How Forces Affect Motion — Detailed Notes
Chapter 4 described motion; this chapter asks what causes changes in motion. The answer is force, and the behaviour of objects under forces is captured by Newton’s three laws of motion.
6.1 The concept of force
A force can set a resting object moving, change the speed or direction of a moving object, or change its shape. Force is a vector — it needs magnitude and direction. SI unit: the newton (N). Its magnitude can be measured with a spring balance (weight = the gravitational pull of the Earth on the object).
6.2 Balanced & unbalanced forces
Usually several forces act together. Balanced forces — equal in magnitude, opposite in direction → net force = 0 (object does not change its motion; e.g. tug of war that does not move). Unbalanced forces → a non-zero net force acts. Forces in the same direction add; forces in opposite directions subtract, with the net force along the larger force.
6.3 Force of friction
Friction acts between surfaces in contact, opposite to the direction of motion (or attempted motion). An object moves only when the applied force exceeds friction. On a moving object, friction gradually brings it to rest unless force is continuously applied. Friction depends on the nature of the surfaces (rough → more friction; smooth/polished → less). Other forces on a resting object: weight (down) balanced by the normal force (up, from the surface).
6.4 Newton’s First Law (Law of Inertia)
An object at rest stays at rest, and an object in motion keeps moving with constant velocity, unless a net force acts on it. If net force = 0, acceleration = 0. Inertia is the tendency of objects to resist change in their state of rest or uniform motion (Galileo’s thought experiments; Newton, 1687).
6.5 Newton’s Second Law
A net force produces acceleration in its own direction; the acceleration is proportional to the force and inversely proportional to the mass. Mathematically a = F/m, or F = ma. 1 N = the force that gives a 1 kg mass an acceleration of 1 m s⁻² (1 N = 1 kg m s⁻²). Gravitational force (weight) = F = mg, with g = 9.8 m s⁻² (≈10 for estimates). The fuller form uses momentum (p = mass × velocity): net force = rate of change of momentum. Everyday uses: a fielder pulls hands back while catching (increases time → smaller force), airbags & landing mats (increase time of impact → smaller force → less injury).
6.6 Newton’s Third Law
When one object exerts a force on a second object, the second exerts an equal and opposite force on the first. The two forces are equal in magnitude, opposite in direction, and act on different objects (so they do not cancel). Examples: walking (foot pushes ground back, friction pushes you forward), rowing a canoe, a rocket (expels gas down, is pushed up), recoil of a gun. Equal forces need not give equal accelerations — because a = F/m and the masses differ (e.g. fruit falls to Earth, but the Earth’s huge mass means its acceleration is negligible). Applies to contact and non-contact (magnetic, electrostatic, gravitational) forces.
6.7 Systems of objects
Connected objects can be treated as one system: ignore the internal forces (e.g. the string tension) and use only external forces. For two boxes (m₁, m₂) pulled by force F on a frictionless surface, acceleration a = F/(m₁ + m₂) — just like a single object of mass (m₁ + m₂).
Key Terms
force, newton, balanced/unbalanced forces, net force, friction, normal force, inertia, Newton’s three laws, F = ma, weight (mg), momentum, action–reaction pair, system of objects, tension.
Revise, Reflect, Refine — All NCERT Exercise Answers
1A table is moved across the floor at constant velocity by a horizontal force F. How much is the frictional force on the table?
At constant velocity the acceleration is zero, so by Newton’s first law the net force is zero. The applied force F and friction are the only horizontal forces, so friction must balance F exactly. Force of friction = F (equal in magnitude, opposite in direction).
2Ball on a smooth frictionless surface — choose the correct words.
(i) If no net force is applied, the velocity will remain the same (Newton’s first law). (ii) A net force in the direction of motion → the speed will increase. (iii) A net force opposite to the motion → the speed will decrease.
3Block P has 5 N and 4 N acting in opposite directions; block Q moves with constant velocity. Which statement is correct?
(i) P experiences a net force and Q does not. On P the net force = 5 − 4 = 1 N (unbalanced), so P accelerates. Q moves at constant velocity, so by Newton’s first law the net force on Q is zero.
4Snake-boat race: 95 oarsmen row backwards (boat goes forward), 5 row the opposite way; each applies 200 N. Net force on the boat?
Forward force = 95 × 200 = 19,000 N; opposing force = 5 × 200 = 1,000 N. Net force = 19,000 − 1,000 = 18,000 N, in the forward direction (the direction propelling the boat).
5When a net force acts on an object, it accelerates… (choose the correct option).
(iv) in the direction of the force, with acceleration proportional to the force (and inversely proportional to mass). This is Newton’s second law, a = F/m.
6Position–time graphs of A, B, C, D. A net force acts on which object?
A net force acts on the object whose position–time graph is curved — a curve means the velocity (slope) is changing, i.e. the object is accelerating, which requires a net force. Objects whose graphs are straight lines (constant velocity) or horizontal (at rest) have zero net force. (So the correct choice is the object with the curved p–t graph.)
7A sailor jumps forward from a small boat to the shore. Will the boat move? In which direction and why?
Yes, the boat moves backward. When the sailor pushes the boat to jump forward, by Newton’s third law the boat pushes the sailor forward and the sailor pushes the boat backward with an equal force — so the boat recoils in the opposite direction to the jump (also understood as conservation of momentum).
8During a high jump, a landing mat / sand bed is placed for the athlete. Explain the reason.
The mat/sand increases the time over which the athlete’s velocity is reduced to zero on landing. By Newton’s second law (in momentum form), a longer stopping time means a smaller deceleration and hence a smaller force on the body, reducing the risk of injury — the same principle as airbags and a fielder drawing their hands back.
9A loaded cart collides with an identical empty cart. During the collision…
(iv) the loaded cart and the empty cart exert equal magnitude forces on each other (in opposite directions) — this is Newton’s third law. Their accelerations differ (the empty cart accelerates more because it has less mass), but the mutual forces are equal.
10From the acceleration–mass graph (same force on different masses, Fig. 6.40), plot the force–mass graph.
Since the same force acts throughout, F = ma is constant for every mass (as mass rises, acceleration falls so that the product stays the same). Therefore the force–mass graph is a horizontal straight line (F constant, independent of mass).
11Velocity–time graph of a 10 kg object (Fig. 6.41). Calculate the force acting on it.
Force = mass × acceleration, and acceleration = slope of the velocity–time graph. Read the slope (change in velocity ÷ time) from the graph, then F = 10 kg × a. For example, if the velocity rises from 0 to 30 m s⁻¹ in the first interval and then stays constant, a = (slope) in the sloping part and F = 10 × a there, while F = 0 where the line is flat (constant velocity). (Substitute the slope read off the graph.)
12A 50 g bullet at 100 m s⁻¹ stops after penetrating 50 cm of a block (constant acceleration). Estimate the stopping force.
m = 0.05 kg, u = 100 m s⁻¹, v = 0, s = 0.5 m. From v² = u² + 2as: 0 = 100² + 2a(0.5) → a = −10000/1 = −10,000 m s⁻². Force F = ma = 0.05 × (−10000) = −500 N, i.e. a stopping force of 500 N opposing the motion.
13A footballer kicks a 0.4 kg ball to 108 km h⁻¹ with a force of 800 N. Find the contact time between foot and ball.
u = 0, v = 108 km h⁻¹ = 30 m s⁻¹, m = 0.4 kg, F = 800 N. Acceleration a = F/m = 800/0.4 = 2000 m s⁻². From v = u + at: 30 = 0 + 2000 t → t = 30/2000 = 0.015 s (15 milliseconds).
14A 2 kg object moving at 10 m s⁻¹ meets friction 7 N plus an extra opposing force of 3 N. How far does it travel before stopping?
Total opposing force = 7 + 3 = 10 N. Deceleration a = F/m = −10/2 = −5 m s⁻². From v² = u² + 2as: 0 = 10² + 2(−5)s → s = 100/10 = 10 m. The object travels 10 m before coming to rest.
15A tractor pulls a harrow (mass m₁) with force F giving a₁, and a trolley (m₂) with the same F giving a₂. Find the acceleration when it pulls both together (force F). Ignore friction.
From F = ma: m₁ = F/a₁ and m₂ = F/a₂. Together, a = F/(m₁ + m₂) = F / (F/a₁ + F/a₂) = 1/(1/a₁ + 1/a₂). So a = a₁a₂ / (a₁ + a₂).
16A bar magnet and a compass needle exert equal and opposite forces, yet only the needle moves. Explain.
Although the forces are equal (Newton’s third law), the resulting accelerations are not — because a = F/m. The compass needle has a very small mass and is lightly pivoted, so the same force gives it a large acceleration and it swings; the bar magnet is much heavier (and usually held/resting), so the same force produces negligible acceleration and it appears not to move.