Work, Energy, and | Class 9 Science Notes
By ConceptScroll Team · Published on 17 July 2026 · 5 min read
Work, Energy, and – this guide gives you a concise, exam-ready overview of Work, Energy, and from Class 9 Science, written by ConceptScroll editors and reviewed against the latest NCERT textbook.
Conservation of mechanical energy
The principle of conservation of mechanical energy states that in the absence of external forces like friction, the total mechanical energy (sum of kinetic and potential energy) of an object remains constant.
Consider an object of mass m lifted to a height h and then dropped. At the top (point A), the object has maximum potential energy U = m g h and zero kinetic energy (since velocity u = 0). As it falls, potential energy decreases while kinetic energy increases.
At any intermediate point B during the fall, the height is h', velocity is v, and the energies are:
- Potential energy = m g h'
- Kinetic energy = 1/2 m v²
Using kinematic equations, h' = h - 1/2 g t² and v = g t, the sum of potential and kinetic energy at point B is m g h, same as at the top.
This shows that the loss in potential energy converts into an equal gain in kinetic energy, keeping mechanical energy constant.
An activity with a simple pendulum demonstrates this conservation: the pendulum bob reaches nearly the same height on both sides, showing mechanical energy is conserved except for small losses due to friction and air resistance.
Conservation of mechanical energy simplifies problem-solving by allowing calculation of final velocities or positions without detailed force analysis.
Examples include a child sliding down a slide where potential energy converts to kinetic energy, and escape ramps where kinetic energy is converted to potential energy and work done against friction.
This principle is foundational in physics and engineering, explaining many natural and man-made phenomena.
📊 Diagram: Fig. 7.19: An object falling freely due to gravity; Fig. 7.20: A pendulum; Fig. 7.21: Escape ramp
🧪 Activity: Activity 7.2: Experiment with a simple pendulum to observe conservation of mechanical energy by noting the height reached by the bob during oscillations.
🔗 Connection: This section leads to the study of power and simple machines, which apply work and energy concepts practically.
Frequently asked questions
Which factors determine the energy required to raise a flag from the ground to the top of a tall flagpole using a pulley? Does raising the flag slowly or quickly change the amount of work done? If the speed at which the flag is raised is doubled, how does the power requirement change? Explain your answers.
The energy required to raise a flag depends on the height of the flagpole and the weight (mass) of the flag. The work done is equal to the gravitational potential energy gained by the flag, which is mgh (mass × gravity × height).
Raising the flag slowly or quickly does not change the amount of work done because work depends only on force and displacement, not on time.
If the speed of raising the flag is doubled, the power requirement doubles because power is work done per unit time. Doubling s
A man of mass 60 kg rides a scooter of mass 100 kg. He accelerates the scooter to a velocity v. The next day, his son with a mass of 40 kg joins him as a passenger. If the scooter reaches the same speed on both days in the same time interval, what is the ratio of the fuel of the tank used on the two days? Assume that the energy transfer to the scooter happens entirely due to fuel, and no other losses occur due to air resistance and friction.
First day total mass = 60 + 100 = 160 kg Second day total mass = 60 + 40 + 100 = 200 kg
Kinetic energy on day 1 = (1/2) × 160 × v² = 80 v² Kinetic energy on day 2 = (1/2) × 200 × v² = 100 v²
Since energy transfer is from fuel, fuel used ∝ kinetic energy.
Ratio of fuel used on day 2 to day 1 = 100 v² / 80 v² = 5/4 = 1.25
So, the fuel used on the second day is 1.25 times that on the first day.
On a seesaw with sliding seats, a child is sitting on one side and an adult on the other side. The adult weighs twice that of the child. The seesaw however is balanced. Draw a figure which depicts this situation showing the distances from the fulcrum where the child and the adult are seated.
Let the weight of the child be W and the adult be 2W.
For balance, torque by child = torque by adult
W × d1 = 2W × d2
=> d1 = 2 d2
So, the child sits twice as far from the fulcrum as the adult.
Draw a seesaw with fulcrum in the center, child on one side at distance d1, adult on other side at distance d2 = d1/2.
This shows balance despite adult being heavier.
A ball of mass 2 kg is thrown up with a velocity of 20 m/s. (i) Identify the sign of the work done by gravity on the ball during its upward motion and its downward motion. (ii) If the ball reaches a height of 19.4 m, how much work was done by air resistance (assume g = 10 m/s²).
(i) Work done by gravity during upward motion is negative because gravity acts downward while displacement is upward. Work done by gravity during downward motion is positive because gravity and displacement are in the same direction.
(ii) Initial kinetic energy = (1/2) m v² = 0.5 × 2 × 20² = 400 J
Potential energy at max height = mgh = 2 × 10 × 19.4 = 388 J
Work done by air resistance = Initial KE - Potential Energy = 400 - 388 = 12 J (energy lost due to air resistance)
So, air resistance di
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