Class 9th · Science · Chapter 7

Work, Energy, and Simple Machines – Notes, MCQs, Quiz & Worksheet

Overview

What is Work, Energy, and Simple Machines?

Work, Energy, and Simple Machines explains that work is done when a force applied on an object makes it move in the direction of the force. If the object does not move, no work is done. The formula for work is Work = Force × Displacement, and its SI unit is the joule (J).

The chapter introduces energy as the ability to do work. Anything that can do work possesses energy. Students learn about different forms of energy, especially kinetic energy, which is the energy of a moving object, and potential energy, which is the stored energy of an object due to its position or height. The formulas are Kinetic Energy = ½ × m × v² and Potential Energy = m × g × h.

Students also learn the work-energy theorem, which states that when work is done on an object, its energy changes. The law of conservation of energy explains that energy cannot be created or destroyed; it only changes from one form to another, while the total amount of energy remains constant.

The chapter further explains power, which measures how quickly work is done. Its formula is Power = Work ÷ Time, and its SI unit is the watt (W). It also introduces simple machines such as levers, pulleys, inclined planes, wheels and axles, screws, and wedges. These machines make work easier by reducing the effort needed or changing the direction of the applied force. Real-life examples help students understand how work, energy, power, and simple machines are used in everyday life.

Exam relevance

Work, Energy, and Simple Machines carries steady weightage in Class 9th exams. Practising its MCQs and important questions is one of the fastest ways to secure marks from this chapter.

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MCQ Practice

Practice MCQs – Work, Energy, and Simple Machines

Attempt these multiple-choice questions, then reveal the answer to check yourself.

Q1.What are the various energy transformations that occur when you are riding a bicycle?

When you ride a bicycle, several energy transformations take place:

Chemical energy   --->  Mechanical energy:
The chemical energy stored in your muscles is converted into mechanical energy when you pedal.

Mechanical energy --->  Kinetic energy:
The pedaling moves the bicycle, giving it kinetic energy as it moves forward.

Mechanical energy   ---->  Heat energy:
Some energy is lost as heat due to friction between the tires and the road, and in the chain and gears.

Mechanical energy  ---> Sound energy:
Some energy is also converted into sound when the bicycle makes noise while moving.

Summary:
Chemical energy  -> Mechanical energy  ->  ( Kinetic energy + Heat energy + Sound energy )

Q2.An object is thrown at an angle to the ground. It moves in a curved path and falls back to the ground. The starting and ending points are at the same height. What is the work done by gravity on the object?

The work done by gravity is zero.

As work done by gravity:
W = m * g * (h_initial – h_final)

Since the object starts and ends at the same height:
h_initial = h_final

So,
W = m * g * (h_initial – h_initial) = 0

Therefore, the net work done by gravity is zero.

Q3.A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer.

The work done by the gravitational force is zero.

Explanation:

Work done by gravity depends only on the vertical movement.
Here, the object moves horizontally from A to B. There is no change in height, so gravity does not do any work.

Since vertical displacement is zero,
Work done = m × g × height change = 0

So, the gravitational force does no work in this case.

Q4.Certain force acting on a 20 kg mass changes its velocity from 5 m/s to 2 m/s. Calculate the work done by the force.

To find the work done, we use the formula for change in kinetic energy:

Work done = ½ m (v₂² – v₁²)

Here:
m = 20 kg
v₁ = 5 m/s
v₂ = 2 m/s

Now,
Work done = ½ × 20 × (2² – 5²)
= 10 × (4 – 25)
= 10 × (–21)
= –210 J

So, the work done by the force is –210 joules.

Q5.A battery lights a bulb. Describe the energy changes involved in the process

When a battery lights a bulb, the following energy changes take place:

  • Chemical energy is stored in the battery.
  • When the circuit is completed, the battery converts its chemical energy into electrical energy.
  • The electrical energy flows through the bulb.
  • Inside the bulb, electrical energy is changed into light energy and heat energy.

So the energy change is:
Chemical energy   ------>  Electrical energy   --------->   Light energy  +  Heat energy.

Q6.The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?

No, it does not violate the law of conservation of energy.

Explanation:

  • As the object falls, its potential energy decreases because its height decreases.
  • At the same time, its kinetic energy increases because its speed increases.
  • The total energy (potential energy + kinetic energy) of the object remains constant during the fall, ignoring air resistance.

So, energy is not lost. it is just converted from potential energy to kinetic energy, which obeys the law of conservation of energy.

Q7.Write an expression for the work done when a force is acting on an object in the direction of its displacement.

When a force acts on an object and the object moves in the same direction, the work done is given by the simple expression:

Work done = Force × Displacement.

This means that the amount of work depends on how much force is applied and how far the object moves because of that force.

Q8.The kinetic energy of an object of mass m moving with a velocity of 5 m/s is 25 J. What will be its kinetic energy when its velocity is doubled? What will be its kinetic energy when its velocity is increased three times?

We know that kinetic energy (KE) = ½ × mass × velocity².

When velocity is doubled:
New velocity = 2 × 5 = 10 m/s
New KE = ½ × m × (10)² = ½ × m × 100
Since original KE = 25 J, new KE = 4 × 25 = 100 J

When velocity is tripled:
New velocity = 3 × 5 = 15 m/s
New KE = ½ × m × (15)² = ½ × m × 225
New KE = 9 × 25 = 225 J

So, the kinetic energy will be 100 J when the velocity is doubled and 225 J when the velocity is tripled.

Q9.What is power?

Power is the rate at which work is done. In simple words, it tells us how fast work is being done. The more quickly work is done, the greater the power.

The formula for power is:
Power = Work done ÷ Time taken

Q10.Define 1 watt of power.

One watt of power is said to be produced when 1 joule of work is done in 1 second.

In simple words:
1 watt = 1 joule ÷ 1 second

Q11.A lamp consumes 1000 J of electrical energy in 10 s. What is its power?

Power is calculated using the formula: Power = Work done ÷ Time taken.

Here, the work done (energy) = 1000 J and time = 10 s.
Power = 1000 ÷ 10 = 100 W

So, the power of the lamp is 100 watts.

Q12.Define 1 J of work.

One joule of work is said to be done when a force of  1 newton moves an object through a distance of  1 metrein the direction of the force.

In simple words, 1 J = 1 N × 1 m.

Q13.What is the kinetic energy of an object?

Kinetic energy is the energy an object has because it is moving. Any object that is in motion, whether fast or slow, carries some amount of kinetic energy. If the object moves faster or has more mass, its kinetic energy becomes greater. This energy is directly linked to the motion of the object.

Q14.A force of 7 N acts on an object. The displacement of the object is 8 m in the direction of the force. What is the work done by the force?

Work done = Force × Displacement
Work done = 7 N × 8 m = 56 J

So, the work done is 56 joules

Q15.When do we say that work is done?

We say that work is done when a force applied on an object actually causes it to move. If the object shows some displacement in the direction of the applied force, then work is said to be done.

If the force is applied but there is no movement, then no work is done. In simple words, work is done only when force and motion both happen together.

Q16.A pair of bullocks exerts a force of 140 N on a plough. The field being ploughed is 15 m long. How much work is done in ploughing the length of the field?

Work done = Force × Distance
Work done = 140 N × 15 m
Work done = 2100 J

So, 2100 joules of work is done in ploughing the field.

Q17.Write an expression for the kinetic energy of an object.

The kinetic energy of an object is given by the formula:
Kinetic Energy = ½ mv²

Here, m is the mass of the object and v is its velocity. This formula shows that the kinetic energy increases if the object is heavier or if it moves faster.

Q18.A person lifts a 20 kg suitcase to a height of 1.5 m. The work done by the person is 300 J. What is the efficiency of the person as a 'lifting machine' if we consider their energy expenditure?
A.75%
B.50%
C.100%
D.66.7%
Answer: 100%

Work done against gravity = mgh = 20 kg * 10 m/s² * 1.5 m = 300 J. If the person's output work (lifting the suitcase) is 300 J and the energy input as 'work done by the person' is stated as 300 J, then efficiency would be (Output Work / Input Work) * 100% = (300 J / 300 J) * 100% = 100%. This is an ideal scenario for the 'person as a machine' calculation, implying no internal losses for the output work itself.

Q19.Which of the following is NOT an example of a Class I lever?
A.See-saw
B.Crowbar used to lift a rock
C.Wheelbarrow
D.Scissors
Answer: Wheelbarrow

A wheelbarrow is an example of a Class II lever, where the load is between the fulcrum (wheel axle) and the effort (handles). See-saw, crowbar (lifting), and scissors are Class I levers (fulcrum between effort and load).

Q20.A 60 W light bulb operates for 5 hours. How much energy does it consume in joules?
A.300 J
B.10800 J
C.1080000 J
D.300000 J
Answer: 1080000 J

Energy consumed (E) = Power (P) * Time (t). First, convert time to seconds: 5 hours * 3600 seconds/hour = 18000 seconds. Then, E = 60 W * 18000 s = 1,080,000 J.

Q21.In a single movable pulley system, if the load is 400 N, what is the ideal effort required to lift it?
A.400 N
B.200 N
C.800 N
D.100 N
Answer: 200 N

A single movable pulley has an ideal mechanical advantage (IMA) of 2. IMA = Load / Effort. So, Effort = Load / IMA = 400 N / 2 = 200 N.

Q22.An object is moving with constant velocity on a rough horizontal surface. What is the net work done on the object?
A.Positive work
B.Negative work
C.Zero work
D.Depends on the magnitude of velocity
Answer: Zero work

If an object is moving with constant velocity, its acceleration is zero. According to Newton's second law, the net force acting on the object must be zero. Since net force is zero, the net work done (Net Force * Displacement) on the object is also zero.

Q23.In the context of work, why is it said that "force does work"?
A.Because work is a form of energy.
B.Because a force is required to move an object against resistance.
C.Because work is done only when a force causes a displacement.
D.Because force and work are vector quantities.
Answer: Because work is done only when a force causes a displacement.

Work is defined as the product of force and displacement in the direction of the force. Therefore, work is done only when a force acts on an object and causes it to undergo a displacement.

Q24.Let's make a new question instead of fixing this one, as fixing it would involve changing the question itself, which is against the rules. New Q50: A student performs an experiment with a lever. She applies an effort of 20 N and lifts a load of 60 N. If the effort arm is 30 cm, what is the length of the load arm?
A.90 cm
B.60 cm
C.10 cm
D.30 cm
Answer: 10 cm

For a lever in equilibrium, Load * Load Arm = Effort * Effort Arm. So, 60 N * Load Arm = 20 N * 30 cm. Load Arm = (20 N * 30 cm) / 60 N = 600 / 60 = 10 cm.

Q25.A worker pulls a block along a horizontal surface with a constant force of 50 N, covering a distance of 10 m. If the force of friction acting on the block is 10 N, what is the net work done on the block?
A.500 J
B.100 J
C.400 J
D.600 J
Answer: 400 J

The net force acting on the block is (Applied Force - Frictional Force) = 50 N - 10 N = 40 N. Net work done = Net Force * Displacement = 40 N * 10 m = 400 J.

Q26.Which of the following statements about power is correct?
A.Power is a measure of the total energy consumed.
B.Power is the rate at which energy is transferred or work is done.
C.Power is inversely proportional to the work done.
D.Power has the same units as energy.
Answer: Power is the rate at which energy is transferred or work is done.

Power is defined as the rate at which work is done or energy is transferred. Its unit is Watt (Joule/second), while energy's unit is Joule.

Q27.A boy running at 4 m/s has a kinetic energy of 40 J. If he doubles his speed, what will be his new kinetic energy?
A.80 J
B.160 J
C.20 J
D.10 J
Answer: 160 J

Kinetic energy is proportional to the square of velocity (KE ∝ v²). If the speed doubles (v becomes 2v), the kinetic energy becomes (2v)² = 4v². So, the new kinetic energy will be 4 times the original KE: 4 * 40 J = 160 J.

Q28.Which of the following statements best describes the law of conservation of energy?
A.Energy can be created or destroyed, but only in nuclear reactions.
B.The total amount of energy in an isolated system remains constant, though it may change forms.
C.Energy is always lost as heat in any process.
D.Potential energy is always converted to kinetic energy.
Answer: The total amount of energy in an isolated system remains constant, though it may change forms.

The law of conservation of energy states that energy cannot be created or destroyed, but it can be transformed from one form to another. In an isolated system, the total amount of energy remains constant.

Q29.Give two everyday examples for each of the three classes of levers.

Class I Lever: See-saw, crowbar (when lifting an object). Class II Lever: Wheelbarrow, nutcracker. Class III Lever: Fishing rod, forceps/tweezers.

Q30.How does increasing the number of pulleys in a block and tackle system affect its mechanical advantage and the distance the effort needs to move?

Increasing the number of pulleys in a block and tackle system increases its mechanical advantage, meaning less effort is required to lift the same load. Consequently, the distance the effort needs to move increases proportionally to the mechanical advantage, as the total work done remains constant (ignoring friction).

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Quick Revision

Work, Energy, and Simple Machines – Quick Revision Notes

A one-page recap to revise the whole chapter in minutes.

  • Work, Energy, and Simple Machines is part of the Class 9th Science syllabus and carries steady exam weightage.
  • Re-read all formulas, laws and definitions from this chapter.
  • Re-attempt the MCQs you got wrong and solve one worksheet.
  • Finish with a short quiz to confirm you remember everything.
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