Work, Energy, and | Class 9 Science Notes
By ConceptScroll Team · Published on 17 July 2026 · 6 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.
Mechanical Energy
Mechanical energy is the energy possessed by an object due to its motion or position. It is the sum of kinetic energy (energy of motion) and potential energy (energy stored due to position or configuration).
Kinetic energy is the energy an object has because it is moving. For example, a moving bicycle or a rolling ball has kinetic energy. The kinetic energy (K) of an object of mass m moving with velocity v is given by the formula K = 1/2 m v².
This formula is derived using the work-energy theorem and kinematic equations. When a constant force acts on an object, causing it to accelerate from an initial velocity u to final velocity v over displacement s, the work done by the force equals the change in kinetic energy. Using Newton's second law (F = ma) and the equation v² = u² + 2as, the work done W = F × s = m a × s = 1/2 m (v² - u²).
If the object starts from rest (u = 0), its kinetic energy is K = 1/2 m v².
Potential energy is the energy stored in an object due to its position or configuration. For example, a stretched spring, a bent bow, or a raised object have potential energy. This energy can be released to do work, such as a stretched rubber band shooting a projectile or a raised ball falling under gravity.
The gravitational potential energy (U) of an object of mass m at height h above the ground is U = m g h, where g is acceleration due to gravity. This formula is valid near Earth's surface where g is approximately constant.
Mechanical energy is conserved in the absence of external forces like friction. For example, a freely falling object converts potential energy into kinetic energy, keeping the total mechanical energy constant.
Understanding mechanical energy and its conservation helps analyze many physical situations more simply than using forces alone.
📊 Diagram: Fig. 7.10: Different forms of energy; Fig. 7.11: Calculating change in kinetic energy using the work-energy theorem; Fig. 7.13: (a) A slingshot; (b) Shooting an arrow using a bow; Fig. 7.14: Spring (a) in its original shape, (b) in compressed shape, and (c) moving back to its original shape; Fig. 7.15: A system of two (a) magnets, and (b) electric charges; Fig. 7.16: Earth ball system
🧪 Activity: Activity 7.1: Investigate the depth of depressions created by dropping a ball from different heights onto sand to understand potential energy.
🔗 Connection: This section prepares for the study of conservation of mechanical energy and real-life applications.
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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