9th Class Physics Notes 2026 – Chapter 5

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Class 9 Physics Notes – Chapter 5: Work, Energy and Power

Punjab Board | 2026

A. Multiple Choice Questions

5.1. Work done is maximum when the angle between the force \(F\) and the displacement \(d\) is:

(a) 0°
(b) 30°
(c) 60°
(d) 90°

Correct Answer: (a) 0°

Explanation: Work is given by:

\[ W=Fd\cos\theta \]

Since \(\cos 0^\circ=1\), work done is maximum when force and displacement are in the same direction.

5.2. A joule can also be written as:

(a) kg · m · s⁻²
(b) kg · m · s⁻¹
(c) kg · m² · s⁻³
(d) kg · m² · s⁻²

Correct Answer: (d) kg · m² · s⁻²

Explanation: Work = Force × distance.

\[ \begin{aligned} W&=\text{kg}\cdot\text{m}\cdot\text{s}^{-2}\times\text{m}\\ &=\text{kg}\cdot\text{m}^2\cdot\text{s}^{-2} \end{aligned} \]

5.3. The SI unit of power is:

(a) Joule
(b) Newton
(c) Watt
(d) Second

Correct Answer: (c) Watt

Explanation: Power is measured in watts (W), which are equivalent to joules per second (J/s).

5.4. The power of a water pump is 2 kW. The amount of water it can raise in one minute to a height of 5 metres is:

(a) 1000 litres
(b) 1200 litres
(c) 2000 litres
(d) 2400 litres

Correct Answer: (d) 2400 litres

Explanation:

\[ P=\frac{W}{t} \]
\[ W=P\times t=2000\times60=120000\text{ J} \]

Using \(W=mgh\):

\[ 120000=m\times10\times5 \]
\[ m=\frac{120000}{50} \]
\[ m=2400\text{ kg}=2400\text{ litres} \]

5.5. A bullet of mass 0.05 kg has a speed of 300 m s⁻¹. Its kinetic energy will be:

(a) 2250 J
(b) 4500 J
(c) 1500 J
(d) 1125 J

Correct Answer: (a) 2250 J

Explanation:

\[ KE=\frac{1}{2}mv^2 \]
\[ KE=\frac{1}{2}\times0.05\times(300)^2 \]
\[ KE=0.025\times90000 \]
\[ \boxed{KE=2250\text{ J}} \]

5.6. If a car doubles its speed, its kinetic energy will be:

(a) The same
(b) Doubled
(c) Increased to three times
(d) Increased to four times

Correct Answer: (d) Increased to four times

Explanation: Kinetic energy is proportional to the square of speed.

\[ KE\propto v^2 \]

If speed changes from \(v\) to \(2v\), then:

\[ KE’\propto(2v)^2=4v^2 \]

Therefore, kinetic energy becomes four times the original value.

5.7. The energy possessed by a body by virtue of its position is:

(a) Kinetic energy
(b) Potential energy
(c) Chemical energy
(d) Solar energy

Correct Answer: (b) Potential energy

Explanation: Potential energy is energy stored due to the position or configuration of a body.

5.8. The magnitude of momentum of an object is doubled. The kinetic energy of the object will:

(a) Double
(b) Increase to four times
(c) Reduce to one-half
(d) Remain the same

Correct Answer: (b) Increase to four times

Explanation:

\[ KE=\frac{p^2}{2m} \]

If momentum changes from \(p\) to \(2p\):

\[ KE’=\frac{(2p)^2}{2m} \]
\[ KE’=\frac{4p^2}{2m}=4KE \]

Therefore, kinetic energy increases to four times its original value.

5.9. Which of the following is not a renewable energy source?

(a) Hydroelectric energy
(b) Fossil fuels
(c) Wind energy
(d) Solar energy

Correct Answer: (b) Fossil fuels

Explanation: Fossil fuels, including coal, petroleum and natural gas, are exhaustible and non-renewable.

B. Short Answer Questions

5.1. What is the work done on an object that remains at rest when a force is applied to it?

Answer: Zero work is done because the displacement is zero.

\[ d=0 \]
\[ W=F\times0=0 \]

5.2. A slow-moving car may have more kinetic energy than a fast-moving motorcycle. How is this possible?

Answer: Kinetic energy depends on both mass and velocity.

\[ KE=\frac{1}{2}mv^2 \]

Even if a car moves slowly, its much greater mass compared with a motorcycle can result in higher kinetic energy.

5.3. A force \(F_1\) does 5 J of work in 10 s. Another force \(F_2\) does 3 J of work in 5 s. Which force delivers greater power?

Answer:

Power delivered by \(F_1\):

\[ P_1=\frac{W}{t}=\frac{5}{10} \]
\[ P_1=0.5\text{ W} \]

Power delivered by \(F_2\):

\[ P_2=\frac{W}{t}=\frac{3}{5} \]
\[ P_2=0.6\text{ W} \]

Therefore, force \(F_2\) delivers greater power.

5.4. A woman runs up a flight of stairs. The gain in her gravitational potential energy is 4500 J. If she runs up the same stairs with twice the speed, what will be her gain in potential energy?

Answer: Gravitational potential energy depends on mass, gravitational acceleration and height.

\[ PE=mgh \]

Speed does not affect the gain in potential energy. Therefore, the gain remains 4500 J.

5.5. Define work and its SI unit.

Answer: Work is the product of the force applied to an object, its displacement, and the cosine of the angle between the force and displacement.

\[ W=Fd\cos\theta \]

Its SI unit is the joule (J).

5.6. What is the potential energy of a body of mass \(m\) when it is raised through a height \(h\)?

Answer: The gravitational potential energy of a body raised through a height \(h\) is:

\[ PE=mgh \]

Where:

  • \(m\) = mass of the body

  • \(g\) = gravitational acceleration

  • \(h\) = height

5.7. Find an expression for the kinetic energy of a moving body.

Answer: Consider a body of mass \(m\), initially at rest (\(v_i=0\)), accelerated to velocity \(v\) by a force \(F\) over a distance \(s\).

Work done:

\[ W=Fs \]

Since \(F=ma\):

\[ W=mas \]

Using the equation of motion:

\[ v_f^2=v_i^2+2as \]

Since \(v_i=0\) and \(v_f=v\):

\[ v^2=2as \]
\[ as=\frac{v^2}{2} \]

Therefore:

\[ W=m\left(\frac{v^2}{2}\right) \]

Thus, the kinetic energy of the body is:

\[ \boxed{KE=\frac{1}{2}mv^2} \]

5.8. Define the efficiency of a working system. Why can a system not have 100% efficiency?

Answer: Efficiency is the ratio of useful work output to total energy input, expressed as a percentage.

\[ \text{Efficiency}=\frac{\text{Output}}{\text{Input}}\times100\% \]

No real system can have 100% efficiency because some energy is inevitably transferred to unwanted forms, such as heat and sound, and through friction.

5.9. What is power? Define the unit used for it.

Answer: Power is the rate of doing work.

\[ P=\frac{W}{t} \]

Its SI unit is the watt (W). One watt is the power when one joule of work is done in one second.

\[ 1\text{ W}=1\text{ J/s} \]

5.10. Differentiate between renewable and non-renewable energy sources.

Answer:

  • Renewable energy sources: These are naturally replenished. Examples include solar, wind and hydroelectric energy.

  • Non-renewable energy sources: These are finite and exhaustible. Examples include coal, oil and natural gas.

C. Constructed Response Questions

5.1. Can the kinetic energy of a body ever be negative?

Answer: No. Kinetic energy is given by:

\[ KE=\frac{1}{2}mv^2 \]

Mass is positive, and the square of velocity, \(v^2\), cannot be negative. Therefore, kinetic energy cannot be negative.

5.2. Which has greater kinetic energy: an object travelling with velocity \(v\), or an object twice as heavy travelling with velocity \(\frac{1}{2}v\)?

Answer:

For the first object:

\[ KE_1=\frac{1}{2}mv^2 \]

For the second object, mass \(=2m\) and velocity \(=\frac{v}{2}\):

\[ KE_2=\frac{1}{2}(2m)\left(\frac{v}{2}\right)^2 \]
\[ KE_2=m\frac{v^2}{4} \]
\[ KE_2=\frac{1}{4}mv^2 \]

Since:

\[ KE_1=\frac{1}{2}mv^2 \]

Therefore:

\[ KE_2=\frac{1}{2}KE_1 \]

The first object has greater kinetic energy.

5.3. A car is moving along a curved road at constant speed. Does its kinetic energy change?

Answer: No. Kinetic energy is a scalar quantity that depends on the magnitude of velocity, or speed.

\[ KE=\frac{1}{2}mv^2 \]

Since the speed remains constant, kinetic energy also remains constant, even though the direction of motion changes.

5.4. Comment on the statement: “An object has one joule of potential energy.”

Answer: It means that the object possesses one joule of potential energy relative to a chosen reference level. This can correspond to the work done against gravity to raise the object to its current position.

For example, lifting a mass of 100 g through a height of 1 metre requires approximately one joule of work, using \(g\approx10\text{ m/s}^2\).

5.5. While driving on a motorway, the tyre of a vehicle sometimes bursts. What may be its cause?

Answer: Possible causes include:

  • High temperatures caused by friction.

  • Overloading the vehicle.

  • Underinflation, which causes excessive tyre flexing and heating.

  • High-speed driving on hot pavement.

5.6. While playing cricket on a street, the ball smashes a window pane. Describe the energy changes in this event.

Answer: Chemical energy in the batter’s muscles is converted into the kinetic energy of the bat and then into the kinetic energy of the moving ball. Upon impact, the ball’s kinetic energy is converted into sound energy, thermal energy and the energy involved in the deformation and fracture of the glass.

5.7. A man rowing a boat upstream is at rest with respect to the shore. Is he doing work?

Answer: With respect to the shore, the mechanical work done on the man is zero if his displacement relative to the shore is zero.

\[ W=F\times0=0 \]

However, he expends muscular and chemical energy while rowing against the water current.

5.8. A cyclist goes downhill from the top of a steep hill without pedalling and reaches the top of the next hill. (i) Draw a diagram. (ii) Analyse this event in terms of potential and kinetic energy.

Answer:

(i) Diagram

The diagram should show a cyclist moving down one hill and up the next hill.

(ii) Energy changes

At the top of the first hill, the cyclist has maximum gravitational potential energy. As the cyclist moves downhill, potential energy decreases while kinetic energy increases, reaching its maximum at the bottom.

As the cyclist coasts up the next hill, kinetic energy is converted back into potential energy. The cyclist slows down and may come to a stop at the top if energy losses are neglected and the second hill reaches the same height.

D. Comprehensive Questions

5.1. What is meant by kinetic energy? State its unit. Describe how it is determined.

Answer: Kinetic energy is the energy possessed by a body due to its motion.

\[ KE=\frac{1}{2}mv^2 \]

Its SI unit is the joule (J). It is determined by calculating the work done in accelerating the body from rest to velocity \(v\).

5.2. State the law of conservation of energy. Explain it with the help of an example of a body falling from a certain height in terms of its potential energy and kinetic energy.

Answer: The law of conservation of energy states that energy cannot be created or destroyed; it can only be converted from one form to another.

For a freely falling body, gravitational potential energy decreases as the body falls, while kinetic energy increases. The total mechanical energy, \(PE+KE\), remains constant throughout the fall, assuming no energy losses.

5.3. Differentiate between renewable and non-renewable sources of energy. Give three examples for each.

Answer:

  • Renewable sources: These are naturally replenished. Examples: solar energy, wind energy and hydroelectric energy.

  • Non-renewable sources: These are finite and exhaustible. Examples: coal, petroleum and natural gas.

5.4. Explain what is meant by the efficiency of a machine. How is it calculated? Why is there a limit to the efficiency of a machine?

Answer: Efficiency is the ratio of useful output energy to total input energy, expressed as a percentage.

\[ \text{Efficiency}=\frac{\text{Output energy}}{\text{Input energy}}\times100\% \]

The efficiency of a real machine is below 100% because some energy is transferred to unwanted forms, such as heat and sound, and lost through friction.

5.5. Describe the process of electricity generation by drawing a block diagram for: (i) Hydroelectric power generation (ii) Fossil fuels.

Answer:

(i) Hydroelectric power generation

Water reservoir → Penstock → Turbine → Generator → Electrical energy

(ii) Electricity generation using fossil fuels

Fuel combustion → Boiler (steam) → Turbine → Generator → Electrical energy

E. Numerical Problems

5.1. A force of 20 N acting at an angle of 60° to the horizontal is used to pull a box through a distance of 3 m across a floor. How much work is done?

Answer:

Given:

  • \(F=20\text{ N}\)

  • \(\theta=60^\circ\)

  • \(d=3\text{ m}\)

Using:

\[ W=Fd\cos\theta \]
\[ W=20\times3\times\cos60^\circ \]
\[ W=60\times0.5 \]
\[ \boxed{W=30\text{ J}} \]

5.2. A body moves a distance of 5 metres in a straight line under the action of a force of 8 newtons. If the work done is 20 joules, find the angle which the force makes with the direction of motion of the body.

Answer:

Given:

  • \(d=5\text{ m}\)

  • \(F=8\text{ N}\)

  • \(W=20\text{ J}\)

Using:

\[ W=Fd\cos\theta \]
\[ 20=8\times5\times\cos\theta \]
\[ 20=40\cos\theta \]
\[ \cos\theta=\frac{20}{40}=0.5 \]
\[ \boxed{\theta=60^\circ} \]

5.3. An engine raises 100 kg of water through a height of 80 m in 25 s. What is the power of the engine?

Answer:

Given:

  • \(m=100\text{ kg}\)

  • \(h=80\text{ m}\)

  • \(t=25\text{ s}\)

  • \(g=10\text{ m/s}^2\)

Work done:

\[ W=mgh \]
\[ W=100\times10\times80 \]
\[ W=80000\text{ J} \]

Power:

\[ P=\frac{W}{t}=\frac{80000}{25} \]
\[ \boxed{P=3200\text{ W}} \]

5.4. A body of mass 20 kg is at rest. A 40 N force acts on it for 5 seconds. What is the kinetic energy of the body at the end of this time?

Answer:

Given:

  • \(m=20\text{ kg}\)

  • \(v_i=0\)

  • \(F=40\text{ N}\)

  • \(t=5\text{ s}\)

Acceleration:

\[ a=\frac{F}{m}=\frac{40}{20} \]
\[ a=2\text{ m/s}^2 \]

Final velocity:

\[ v_f=v_i+at \]
\[ v_f=0+(2\times5)=10\text{ m/s} \]

Kinetic energy:

\[ KE=\frac{1}{2}mv^2 \]
\[ KE=\frac{1}{2}\times20\times10^2 \]
\[ \boxed{KE=1000\text{ J}} \]

5.5. A ball of mass 160 g is thrown vertically upward. The ball reaches a height of 20 m. Find the potential energy gained by the ball at this height.

Answer:

Given:

  • \(m=160\text{ g}=0.16\text{ kg}\)

  • \(h=20\text{ m}\)

  • \(g=10\text{ m/s}^2\)

Potential energy:

\[ PE=mgh \]
\[ PE=0.16\times10\times20 \]
\[ \boxed{PE=32\text{ J}} \]

5.6. A 0.14 kg ball is thrown vertically upward with an initial velocity of 35 m s⁻¹. Find the maximum height reached by the ball.

Answer:

Given:

  • \(m=0.14\text{ kg}\)

  • \(v_i=35\text{ m/s}\)

  • \(v_f=0\)

  • \(g=10\text{ m/s}^2\)

Using the equation of motion:

\[ v_f^2=v_i^2-2gh \]
\[ 0=35^2-2(10)h \]
\[ 20h=1225 \]
\[ h=\frac{1225}{20} \]
\[ \boxed{h=61.25\text{ m}} \]

5.7. A girl is swinging on a swing. At the lowest point of her swing, she is 1.2 m from the ground, and at the highest point she is 2.0 m from the ground. What is her maximum velocity and where?

Answer:

Height difference:

\[ \Delta h=2.0-1.2=0.8\text{ m} \]

By conservation of energy, the loss in potential energy equals the gain in kinetic energy:

\[ mg\Delta h=\frac{1}{2}mv^2 \]

Cancelling \(m\) and rearranging:

\[ v=\sqrt{2g\Delta h} \]
\[ v=\sqrt{2\times10\times0.8} \]
\[ v=\sqrt{16} \]
\[ \boxed{v=4\text{ m/s}} \]

The maximum velocity occurs at the lowest point of the swing.

5.8. A person pushes a lawn mower with a force of 50 N making an angle of 45° with the horizontal. If the mower is moved through a distance of 20 m, how much work is done?

Answer:

Given:

  • \(F=50\text{ N}\)

  • \(\theta=45^\circ\)

  • \(d=20\text{ m}\)

Using:

\[ W=Fd\cos\theta \]
\[ W=50\times20\times\cos45^\circ \]
\[ W=1000\times0.7071 \]
\[ \boxed{W\approx707\text{ J}} \]

5.9. Calculate the work done in: (i) Pushing a 5 kg box up a frictionless inclined plane 10 m long that makes an angle of 30° with the horizontal. (ii) Lifting the box vertically from the ground to the top of the inclined plane.

Answer:

(i) Work done along the inclined plane

Given:

  • \(m=5\text{ kg}\)

  • \(g=10\text{ m/s}^2\)

  • \(d=10\text{ m}\)

  • \(\theta=30^\circ\)

The force required along the frictionless incline is:

\[ F=mg\sin\theta \]
\[ F=5\times10\times\sin30^\circ \]
\[ F=50\times0.5=25\text{ N} \]

Work done:

\[ W=Fd=25\times10 \]
\[ \boxed{W=250\text{ J}} \]

(ii) Work done in lifting the box vertically

Height of the incline:

\[ h=10\sin30^\circ=5\text{ m} \]

Work done:

\[ W=mgh \]
\[ W=5\times10\times5 \]
\[ \boxed{W=250\text{ J}} \]

5.10. A box of mass 10 kg is pushed up along a ramp 15 m long with a force of 80 N. If the box rises through a height of 5 m, what is the efficiency of the system?

Answer:

Given:

  • \(m=10\text{ kg}\)

  • \(h=5\text{ m}\)

  • Applied force \(=80\text{ N}\)

  • Distance \(=15\text{ m}\)

  • \(g=10\text{ m/s}^2\)

Output work:

\[ W_{\text{out}}=mgh \]
\[ W_{\text{out}}=10\times10\times5=500\text{ J} \]

Input work:

\[ W_{\text{in}}=Fd=80\times15=1200\text{ J} \]

Efficiency:

\[ \text{Efficiency}=\frac{W_{\text{out}}}{W_{\text{in}}}\times100\% \]
\[ \text{Efficiency}=\frac{500}{1200}\times100\% \]
\[ \boxed{\text{Efficiency}=41.7\%} \]

5.11. A force of 600 N acts on a box to push it 5 m in 15 s. Calculate the power.

Answer:

Given:

  • \(F=600\text{ N}\)

  • \(d=5\text{ m}\)

  • \(t=15\text{ s}\)

Work done:

\[ W=Fd=600\times5=3000\text{ J} \]

Power:

\[ P=\frac{W}{t}=\frac{3000}{15} \]
\[ \boxed{P=200\text{ W}} \]

5.12. A 40 kg boy runs up stairs 10 m high in 8 s. What power does he develop?

Answer:

Given:

  • \(m=40\text{ kg}\)

  • \(h=10\text{ m}\)

  • \(t=8\text{ s}\)

  • \(g=10\text{ m/s}^2\)

Work done:

\[ W=mgh \]
\[ W=40\times10\times10=4000\text{ J} \]

Power:

\[ P=\frac{W}{t}=\frac{4000}{8} \]
\[ \boxed{P=500\text{ W}} \]

5.13. A force \(F\) acts through a distance \(L\) on a body. The force is then increased to \(2F\), which acts through a further distance of \(2L\). Sketch a force-displacement graph and calculate the total work done.

Answer: Work done is equal to the area under the force-displacement graph.

Part 1:

Force \(=F\), displacement \(=L\).

\[ W_1=F\times L=FL \]

Part 2:

Force \(=2F\), additional displacement \(=2L\).

\[ W_2=2F\times2L=4FL \]

Total work done:

\[ W_{\text{total}}=W_1+W_2 \]
\[ W_{\text{total}}=FL+4FL \]
\[ \boxed{W_{\text{total}}=5FL} \]

The force-displacement graph should show force \(F\) over the first distance \(L\), followed by force \(2F\) over the next distance \(2L\). The total area under the graph is \(5FL\).

 

Notes by Sir Dilawar

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