- Work by a constant force equals force times displacement component along the force: W=Fs cosθ.
- Work is positive when force aids displacement and negative when it opposes displacement.
- Work is zero if displacement is zero or force is perpendicular to displacement.
- Energy is a transferable quantity measured in joules.
- Kinetic energy of a mass m moving at speed v is K=½mv².
- Near Earth's surface, change in potential energy is ΔU=mgΔh.
- Net work on a body equals change in kinetic energy.
- Energy changes form but total energy of an isolated system is conserved.
- Average power equals work divided by time: P=W/t, measured in watts.
- A simple machine changes the magnitude or direction of a force, trading force against distance.
- Mechanical advantage MA=load force/effort force; distinguish actual from ideal values.
- Velocity ratio VR=distance moved by effort/distance moved by load.
Use these notes alongside your current textbook. Check important formulas, worked answers and syllabus coverage with your teacher or official learning material.
Work, Energy, and Simple Machines — Class 9 Science, Exploration (2026–27)
Chapter 7. This original study guide explains core concepts with examples and worked practice. It is not a page-by-page verified transcription of NCERT activities, diagrams or numbered exercises. Refer to the prescribed textbook for exact official question wording.
Learning outcomes
Explain the concepts accurately, solve numerical or reasoning problems, choose suitable methods, and verify answers with correct units and assumptions.
Scientific work
Work by a constant force equals force times displacement component along the force: W=Fs cosθ.
Example and interpretation
A 10 N force moving a box 3 m in its direction does 30 J of work.
Positive and negative work
Work is positive when force aids displacement and negative when it opposes displacement.
Example and interpretation
Friction does negative work on a sliding box.
Zero work
Work is zero if displacement is zero or force is perpendicular to displacement.
Example and interpretation
Carrying a bag horizontally at constant height involves zero work by the upward supporting force.
Energy and units
Energy is a transferable quantity measured in joules.
Example and interpretation
A raised object stores gravitational potential energy relative to a reference level.
Kinetic energy
Kinetic energy of a mass m moving at speed v is K=½mv².
Example and interpretation
For m=2 kg,v=3 m/s, K=9 J.
Gravitational potential energy
Near Earth's surface, change in potential energy is ΔU=mgΔh.
Example and interpretation
Lifting 2 kg by 5 m with g=10 gives 100 J increase.
Work–energy theorem
Net work on a body equals change in kinetic energy.
Example and interpretation
If net work is 20 J, kinetic energy rises by 20 J.
Conservation of energy
Energy changes form but total energy of an isolated system is conserved.
Example and interpretation
As a falling object loses gravitational potential energy, kinetic energy rises if air resistance is negligible.
Power
Average power equals work divided by time: P=W/t, measured in watts.
Example and interpretation
200 J in 10 s corresponds to 20 W.
Simple machines
A simple machine changes the magnitude or direction of a force, trading force against distance.
Example and interpretation
An inclined plane reduces effort but increases travel distance.
Mechanical advantage
Mechanical advantage MA=load force/effort force; distinguish actual from ideal values.
Example and interpretation
Load 300 N with effort 100 N gives MA=3.
Velocity ratio
Velocity ratio VR=distance moved by effort/distance moved by load.
Example and interpretation
Effort moves 6 m while load rises 2 m, so VR=3.
Efficiency
Efficiency=(useful output work/input work)×100%; actual machines are typically below 100%.
Example and interpretation
Useful work 80 J from 100 J input gives 80% efficiency.
Levers
A lever rotates about a fulcrum; moments depend on force and perpendicular distance.
Example and interpretation
A 20 N effort at 0.5 m balances 10 N at 1 m when torques oppose.
Pulleys
Fixed pulleys can change force direction; movable pulley arrangements may reduce required effort.
Example and interpretation
An ideal single fixed pulley has mechanical advantage 1.
Inclined planes
Ideal effort along a frictionless plane is reduced in proportion to height/length.
Example and interpretation
Raise 100 N through 1 m using a 5 m frictionless incline: ideal effort 20 N.
Friction and losses
Friction converts mechanical energy to thermal energy and lowers machine efficiency.
Example and interpretation
A rough pulley requires more effort than an ideal pulley.
Energy accounting
Specify the system, reference level and losses before applying conservation.
Example and interpretation
For a falling object with air resistance, mechanical energy decreases but total energy remains conserved.
Common errors
Work is not simply force×time; power is not energy; mechanical advantage and efficiency are different quantities.
Example and interpretation
A machine may have MA greater than 1 without producing more output work than input.
Chapter summary table
| Topic | Essential explanation | Example |
|---|---|---|
| Scientific work | Work by a constant force equals force times displacement component along the force: W=Fs cosθ. | A 10 N force moving a box 3 m in its direction does 30 J of work. |
| Positive and negative work | Work is positive when force aids displacement and negative when it opposes displacement. | Friction does negative work on a sliding box. |
| Zero work | Work is zero if displacement is zero or force is perpendicular to displacement. | Carrying a bag horizontally at constant height involves zero work by the upward supporting force. |
| Energy and units | Energy is a transferable quantity measured in joules. | A raised object stores gravitational potential energy relative to a reference level. |
| Kinetic energy | Kinetic energy of a mass m moving at speed v is K=½mv². | For m=2 kg,v=3 m/s, K=9 J. |
| Gravitational potential energy | Near Earth's surface, change in potential energy is ΔU=mgΔh. | Lifting 2 kg by 5 m with g=10 gives 100 J increase. |
| Work–energy theorem | Net work on a body equals change in kinetic energy. | If net work is 20 J, kinetic energy rises by 20 J. |
| Conservation of energy | Energy changes form but total energy of an isolated system is conserved. | As a falling object loses gravitational potential energy, kinetic energy rises if air resistance is negligible. |
| Power | Average power equals work divided by time: P=W/t, measured in watts. | 200 J in 10 s corresponds to 20 W. |
| Simple machines | A simple machine changes the magnitude or direction of a force, trading force against distance. | An inclined plane reduces effort but increases travel distance. |
| Mechanical advantage | Mechanical advantage MA=load force/effort force; distinguish actual from ideal values. | Load 300 N with effort 100 N gives MA=3. |
| Velocity ratio | Velocity ratio VR=distance moved by effort/distance moved by load. | Effort moves 6 m while load rises 2 m, so VR=3. |
| Efficiency | Efficiency=(useful output work/input work)×100%; actual machines are typically below 100%. | Useful work 80 J from 100 J input gives 80% efficiency. |
| Levers | A lever rotates about a fulcrum; moments depend on force and perpendicular distance. | A 20 N effort at 0.5 m balances 10 N at 1 m when torques oppose. |
| Pulleys | Fixed pulleys can change force direction; movable pulley arrangements may reduce required effort. | An ideal single fixed pulley has mechanical advantage 1. |
| Inclined planes | Ideal effort along a frictionless plane is reduced in proportion to height/length. | Raise 100 N through 1 m using a 5 m frictionless incline: ideal effort 20 N. |
| Friction and losses | Friction converts mechanical energy to thermal energy and lowers machine efficiency. | A rough pulley requires more effort than an ideal pulley. |
| Energy accounting | Specify the system, reference level and losses before applying conservation. | For a falling object with air resistance, mechanical energy decreases but total energy remains conserved. |
| Common errors | Work is not simply force×time; power is not energy; mechanical advantage and efficiency are different quantities. | A machine may have MA greater than 1 without producing more output work than input. |
Original practice exercise 1 — Solved applications
- Question: A 25 N horizontal force moves a box 4 m horizontally. Find work.
Solution: W=Fs=25×4=100 J.
- Question: A 3 kg object moves at 4 m/s. Find kinetic energy.
Solution: K=½×3×4²=24 J.
- Question: Find potential-energy gain when a 5 kg load rises 2 m, taking g=10.
Solution: ΔU=mgh=5×10×2=100 J.
- Question: A machine transfers 600 J in 20 s. Find average power.
Solution: P=600/20=30 W.
- Question: Input work is 250 J and useful output is 175 J. Find efficiency.
Solution: Efficiency=(175/250)×100=70%.
- Question: A load of 400 N is lifted using 100 N effort. Find mechanical advantage.
Solution: MA=400/100=4.
- Question: Effort moves 8 m while load rises 2 m. Find velocity ratio.
Solution: VR=8/2=4.
- Question: If MA=3 and VR=4, find efficiency.
Solution: Efficiency=(MA/VR)×100=75%.
- Question: Why can a machine reduce force without reducing work in the ideal case?
Solution: It increases the distance over which effort acts; ideal input work equals output work.
- Question: A 12 N frictional force opposes 5 m motion. Find work by friction.
Solution: W=−12×5=−60 J.
Original practice exercise 2 — Concept questions with answers
- Question: Explain scientific work.
Answer: Work by a constant force equals force times displacement component along the force: W=Fs cosθ.
- Question: Explain positive and negative work.
Answer: Work is positive when force aids displacement and negative when it opposes displacement.
- Question: Explain zero work.
Answer: Work is zero if displacement is zero or force is perpendicular to displacement.
- Question: Explain energy and units.
Answer: Energy is a transferable quantity measured in joules.
- Question: Explain kinetic energy.
Answer: Kinetic energy of a mass m moving at speed v is K=½mv².
- Question: Explain gravitational potential energy.
Answer: Near Earth's surface, change in potential energy is ΔU=mgΔh.
- Question: Explain work–energy theorem.
Answer: Net work on a body equals change in kinetic energy.
- Question: Explain conservation of energy.
Answer: Energy changes form but total energy of an isolated system is conserved.
- Question: Explain power.
Answer: Average power equals work divided by time: P=W/t, measured in watts.
- Question: Explain simple machines.
Answer: A simple machine changes the magnitude or direction of a force, trading force against distance.
- Question: Explain mechanical advantage.
Answer: Mechanical advantage MA=load force/effort force; distinguish actual from ideal values.
- Question: Explain velocity ratio.
Answer: Velocity ratio VR=distance moved by effort/distance moved by load.
Original practice exercise 3 — Higher-order reasoning
- Identify an everyday situation that demonstrates two different chapter concepts. Explain how they apply.
- Compare two methods, models or quantities introduced in the chapter and identify their limitations.
- Design a safe observation or calculation that tests one claim, listing variables and assumptions.
- Construct a multistep numerical or classification problem and verify its answer.
- Identify a common misconception and correct it using the chapter principles.
Frequently asked questions
A 25 N horizontal force moves a box 4 m horizontally. Find work.
W=Fs=25×4=100 J.
A 3 kg object moves at 4 m/s. Find kinetic energy.
K=½×3×4²=24 J.
Find potential-energy gain when a 5 kg load rises 2 m, taking g=10.
ΔU=mgh=5×10×2=100 J.
A machine transfers 600 J in 20 s. Find average power.
P=600/20=30 W.
Input work is 250 J and useful output is 175 J. Find efficiency.
Efficiency=(175/250)×100=70%.
A load of 400 N is lifted using 100 N effort. Find mechanical advantage.
MA=400/100=4.
Effort moves 8 m while load rises 2 m. Find velocity ratio.
VR=8/2=4.
If MA=3 and VR=4, find efficiency.
Efficiency=(MA/VR)×100=75%.
Why can a machine reduce force without reducing work in the ideal case?
It increases the distance over which effort acts; ideal input work equals output work.
A 12 N frictional force opposes 5 m motion. Find work by friction.
W=−12×5=−60 J.
Explain scientific work.
Work by a constant force equals force times displacement component along the force: W=Fs cosθ.
Explain positive and negative work.
Work is positive when force aids displacement and negative when it opposes displacement.
Explain zero work.
Work is zero if displacement is zero or force is perpendicular to displacement.
Explain energy and units.
Energy is a transferable quantity measured in joules.
Explain kinetic energy.
Kinetic energy of a mass m moving at speed v is K=½mv².
Revision checklist
- Explain and apply scientific work.
- Explain and apply positive and negative work.
- Explain and apply zero work.
- Explain and apply energy and units.
- Explain and apply kinetic energy.
- Explain and apply gravitational potential energy.
- Explain and apply work–energy theorem.
- Explain and apply conservation of energy.
- Explain and apply power.
- Explain and apply simple machines.
- Explain and apply mechanical advantage.
- Explain and apply velocity ratio.
- Explain and apply efficiency.
- Explain and apply levers.
- Explain and apply pulleys.
- Explain and apply inclined planes.
- Explain and apply friction and losses.
- Explain and apply energy accounting.
- Explain and apply common errors.
Official textbook
NCERT ePathshala: Exploration. The practice exercises here are original and are not claimed to be the complete official NCERT exercise answer key.
Frequently asked questions
A 25 N horizontal force moves a box 4 m horizontally. Find work.
W=Fs=25×4=100 J.
A 3 kg object moves at 4 m/s. Find kinetic energy.
K=½×3×4²=24 J.
Find potential-energy gain when a 5 kg load rises 2 m, taking g=10.
ΔU=mgh=5×10×2=100 J.
A machine transfers 600 J in 20 s. Find average power.
P=600/20=30 W.
Input work is 250 J and useful output is 175 J. Find efficiency.
Efficiency=(175/250)×100=70%.
A load of 400 N is lifted using 100 N effort. Find mechanical advantage.
MA=400/100=4.
Effort moves 8 m while load rises 2 m. Find velocity ratio.
VR=8/2=4.
If MA=3 and VR=4, find efficiency.
Efficiency=(MA/VR)×100=75%.
A 12 N frictional force opposes 5 m motion. Find work by friction.
W=−12×5=−60 J.
Explain scientific work.
Work by a constant force equals force times displacement component along the force: W=Fs cosθ.
Explain positive and negative work.
Work is positive when force aids displacement and negative when it opposes displacement.
Explain zero work.
Work is zero if displacement is zero or force is perpendicular to displacement.
Explain energy and units.
Energy is a transferable quantity measured in joules.
Explain kinetic energy.
Kinetic energy of a mass m moving at speed v is K=½mv².
