ScienceClass 9Motion

Motion | Class 9 Science Notes

By ConceptScroll Team · Published on 17 July 2026 · 5 min read

Motion | Class 9 Science Notes

Motion – this guide gives you a concise, exam-ready overview of Motion from Class 9 Science, written by ConceptScroll editors and reviewed against the latest NCERT textbook.

7.1 Work Done by a Constant Force

Work is a fundamental concept in physics defined as the product of the force applied on an object and the displacement of the object in the direction of the force. To understand work done by a constant force, consider lifting a wheat bag of mass 5 kg from the floor to a height of 1 meter. The gravitational force acting downward on the bag is mg, where m is the mass and g is the acceleration due to gravity. To lift the bag slowly, you must apply an upward force equal in magnitude to mg. The force you apply and the displacement of the bag are in the same direction (upwards), so work is done on the bag.

If you lift three such bags one after another to the same height, you do three times the work compared to lifting one bag. Alternatively, lifting all three bags together requires applying a force three times larger but over the same distance, again resulting in three times the work. Similarly, lifting a single bag to a height of 3 meters requires three times more work than lifting it to 1 meter. These observations lead to the scientific definition of work done by a constant force:

work done = force applied × displacement in the direction of the force.

This definition applies regardless of whether the force and displacement are vertical, horizontal, or at any angle (angle considerations are introduced in higher classes). For example, if a constant force F acts on an object causing displacement s in the direction of the force, the work done W is W = F × s.

Work is a scalar quantity and can be positive or negative depending on the direction of force relative to displacement. The SI unit of work is the joule (J), defined as the work done when a force of 1 newton displaces an object by 1 meter in the direction of the force. Since 1 newton equals 1 kg·m/s², 1 joule equals 1 kg·m²/s².

When the force is not constant, work done can be calculated as the area under the force-displacement graph between the initial and final positions. Work done is zero if either the force is zero or the displacement is zero. For instance, pushing a rigid wall does not cause displacement, so no work is done on the wall, even though you feel tired due to internal energy expenditure.

If the force acts perpendicular to the displacement, the work done by that force is zero. For example, when carrying a box horizontally, the upward force applied balances the weight, but since displacement is horizontal and force is vertical, no work is done by the upward force on the box.

Work done can be positive or negative. Positive work occurs when force and displacement are in the same direction, such as pushing a wheelchair forward. Negative work occurs when force and displacement are opposite, such as a goalkeeper applying force opposite to the motion of a football to stop it.

Examples include a girl lifting and lowering a dumbbell (positive work when lifting, negative when lowering) and a goalkeeper stopping a ball (negative work).

📊 Diagram: Fig. 7.2: Lifting bags to a height; Fig. 7.3: Work done by a force while displacing an object in (a) horizontal direction, and (b) vertical direction

🔗 Connection: This section lays the foundation for understanding work-energy relations, leading to the work-energy theorem and forms of energy discussed in subsequent sections.

Frequently asked questions

The numerical ratio of displacement to distance for moving object is:

equals to 1 or less than 1

A particle is moving in a circular path of radius r. The displacement after half circle would be:

2r

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 mass of the flag, the height of the flagpole, and the gravitational acceleration (E = mgh). The speed at which the flag is raised does not change the amount of work done because work depends on force and displacement, not on time. However, power is the rate of doing work, so if the speed is doubled, the power requirement also doubles.

A man of mass 60 kg rides a scooter of mass 100 kg. He accelerates the scooter to a velocity ν. 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.

Let the velocity be ν.

Day 1 total mass = 60 + 100 = 160 kg Kinetic energy on Day 1 = (1/2) × 160 × ν² = 80ν²

Day 2 total mass = 60 + 40 + 100 = 200 kg Kinetic energy on Day 2 = (1/2) × 200 × ν² = 100ν²

Ratio of fuel used = Energy on Day 2 / Energy on Day 1 = 100ν² / 80ν² = 5/4 = 1.25

So, the fuel used on Day 2 is 1.25 times that on Day 1.

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