A child pushes a swing from rest. The work done by the child on the swing is positive because:

the force and displacement are in the same direction

the force and displacement are in opposite directions

Assertion (A): A girl carrying a box horizontally at constant speed does no work on the box against gravity. Reason (R): The force applied by the girl against gravity is vertically upward, while the displacement of the box is horizontal, making them perpendicular.

Both A and R are true and R is the correct explanation of A.

Both A and R are true but R is not the correct explanation of A.

Define work done by a constant force acting on an object. Write its mathematical expression and SI unit.

Work done by a constant force is the product of the force and the displacement of the object in the direction of the force. Mathematically, W = F × s. Its SI unit is joule (J).

A horizontal force of 15 N pushes a box along a floor for 4 m in the direction of the force. Then the force is reversed and the box is pushed back for 2 m under the same magnitude of force. Calculate the total work done by the force on the box.

Work during forward push: W1 = F × s1 = 15 N × 4 m = 60 J. Work during backward push (force opposite to displacement): W2 = F × (– s2) = 15 N × (–2 m) = –30 J. Total work = W1 + W2 = 60 J + (–30 J) = 30 J.

Derive the relation for the work done by a force on an object of mass m that changes its velocity from u to v. Hence, show that this work is equal to ½ m(v² – u²).

Let a constant force F act on the object, producing an acceleration a. By Newton’s second law, F = m a. If the object undergoes displacement s in the direction of force, work W = F s = m a s. Using the equation of motion, v² = u² + 2 a s, we get a s = (v² – u²)/2. Substituting, W = m × (v² – u²)/2 = ½ m (v² – u²). This is the required relation.

A constant horizontal force of 40 N pushes a box of mass 10 kg across a rough floor through a distance of 5 m. A frictional force of 10 N acts opposite to the motion. (a) Calculate the work done by the applied force. (b) Calculate the work done by friction. (c) Determine the net work done on the box. (d) Using the work-energy theorem, state the change in kinetic energy of the box.

(a) 40 N × 5 m = 200 J. (b) –10 N × 5 m = –50 J. (c) Net work = 150 J. (d) Change in kinetic energy = +150 J.

A stone of mass 2 kg is dropped from a height of 20 m. Take g = 10 m/s². (a) Use the work-energy theorem to find the speed of the stone just before it hits the ground. (b) The stone then penetrates 0.5 m into soft soil and comes to rest; find the average resistive force exerted by the soil. (c) Describe the energy transformation throughout the whole event.

(a) Work by gravity = mgh = 400 J. Initial KE=0 → final KE=400 J → v=20 m/s. (b) Resistive force F does work –F×0.5; change in KE = 0–400 J → –0.5F = –400 → F=800 N. (c) Gravitational PE → KE during fall; then KE → work against soil (heat + deformation).

Class 9 Science Chapter 7: Work, Energy and Simple Machines — Important Questions & Sample Paper

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Yes — this page has 44+ original Class 9 Science Chapter 7 (“Work, Energy and Simple Machines”) important questions with answers (Multiple Choice (MCQ), Assertion–Reason, Short Answer, Short Answer, Long Answer, Case Study). Practise them free, or generate a full CBSE board-pattern sample paper (80 marks) and export it to PDF or Word — in English & Hindi, for 2026-27.

Chapter 7 'Work, Energy and Simple Machines' builds on the basics of force and motion to introduce the scientific concept of work. Work is defined as the product of force and displacement in the direction of the force (W = F × s × cosθ). Students learn to distinguish when work is positive, negative, or zero through everyday examples such as pushing a swing or lifting a box. The work-energy theorem—stating that net work done equals the change in kinetic energy (W_net = ΔK = ½mv² − ½mu²)—is a central theme, enabling calculations of speed or work without directly using kinematics. Simple machines like levers, pulleys, and inclined planes are explored, with emphasis on mechanical advantage and efficiency. In CBSE exams, this chapter often features numerical problems: calculating work done by constant or frictional forces, using force-displacement graphs, and applying the work-energy theorem to determine final speeds or braking work. Conceptual questions test understanding of when work is zero and the physical meaning of positive versus negative work. Teachers can use qpaper.in’s vast question bank to generate practice papers that target these specific question types, helping students master both calculation techniques and conceptual clarity.

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Science — Work, Energy and Simple Machines

Class 9Time: 3 hrsMax Marks: 80

SECTION A

1.
A box is pushed along a rough floor. The friction force between the box and the floor is 12 N. If the box moves 1.5 m before coming to rest, the work done by friction on the box is:
(a) 18 J(b) -18 J(c) 8 J(d) -8 J
1
2.
A bullet of mass 25 g moving with a velocity of 200 m/s hits a wooden target and comes to rest after penetrating a distance of 40 cm. The average resistive force exerted by the target on the bullet is roughly:
(a) 1250 N(b) 2500 N(c) 5000 N(d) 1000 N
1
3.
A construction worker lifts a pile of 8 bricks, each of mass 2.5 kg, to a height of 4 m. The work done by the worker against gravity is (take g = 10 m s⁻²):
(a) 80 J(b) 200 J(c) 800 J(d) 2000 J
1

+ 41 more questions in the full paper

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In a CBSE exam, this chapter typically contributes questions across the following types. The last column shows how many original questions of each type we have ready in our bank for this chapter:

Question type	Marks each	In our bank
Multiple Choice (MCQ)	1 mark	13
Assertion–Reason	1 mark	6
Short Answer	2 marks	8
Short Answer	3 marks	6
Long Answer	5 marks	5
Case Study	4 marks	6

44 original, exam-style questions in our bank for this chapter — with answers.

Important & sample questions (with answers)

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Q1. A box is pushed along a rough floor. The friction force between the box and the floor is 12 N. If the box moves 1.5 m before coming to rest, the work done by friction on the box is:
1 mark
Multiple Choice (MCQ)
(A) 18 J(B) -18 J(C) 8 J(D) -8 J
▸ Answer▾ Answer
-18 J
Q2. A bullet of mass 25 g moving with a velocity of 200 m/s hits a wooden target and comes to rest after penetrating a distance of 40 cm. The average resistive force exerted by the target on the bullet is roughly:
1 mark
Multiple Choice (MCQ)
(A) 1250 N(B) 2500 N(C) 5000 N(D) 1000 N
▸ Answer▾ Answer
1250 N
Q3. A construction worker lifts a pile of 8 bricks, each of mass 2.5 kg, to a height of 4 m. The work done by the worker against gravity is (take g = 10 m s⁻²):
1 mark
Multiple Choice (MCQ)
(A) 80 J(B) 200 J(C) 800 J(D) 2000 J
▸ Answer▾ Answer
800 J
Q4. A child pushes a swing from rest. The work done by the child on the swing is positive because:
1 mark
Multiple Choice (MCQ)
(A) the force and displacement are in the same direction(B) the force and displacement are in opposite directions(C) no displacement occurs(D) the swing moves on its own
▸ Answer▾ Answer
the force and displacement are in the same direction
Q5. Assertion (A): A girl carrying a box horizontally at constant speed does no work on the box against gravity. Reason (R): The force applied by the girl against gravity is vertically upward, while the displacement of the box is horizontal, making them perpendicular.
1 mark
Assertion–Reason
(A) Both A and R are true and R is the correct explanation of A.(B) Both A and R are true but R is not the correct explanation of A.(C) A is true but R is false.(D) A is false but R is true.
▸ Answer▾ Answer
Both A and R are true and R is the correct explanation of A.
Q6. A car of mass 1200 kg accelerates uniformly from rest to 25 m/s. Using the work–energy theorem, find the work done by the net force on the car.
2 marks
Short Answer
▸ Answer▾ Answer
3,75,000 J (or 375 kJ)
Q7. Define work done by a constant force acting on an object. Write its mathematical expression and SI unit.
2 marks
Short Answer
▸ Answer▾ Answer
Work done by a constant force is the product of the force and the displacement of the object in the direction of the force. Mathematically, W = F × s. Its SI unit is joule (J).
Q8. A horizontal force of 15 N pushes a box along a floor for 4 m in the direction of the force. Then the force is reversed and the box is pushed back for 2 m under the same magnitude of force. Calculate the total work done by the force on the box.
3 marks
Short Answer
▸ Answer▾ Answer
Work during forward push: W1 = F × s1 = 15 N × 4 m = 60 J. Work during backward push (force opposite to displacement): W2 = F × (– s2) = 15 N × (–2 m) = –30 J. Total work = W1 + W2 = 60 J + (–30 J) = 30 J.
Q9. Derive the relation for the work done by a force on an object of mass m that changes its velocity from u to v. Hence, show that this work is equal to ½ m(v² – u²).
3 marks
Short Answer
▸ Answer▾ Answer
Let a constant force F act on the object, producing an acceleration a. By Newton’s second law, F = m a. If the object undergoes displacement s in the direction of force, work W = F s = m a s. Using the equation of motion, v² = u² + 2 a s, we get a s = (v² – u²)/2. Substituting, W = m × (v² – u²)/2 = ½ m (v² – u²). This is the required relation.
Q10. A constant horizontal force of 40 N pushes a box of mass 10 kg across a rough floor through a distance of 5 m. A frictional force of 10 N acts opposite to the motion. (a) Calculate the work done by the applied force. (b) Calculate the work done by friction. (c) Determine the net work done on the box. (d) Using the work-energy theorem, state the change in kinetic energy of the box.
5 marks
Long Answer
▸ Answer▾ Answer
(a) 40 N × 5 m = 200 J. (b) –10 N × 5 m = –50 J. (c) Net work = 150 J. (d) Change in kinetic energy = +150 J.
Q11. A stone of mass 2 kg is dropped from a height of 20 m. Take g = 10 m/s². (a) Use the work-energy theorem to find the speed of the stone just before it hits the ground. (b) The stone then penetrates 0.5 m into soft soil and comes to rest; find the average resistive force exerted by the soil. (c) Describe the energy transformation throughout the whole event.
5 marks
Long Answer
▸ Answer▾ Answer
(a) Work by gravity = mgh = 400 J. Initial KE=0 → final KE=400 J → v=20 m/s. (b) Resistive force F does work –F×0.5; change in KE = 0–400 J → –0.5F = –400 → F=800 N. (c) Gravitational PE → KE during fall; then KE → work against soil (heat + deformation).
Q12. A porter carries a 20 kg suitcase on his head. First, he walks horizontally for 100 m without changing his height. Then, he climbs a staircase of 5 m height. (g = 10 m/s²)
4 marks
Case Study
1. (i) Calculate the work done by the porter on the suitcase while walking horizontally.1 mark
2. (ii) How much work does he do in lifting the suitcase along the staircase?2 marks
3. (iii) What is the total work done by the porter on the suitcase during the entire journey?1 mark
▸ Answer▾ Answer
(a) 0 J (b) 1000 J (c) 1000 J

Frequently asked questions

What is the work-energy theorem for Class 9?

The work-energy theorem states that the net work done by all forces acting on an object equals the change in its kinetic energy. For an object of mass m moving from initial speed u to final speed v, it is expressed as W_net = ½mv² − ½mu². In problems, it is often used to find the final speed when the net work is known (e.g., work done by a constant force or by brakes), without using equations of motion.

When is work done by a force zero?

Work done is zero if (i) there is no displacement (e.g., pushing a wall), (ii) the force is perpendicular to the displacement (e.g., centripetal force in circular motion, or the force of gravity on a horizontally moving object), or (iii) the net force and displacement are zero over the path. The chapter uses examples like a man carrying a box horizontally at constant speed to illustrate this.

How do simple machines help in doing work?

Simple machines such as levers, pulleys, inclined planes, and screws make work easier by changing the magnitude or direction of the applied force. They provide mechanical advantage (MA = Load/Effort), allowing a smaller effort to lift a larger load, though the total work done remains the same (ignoring friction). Efficiency is the ratio of useful work output to total work input, often expressed as a percentage.

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