A. Work is zero if displacement is zero
  • B. Work is positive when force and displacement are in the same direction
  • C. Work is always positive regardless of force direction ✓
  • D. Work is zero when force is perpendicular to displacement

Answer: C — Work can be positive, negative, or zero depending on the relative direction of force and displacement; it is not always positive.

Q3. A student pushes a wall with all their strength but the wall does not move. Which statement correctly explains why no work is done?

  • A. The force applied is too small
  • B. There is no displacement, so work done is zero ✓
  • C. The wall is too heavy to move
  • D. Work requires infinite force to move an immovable object

Answer: B — Since displacement s = 0, work W = F × 0 = 0, regardless of how large the force is.

Q4. Ramesh observes that lifting 5 bags to a height of 1 m requires much more effort than lifting 1 bag to the same height. Which concept explains this observation?

  • A. Work depends only on distance, not on force
  • B. Work is the product of force applied and displacement in the direction of force ✓
  • C. Lifting bags does not involve any work
  • D. The number of bags does not affect the work done

Answer: B — Lifting 5 bags requires 5 times greater force; since W = F × s and force increases 5-fold while displacement stays same, work increases 5-fold.

Q5. A girl carries a heavy box while walking 10 m horizontally. She applies an upward force equal to the box's weight. How much work is done by this upward force?

  • A. Work = weight × 10 m
  • B. Work = weight × height of the box
  • C. Work = 0 because force is perpendicular to displacement ✓
  • D. Work = weight × distance walked

Answer: C — The upward force is perpendicular to the horizontal displacement, so no component of force acts in the direction of motion; work = 0.

Q6. Which graph correctly represents the relationship between work done and displacement when a constant force of 5 N is applied?

  • A. A horizontal line at 5 on the y-axis
  • B. A straight line passing through origin with slope 5 ✓
  • C. A curve that decreases with increasing displacement
  • D. A vertical line at displacement = 5

Answer: B — Since W = 5 × s, work increases linearly with displacement, giving a straight line through origin with slope equal to the constant force (5).

Q7. A stone is lifted vertically upward by 4 m. The work done is 200 J. What is the force applied to lift the stone?

  • A. 25 N
  • B. 50 N ✓
  • C. 800 N
  • D. 0.02 N

Answer: B — Using W = F × s, we get 200 J = F × 4 m, so F = 200/4 = 50 N.

Q8. A boy pulls a toy in a horizontal direction while walking. Does the force applied by his hand do positive work on the toy?

  • A. No, because the toy is on the ground
  • B. Yes, because force and displacement are in the same direction ✓
  • C. No, because the toy's weight acts downward
  • D. Work cannot be done on toys, only on machines

Answer: B — The pulling force is in the same direction as the toy's horizontal displacement, so work done is positive: W = +F × s.

Q9. Why do muscles feel tired even when you push a rigid wall that does not move, even though scientifically no work is done?

  • A. The wall absorbs the work energy
  • B. Muscles contract and relax repeatedly, consuming internal body energy ✓
  • C. No force is actually applied to the wall
  • D. Work is being done but cannot be measured

Answer: B — Muscle fatigue comes from repeated contraction and biochemical energy use within the muscle, not from work done on external objects.

Q10. The area under a force-displacement graph represents which quantity?

  • A. The velocity of the object
  • B. The work done by the force ✓
  • C. The mass of the object
  • D. The power applied to the object

Answer: B — In a force-displacement graph, the area under the curve equals work done: W = F × s, which is the area of the rectangle or area under any force curve.

Flashcards

Define work done by a constant force in physics.

Work done = Force applied × Displacement in the direction of force.

What is 1 joule of work?

Work done when a force of 1 newton moves an object 1 metre in the direction of the force.

When is work done equal to zero?

Work is zero when there is no displacement, no force applied, or force is perpendicular to displacement.

How does lifting 3 bags differ from lifting 1 bag to the same height?

Lifting 3 bags requires 3 times the force and 3 times the work compared to lifting 1 bag.

Give an example where force is applied but work done is zero.

Pushing a wall: you apply force but the wall does not move, so no work is done.

What is the relationship between force, displacement, and work on a force-displacement graph?

Work done equals the area under the force-displacement graph between initial and final positions.

Is work positive or negative when lifting an object upward?

Work is positive because the applied force and displacement are both in the same direction (upward).

A girl carries a box while walking horizontally. Is work done by the upward force?

No, work is zero because the upward force is perpendicular to the horizontal displacement.

Why do muscles get tired even when pushing a rigid wall with no work done?

Muscles repeatedly contract and expand, using internal energy from the body, even though no external work is done.

Express 1 joule in terms of kilogram, metre, and second.

1 joule = 1 kg m² s⁻² (since 1 N = 1 kg m s⁻²).

Important Board Questions

Define work done by a constant force. Write the SI unit of work and express it in fundamental units. [2 marks]

State the formula W = F × s. 1 joule = 1 N × 1 m; express in kg, m, s: 1 J = 1 kg m² s⁻².

A student lifts a 10 kg bag from the ground to a shelf 2 m high. Explain why the work done by the student is not zero even though the bag returns to rest. Is the work positive or negative? [3 marks]

Work = force (mg) × displacement (2 m upward). Since force and displacement are in the same direction, work is positive. The bag's final motion state does not affect work done during the lifting process.

Compare the work done in the following scenarios: (a) Lifting 1 bag to a height of 1 m, (b) Lifting 2 bags to a height of 1 m, (c) Lifting 1 bag to a height of 2 m. Explain using the work formula why the work differs in each case and provide a real-life example showing how machines use this principle to reduce human effort. [5 marks]

Use W = F × s to show: (a) W = mg × 1, (b) W = 2mg × 1 = 2W, (c) W = mg × 2 = 2W. Recognize that increasing force or displacement increases work proportionally. Connect to simple machines like levers or pulleys that reduce the force needed by increasing displacement.

Assertion-Reason Questions for Work, Energy, and Simple Machines

Q1. Assertion (A):

Work done on an object by a constant force is always positive when the force and displacement are in the same direction.

Reason (R):

According to the definition W = F × s, the product of force magnitude and displacement magnitude gives a positive value when both are in the same direction.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true. When force and displacement are in the same direction, W = F × s yields a positive value because both F and s are positive. The reason correctly explains why the assertion is true. This is a straightforward application of the work formula and the definition of positive work.

Q2. Assertion (A):

A person pushing against a rigid wall without moving it has done zero work on the wall in the scientific sense.

Reason (R):

Work is defined as W = F × s, and since displacement s = 0, the work done is zero, even though the person's muscles use energy and feel tired.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true, and the reason is the correct explanation. The assertion correctly states that work done is zero, and the reason correctly explains this using the fundamental definition of work. The additional statement about feeling tired despite doing no work emphasizes the distinction between biological energy expenditure and mechanical work done on an external object.

Q3. Assertion (A):

When a girl carries a heavy box while walking horizontally, the upward force she applies on the box does positive work on the box.

Reason (R):

The force applied is upward and the displacement of the box is horizontal, making the force and displacement perpendicular to each other.

Answer: D — A is false but R is true — The assertion is false. The upward force actually does zero work, not positive work, because force and displacement are perpendicular. However, the reason is true—the force is indeed perpendicular to the displacement. Since perpendicular force and displacement result in zero work (not positive), the assertion contradicts what the reason correctly states. This tests understanding that perpendicularity means zero work, not positive work.

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Case-Based Questions (CBQ) for Work, Energy, and Simple Machines

Case Study:

A construction worker needs to lift materials to the top of a building. He can either (1) lift 50 kg bags one at a time to a height of 10 m, or (2) use a pulley system to lift three 50 kg bags together to the same height. The worker wonders which method requires doing more work. Taking g = 10 m/s², a single bag has a weight of 500 N. In method 1, the worker lifts one bag at a time, requiring 5 separate lifts. In method 2, he applies a larger force to lift all three bags in one operation.

Q1. What is the work done in method 1 to lift all five 50 kg bags to 10 m height?

  • A. 25,000 J
  • B. 50,000 J ✓
  • C. 75,000 J
  • D. 100,000 J

Answer: B

Q2. What is the work done in method 2 to lift three 50 kg bags to 10 m height?

  • A. 15,000 J ✓
  • B. 30,000 J
  • C. 45,000 J
  • D. 60,000 J

Answer: A

Q3. If the worker lifts only one bag using method 2 (applying a force equal to the bag's weight), what is the work done?

  • A. 500 J
  • B. 5,000 J ✓
  • C. 50,000 J
  • D. 500,000 J

Answer: B

Q4. How much more work is required to lift 5 bags individually compared to lifting 3 bags together (assuming both lifted to 10 m)?

  • A. 5,000 J more
  • B. 10,000 J more
  • C. 20,000 J more ✓
  • D. 100,000 J more

Answer: C

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