CLASS 9 SCIENCE · CHAPTER 7 · STUDY NOTES
Work, Energy and Simple Machines
Understand work, energy and simple machines through short explanations, labelled models and original practice. Read one topic, explain it in your own words, then check.
EXPLORE
Work needs a displacement
Name the force and the object before calculating.Follow displacement along the force
- Work by a constant force
- Force × displacement in the force’s direction; unit joule (J).
- Joule
- 1 J=1 N×1 m.
A constant 8 N rightward pull moving a trolley 3 m right does 24 J of work. A strong push on a rigid wall does no mechanical work on that wall if it does not move. Your body can still use energy and feel tired.
For a force at an angle, use only the displacement component along the force. Work is a scalar: it has no spatial direction, but can be positive, negative or zero.
Convert before multiplying
Constant force parallel to displacement.
W=8 N×3 m=24 JA displacement of 300 cm must first become 3 m.
Take a moment. What do you think?
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Question
Does every force acting on an object do nonzero work?
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Direction controls the sign
Positive helps the motion’s energy; negative can remove it.Compare the directions
- Positive / negative work
- Force component along / opposite to displacement gives positive / negative work.
- Zero work by a force
- The displacement along that force is zero.
Gravity does negative work on a rising ball and positive work on a falling ball. A braking force usually does negative work on a forward-moving vehicle.
An upward supporting force does zero work on a box carried horizontally at constant height in the ideal model. Do not conclude that carrying uses no bodily energy. The sign belongs to a particular force’s work.
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Question
For a box moving horizontally, why is work by gravity zero?
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Area can measure work
Read displacement along the horizontal axis.Different forces, different work areas
Force along displacement (N)
Displacement (m)
- Constant 6 N · Area 6×4=24 J.
- Increasing from 0 to 6 N · Area ½×4×6=12 J.
Assumptions: Two alternative original force models, each acting through 4 m. Compare the rectangle and triangle areas.
Read the plotted values
Constant 6 N
| Displacement (m) | Force along displacement (N) |
|---|---|
| 0 | 6 |
| 4 | 6 |
Increasing from 0 to 6 N
| Displacement (m) | Force along displacement (N) |
|---|---|
| 0 | 0 |
| 4 | 6 |
Two alternative original force models, each acting through 4 m. Compare the rectangle and triangle areas.
- Force–displacement graph
- A graph whose signed area gives work for the plotted force component along displacement.
For constant 6 N through 4 m, the rectangular area is 24 N m=24 J. If force grows linearly from 0 to 6 N over 4 m, the triangular area is 12 J.
This is different from velocity–time area, which gives displacement. A negative force component contributes negative work; a force that merely falls in size is not automatically negative.
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Question
A force stays positive but decreases with displacement. Must its work be negative?
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Energy can move and change form
Track where it starts and where it goes.A watermill connects several energy stores
- Model
- A simplified gharat/panchakki model: gravitational potential energy is transferred into flowing water and rotation; some also becomes heat and sound.
- Connecting shaft
- The horizontal line from wheel to grinding stone represents a mechanical connection. It is not a path for flowing water.
- Energy
- A conserved physical quantity that can be stored and transferred; measured in joules.
- Mechanical energy
- Kinetic energy plus potential energy in the chosen mechanical model.
Food stores chemical energy; muscles can transfer some through mechanical work. A heater transfers electrical energy into thermal energy. Vibrating objects transfer energy through sound.
Work is one transfer route, not the only one: heat and radiation also transfer energy. Energy does not vanish when a machine slows; it can spread into thermal energy and sound. Conservation of total energy applies to an isolated system, not just any object with zero net force.
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Question
When friction slows a moving object, has its energy disappeared?
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Speed enters twice
Doubling speed quadruples kinetic energy at the same mass.Kinetic energy curves upward with speed
Kinetic energy (J)
Speed (m/s)
- Mass 2 kg · K=v² joules for this mass.
Assumptions: Original 2 kg model. The square of speed makes this a curve, not a straight proportional line.
Read the plotted values
Mass 2 kg
| Speed (m/s) | Kinetic energy (J) |
|---|---|
| 0 | 0 |
| 1 | 1 |
| 2 | 4 |
| 3 | 9 |
| 4 | 16 |
| 5 | 25 |
| 6 | 36 |
Original 2 kg model. The square of speed makes this a curve, not a straight proportional line.
- Kinetic energy
- Energy associated with motion: K=½mv² in the everyday-speed model.
For an original 2 kg trolley moving at 3 m/s, K=½×2×3²=9 J. At 6 m/s, K=36 J. Doubling mass at the same speed instead doubles K.
Speed must be in m/s for SI calculations. K has no direction and cannot be negative. Reversing velocity while keeping its magnitude does not change K.
Square the converted speed
Mass 2 kg; speed 3 m/s.
K=½×2×3²=9 JDo not square a value still in km/h and report joules.
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Question
Two equal-mass objects have speeds v and 3v. What is the kinetic-energy ratio?
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Add all work to find the kinetic-energy change
Net work is not the work of only the helpful force.Keep an energy account
- Work–kinetic-energy theorem
- The total work of the forces on a particle equals its change in kinetic energy: Wnet=Kfinal−Kinitial.
For a constant net force, combine Fnet=ma and v²−u²=2as to obtain Wnet=½mv²−½mu². The theorem also holds more broadly than that simple derivation.
An original trolley starts with 4 J. A push does +18 J and resistance does −6 J. Net work is +12 J, so final K=16 J. For mass 2 kg, v=4 m/s.
Potential energy is accounted for differently when you use a system-energy equation. Avoid adding gravitational work and the same potential-energy change twice.
- 1
Find initial K=½mu².
- 2
Add the work of all forces with signs.
- 3
Use Kfinal=Kinitial+Wnet.
- 4
Use v=√(2Kfinal/m) if speed is requested.
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Question
Why can a positive pushing work fail to equal the kinetic-energy gain?
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Position and deformation can store energy
A spring and a raised object store energy differently.Height determines gravitational energy
- Potential energy
- Energy associated with configuration or relative position in an interacting system.
- Gravitational potential energy near Earth
- U=mgh relative to a chosen zero height, when g is approximately constant.
Stretching or compressing a spring stores elastic potential energy. Separating attracting magnets can change the magnetic potential energy of their system. A raised object and Earth have gravitational potential energy.
Taking ground level as zero, an original 3 kg object at 2 m has U=3×10×2=60 J using g=10 m/s². Only changes in U affect the calculation; choose the reference consistently. The same final height gives the same ΔU whichever route is taken.
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Question
Does an object at rest always have no energy?
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