Laws of Motion | Class 11 Physics Notes
By ConceptScroll Team · Published on 17 July 2026 · 5 min read
Laws of Motion – this guide gives you a concise, exam-ready overview of Laws of Motion from Class 11 Physics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.
4.9 Common forces in mechanics
This section discusses various forces commonly encountered in mechanics. Gravitational force is pervasive, acting at a distance on all objects due to Earth and celestial bodies. Other forces are contact forces, arising from direct interaction between bodies, such as normal reaction, friction, tension, and spring force. Normal reaction acts perpendicular to surfaces in contact, while friction acts parallel and opposes relative motion. The section explains static friction, which opposes impending motion up to a maximum value proportional to the normal force, and kinetic friction, which opposes actual sliding motion and is usually less than static friction. The laws of friction are empirical and approximately true. Rolling friction is also introduced, which is much smaller than sliding friction and arises due to deformation at contact points. The section highlights the dual role of friction: it causes energy dissipation but is essential for activities like walking and vehicle motion. Methods to reduce friction, such as lubricants, ball bearings, and air cushions, are described.
📊 Diagram: See figure_13: Fig. 4.9 Some examples of contact forces in mechanics.; See figure_14: Fig. 4.10 Static and sliding friction: (a) Impending motion of the body is opposed by static friction. When external force exceeds the maximum limit of static friction, the body begins to move. (b) Once the body is in motion, it is subject to sliding or kinetic friction which opposes relative motion between the two surfaces in contact. Kinetic friction is usually less than the maximum value of static friction.; See figure_15: Answer The forces acting on a block of mass m at rest on an inclined plane are (i) the weight mg acting vertically downwards (ii) the normal force N of the plane on the block, and (iii) the static frictional force f_s opposing the impending motion.; See figure_16: (a); See figure_17: Fig. 4.13 Some ways of reducing friction. (a) Ball bearings placed between moving parts of a machine. (b) Compressed cushion of air between surfaces in relative motion.
🔗 Connection: This section prepares for the study of circular motion, where friction and other forces provide centripetal acceleration.
Frequently asked questions
4.1 Give the magnitude and direction of the net force acting on (a) a drop of rain falling down with a constant speed, (b) a cork of mass 10g floating on water, (c) a kite skillfully held stationary in the sky, (d) a car moving with a constant velocity of 30 km/h on a rough road, (e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields.
Solution: (a) Since the drop of rain is falling with constant speed, acceleration is zero. Hence, net force = 0. (b) Cork floating on water is at rest, so net force = 0. (c) Kite held stationary means velocity is zero and not changing, so net force = 0. (d) Car moving with constant velocity means acceleration is zero, so net force = 0. (e) Electron moving at high speed in free space with no fields or objects experiences no force, so net force = 0.
4.2 A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble, (a) during its upward motion, (b) during its downward motion, (c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of 45° with the horizontal direction? Ignore air resistance.
Solution: Mass, m = 0.05 kg Acceleration due to gravity, g = 10 m/s²
(a) During upward motion, the only force acting is gravity downward. Net force = mg downward = 0.05 × 10 = 0.5 N downward.
(b) During downward motion, gravity acts downward. Net force = mg downward = 0.5 N downward.
(c) At the highest point, velocity is zero but acceleration due to gravity still acts downward. Net force = mg downward = 0.5 N downward.
If thrown at 45°, the vertical component of motion is similar, so net for
4.3 Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg (a) just after it is dropped from the window of a stationary train, (b) just after it is dropped from the window of a train running at a constant velocity of 36 km/h, (c) just after it is dropped from the window of a train accelerating with 1 m/s², (d) lying on the floor of a train which is accelerating with 1 m/s², the stone being at rest relative to the train. Neglect air resistance throughout.
Solution: Mass, m = 0.1 kg Acceleration due to gravity, g = 10 m/s²
(a) Dropped from stationary train: Net force = mg downward = 0.1 × 10 = 1 N downward.
(b) Dropped from train moving at constant velocity: Velocity constant means no acceleration, so net force = mg downward = 1 N downward.
(c) Dropped from accelerating train: Just after dropping, stone has horizontal velocity equal to train's velocity but no horizontal acceleration. Net force = mg downward = 1 N downward.
(d) Stone lying on a
4.4 One end of a string of length l is connected to a particle of mass m and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed v the net force on the particle (directed towards the centre) is: (i) T, (ii) T - \frac{mv^2}{l}, (iii) T + \frac{mv^2}{l}, (iv) 0 T is the tension in the string. [Choose the correct alternative].
Answer: (i) T
Explanation: The net force acting on the particle moving in a circle is the centripetal force directed towards the centre. This force is provided by the tension T in the string. Hence, net force = T.
Options (ii) and (iii) incorrectly add or subtract the centripetal force term from T. Option (iv) zero force is incorrect as centripetal force is required for circular motion.
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