This MCQ module is based on: Work and Energy
TOPIC 23 OF 50
Work and Energy
🎓 Class 9
Science
CBSE
Theory
Ch 7 — Work, Energy, and Simple Machines
⏱ ~13 min
🌐 Language: [gtranslate]
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Introduction: When Does a Push Count as Work?
In everyday language, the word work is used loosely. A student preparing for an exam says she has been "working hard" all evening. A coolie at the railway station who carries a trunk on his head from one platform to another also says he has done "a lot of work". Both seem perfectly reasonable in casual conversation — yet, surprisingly, in the language of physics one of them has done zero work. To find out why, we have to set aside everyday meaning and adopt a precise scientific definition.
Once we get the idea of work right, the closely linked idea of energy almost takes care of itself: energy is simply the capacity to do work. In this part we will define work mathematically, study positive, negative and zero work, and meet the two most important mechanical forms of energy — kinetic and potential.
Key Idea: In physics, work is done only when a force acts on a body and the body moves in the direction of (or opposite to) the force. No movement → no work. Movement perpendicular to the force → also no work.
7.1 The Scientific Concept of Work
Two conditions must be satisfied at the same time for work to be done on a body:
- A force must act on the body.
- The body must undergo a displacement in the direction of (or having a component along) the force.
If either condition fails, work done = 0 — even if the body or the person feels tired.
Defining work mathematically
Suppose a constant force \(F\) acts on a body and the body is displaced by \(s\) along the direction of the force. The work done is defined as
\[ W \;=\; F \times s \]
If the force makes an angle \(\theta\) with the direction of displacement, only the component of force along the displacement is effective. The general expression therefore becomes
\[ W \;=\; F\,s\,\cos\theta \]
The SI unit of work is the joule (J), where
\[ 1\;\text{J} \;=\; 1\;\text{N}\times 1\;\text{m} \;=\; 1\;\text{kg}\,\text{m}^{2}\,\text{s}^{-2} \]
Definition. One joule of work is done on a body when a force of one newton displaces it by one metre in the direction of the force.
Sign of work — positive, negative and zero
The angle \(\theta\) between force and displacement decides the sign of work:
- Positive work (\(0° \le \theta < 90°\)): Force has a component along the displacement. Example — a horse pulling a cart forward; gravity acting on a falling stone.
- Zero work (\(\theta = 90°\) or \(s = 0\) or \(F = 0\)): Force is perpendicular to displacement, OR there is no displacement at all. Example — gravity does no work on a satellite in a circular orbit; a coolie carrying a load horizontally on his head.
- Negative work (\(90° < \theta \le 180°\)): Force opposes the displacement. Example — friction acting on a sliding box; gravity acting on a stone thrown upward.
🧱 Three Kinds of Work — Click each scene to compare L4 Analyse
The same block, the same displacement — only the direction of the applied force is different. Click each scene to analyse why the work done is positive, zero or negative.
Click any of the three scenes above to analyse the sign of the work done.
A surprising case — the porter on the platform
A porter (coolie) carries a 20 kg trunk on his head and walks 50 m horizontally on a level platform. He is sweating, panting, and clearly exhausted — yet, scientifically, the work he does against gravity is zero. Why? Because gravity pulls the trunk straight down, while the porter's displacement is horizontal. The angle between the gravitational force on the trunk and the displacement is 90°, so
\[ W = F\,s\,\cos 90° = F\,s\,(0) = 0 \]
The porter's muscles certainly do internal work (which is why he gets tired and his chemical energy is consumed), but in the strict sense of "work done by gravity on the trunk through this displacement", the answer is zero. This is one of the most counter-intuitive results students meet in Class 9 physics.
7.2 Worked Numericals on Work
Example 1 — Pushing a box on a level floor L3
A boy applies a horizontal force of 25 N to push a box through a distance of 4 m on a level floor. Find the work done.
Given: F = 25 N, s = 4 m, θ = 0°.
Formula: \( W = F\,s\,\cos\theta \).
Substitute: \( W = 25 \times 4 \times \cos 0° = 25 \times 4 \times 1 = 100 \) J.
Answer: Work done = 100 J (positive).
Example 2 — Lifting a bucket against gravity L3
A woman lifts a bucket of water of mass 5 kg vertically through a height of 2 m at constant speed. (Take g = 10 m s⁻².) Find the work done by her muscular force.
Force needed to lift at constant speed = weight = mg = 5 × 10 = 50 N (upward).
Displacement = 2 m (upward); θ = 0°.
\( W = F s \cos 0° = 50 \times 2 \times 1 = 100 \) J.
Answer: Work done by the woman = +100 J. Work done by gravity on the bucket = −100 J (equal and opposite).
Example 3 — Friction acting on a sliding crate L3
A crate is dragged 6 m along a floor. The frictional force opposing the motion is 12 N. Find the work done by friction.
F = 12 N, s = 6 m, θ between friction and displacement = 180°.
\( W_f = 12 \times 6 \times \cos 180° = 12 \times 6 \times (-1) = -72 \) J.
Answer: Work done by friction = −72 J. Negative sign shows energy is being taken away from the crate (mostly turned into heat).
Example 4 — Pulling a luggage at an angle L4
A child pulls a small suitcase by a strap that makes 60° with the horizontal. The pull is 50 N and the suitcase moves 8 m horizontally. Find the work done.
F = 50 N, s = 8 m, θ = 60°, cos 60° = 0.5.
\( W = 50 \times 8 \times 0.5 = 200 \) J.
Answer: Work done = 200 J. Notice that only the horizontal component (F cos θ) contributes; the vertical component does no work because there is no vertical displacement.
7.3 Activity — Identifying Work in Daily Situations
Activity 7.1 — Spot the WorkL4 Analyse
Predict first: A boy pushes hard against a brick wall for a full minute and the wall does not move. Has he done any work on the wall in the scientific sense?
- Hold a heavy book on your outstretched palm and stand still for 30 seconds. Did you do any work on the book?
- Now lift the book slowly from waist level to head level. Did you do work on it?
- Walk across the room with the book held still on your head.
- Drop the book carefully onto a cushion. As it falls, which force does work on it — and is that work positive or negative?
Observations: (1) No displacement → no work, even though arm gets tired. (2) Force is upward and displacement is upward → positive work, W = mgh. (3) Force on book is vertical (upward), displacement horizontal → angle 90° → zero work on the book. (4) Gravity acts downward, displacement is downward → positive work done by gravity (it speeds up the book).
Conclusion: "Tiredness" of muscles is NOT a measure of scientific work. Work needs displacement of the body along (or partly along) the line of force.
Conclusion: "Tiredness" of muscles is NOT a measure of scientific work. Work needs displacement of the body along (or partly along) the line of force.
7.4 Energy — The Capacity to Do Work
Wherever work is done, energy is involved. A body that can push, pull, lift, deform, accelerate or heat another body is said to possess energy. A moving cricket ball can knock the stumps over (it does work). A stretched bow can launch an arrow. A dry cell can light a torch. In each case the body has the capacity to do work — that capacity is its energy.
Definition. Energy is the ability of a body to do work. Its SI unit is the same as that of work — the joule (J). A larger unit, the kilojoule, is 1 kJ = 10³ J.
Mechanical energy
The energy a body possesses because of its motion or its position is called mechanical energy. It comes in two flavours — kinetic energy (KE) due to motion and potential energy (PE) due to position or configuration.
7.5 Kinetic Energy
A moving hammer drives a nail; a moving car can push another car aside; flowing water can rotate the blades of a turbine. Anything that moves has the ability to do work — this is its kinetic energy.
Deriving KE = ½ mv²
Consider a body of mass \(m\) initially at rest. A constant force \(F\) accelerates it through a distance \(s\), reaching a final velocity \(v\). Using the equation of motion \(v^2 = u^2 + 2as\) with \(u=0\):
\[ v^2 = 2as \quad\Rightarrow\quad s = \frac{v^2}{2a} \]
The work done by F equals the kinetic energy gained:
\[ \text{KE} = W = F\,s = (ma)\left(\frac{v^2}{2a}\right) = \tfrac{1}{2}\,m\,v^2 \]
Important formula: \( \text{KE} = \tfrac{1}{2}\,m\,v^{2} \). KE is always positive (since v² ≥ 0). It depends on the square of speed — so doubling v makes KE four times larger.
Example 5 — KE of a cricket ball L3
Find the kinetic energy of a 0.16 kg cricket ball moving at 30 m s⁻¹.
\( \text{KE} = \tfrac{1}{2}(0.16)(30)^2 = 0.5 \times 0.16 \times 900 = 72 \) J.
Answer: KE = 72 J.
7.6 Potential Energy
Lift a brick to the edge of a roof. Although it is not moving, it now has the capacity to do work — release it and it will fall, picking up speed and crushing whatever lies below. The energy stored due to its raised position is called gravitational potential energy.
More generally, potential energy (PE) is energy stored because of position or configuration. A stretched bow, a wound spring, and a lifted hammer all have potential energy.
Gravitational PE near the Earth's surface
To raise a body of mass \(m\) slowly through height \(h\) we apply an upward force equal to its weight \(mg\) and produce displacement \(h\) in the same direction. The work done is
\[ W = F\,s = (mg)(h) = mgh \]
This work is stored in the body as gravitational potential energy:
\[ \text{PE} = m\,g\,h \]
where \(g \approx 9.8\) m s⁻² (often taken as 10 m s⁻² in Class 9 problems) and \(h\) is the height above a chosen reference level (usually the ground).
Example 6 — PE of a coconut on a tree L3
A 1.5 kg coconut hangs at a height of 8 m above the ground. Take g = 10 m s⁻². Find its gravitational potential energy with respect to the ground.
\( \text{PE} = mgh = 1.5 \times 10 \times 8 = 120 \) J.
Answer: PE = 120 J. If the coconut falls, this much energy is converted to kinetic energy (ignoring air resistance).
Example 7 — Comparing two stones L4
Stone A (2 kg) is at a height of 5 m. Stone B (5 kg) is at a height of 2 m. Which has greater PE? (g = 10 m s⁻²)
PE(A) = 2 × 10 × 5 = 100 J.
PE(B) = 5 × 10 × 2 = 100 J.
Answer: Both have the same PE — 100 J. Mass and height enter the formula symmetrically.
Quick Recap
| Quantity | Formula | SI Unit | Sign / Range |
|---|---|---|---|
| Work | W = F s cos θ | joule (J) | +, 0, or − |
| Kinetic Energy | KE = ½ m v² | joule (J) | always ≥ 0 |
| Gravitational PE | PE = m g h | joule (J) | ≥ 0 (above reference) |
Competency-Based Questions
Reema observes four situations during her morning walk: (i) a labourer pushing a cart 10 m with 80 N along the direction of motion, (ii) a girl carrying a school bag horizontally for 30 m, (iii) a stone falling 5 m from a balcony, and (iv) a friend pushing a stuck wall for 2 minutes without moving it.
Q1. In which of the four situations is positive work done? L2
(b) In (i) force and displacement are aligned (positive work). In (iii) gravity and displacement are both downward (positive work). In (ii) force on the bag is vertical, displacement horizontal → zero work. In (iv) there is no displacement → zero work.
Q2. Calculate the work done by the labourer in situation (i). L3
W = F s cos θ = 80 × 10 × cos 0° = 800 J.
Q3. A 0.5 kg ball moves at 4 m s⁻¹. Find its kinetic energy. L3
KE = ½ × 0.5 × 4² = ½ × 0.5 × 16 = 4 J.
Q4. True or False: "If the speed of a body is doubled, its kinetic energy also doubles." L2
False. KE depends on v², so doubling v makes KE four times the original value.
Q5. A coolie carries a 25 kg trunk 40 m on a level platform. Why is the work done by him against gravity zero, even though he feels tired? L4
The gravitational force on the trunk acts vertically downward, but the displacement is horizontal. Angle between F and s is 90°, so W = F s cos 90° = 0. The tiredness is due to internal muscular work — chemical energy is consumed inside his body even when no external work is done on the trunk against gravity.
Assertion–Reason Questions
Options: (A) Both A and R are true and R is the correct explanation of A. (B) Both true but R is not the correct explanation. (C) A true, R false. (D) A false, R true.
A: Work done by gravity on a satellite moving in a circular orbit around the Earth is zero.
R: The gravitational force on the satellite is perpendicular to its instantaneous displacement at every point on the circular orbit.
(A) Both true and R correctly explains A. With θ = 90°, cos θ = 0 → W = 0.
A: Kinetic energy of a body can never be negative.
R: Mass and the square of velocity are both always non-negative.
(A) Both true; R explains A. KE = ½ m v² is the product of two non-negative quantities, so KE ≥ 0.
A: When a body falls freely, the work done by gravity is negative.
R: During free fall, gravity acts downward and displacement is also downward.
(D) Assertion is false — work done by gravity during free fall is positive (force and displacement are in the same direction). Reason is true.
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