by How Forces Affect Motion
Why does a football speed up when kicked, a bicycle slow down when brakes are applied, or a rocket rise while gases shoot downward?
These everyday events are explained by force and Newton’s laws of motion. Explore the key ideas, test force and mass with the calculator, and practise all 16 end-of-chapter questions with answers hidden behind toggles.
What This Chapter Covers
Pushes and Pulls
A force can start or stop motion, change speed or direction, or change an object’s shape. Force is a vector quantity, and its SI unit is the newton (N).
Balanced and Unbalanced Forces
Balanced forces give zero net force. An unbalanced net force changes an object’s velocity and produces acceleration.
The Opposing Force
Friction opposes relative motion, or the tendency of surfaces in contact to move relative to one another. It can slow objects down.
Three Laws of Motion
Learn inertia, the relation between net force and acceleration, and the equal-and-opposite forces acting on different objects.
1. Force and Net Force
A force is a push or pull that can change an object’s state of rest, speed, direction, or shape. A spring balance can measure the magnitude of a force.
Forces in the same direction
Fnet = F1 + F2For example, 10 N and 6 N acting to the right give a net force of 16 N to the right.
Forces in opposite directions
|Fnet| = |F1 − F2|For example, 10 N to the right and 6 N to the left give a net force of 4 N to the right.
2. Friction
Friction acts against relative motion or its tendency between surfaces in contact. A toy car usually travels farther over a smooth surface than over a rough surface because friction is often smaller on the smooth surface.
Friction is useful for walking, gripping, and braking, but it can also cause wear and unwanted heating.
3. Newton’s Three Laws of Motion
Newton’s First Law — Law of Inertia
An object at rest remains at rest, and an object in motion continues with constant velocity unless a net external force acts on it.
Inertia is the tendency of an object to resist changes in its state of rest or uniform motion. Objects with greater mass have greater inertia.
Newton’s Second Law
Fnet = ma and a = Fnet / mFor a fixed mass, acceleration increases with net force. For a fixed net force, acceleration decreases as mass increases.
Weight: W = mgWeight is the gravitational force acting on an object. Near Earth’s surface, g is approximately 9.8 m/s²; 10 m/s² is often used for estimates where appropriate.
Newton’s Third Law
When one object exerts a force on another object, the second object simultaneously exerts an equal-magnitude force in the opposite direction on the first object.
Examples include walking, swimming, jumping from a boat, and rocket propulsion. Action and reaction act on different objects, so they do not cancel when analysing either object individually.
Interactive Calculator: Force, Mass and Acceleration
Enter the net force and mass to calculate acceleration using Newton’s Second Law.
Quick Formula Revision
| Concept | Formula / rule |
|---|---|
| Newton’s Second Law | Fnet = ma |
| Acceleration | a = Fnet / m |
| Weight | W = mg |
| Average force from momentum change | Favg = Δp / Δt |
| Net force, same direction | Fnet = F1 + F2 |
| Net force, opposite directions | |Fnet| = |F1 − F2| |
| Connected objects treated as one system | a = Fnet / (m1 + m2) |
Revise, Reflect, Refine — All 16 Exercise Questions
Select Show Answer under any question to reveal its solution. Use the controls to open or close all answers at once.
Q1. Using a horizontal force F, a table is moved across the floor at a constant velocity. How much is the frictional force exerted by the floor on the table?
Constant velocity means acceleration is zero, so the net horizontal force is zero.
F − f = 0 ⇒ f = FAnswer: The frictional force has magnitude F and acts opposite to the direction of motion.
Q2. For a ball moving on a smooth frictionless surface, choose the appropriate option for each statement: (i) no net force is applied; (ii) a net force acts in the direction of motion; (iii) a net force acts opposite to the direction of motion. In each case, does the speed remain the same, increase, or decrease?
(i) The velocity remains the same because acceleration is zero.
(ii) The speed increases because acceleration is in the direction of motion.
(iii) The speed decreases because acceleration is opposite to the direction of motion.
Q3. Two blocks P and Q are on a smooth horizontal surface. Forces of 4 N and 5 N act in opposite directions on P, while Q moves with constant velocity. Which statement is correct? (i) P experiences a net force and Q does not; (ii) P does not experience a net force and Q does; (iii) both experience a net force; (iv) neither experiences a net force.
For block P, the forces are unequal: |5 − 4| = 1 N, so P has a non-zero net force. Block Q moves with constant velocity, so its net force is zero.
Answer: Option (i). P experiences a net force; Q does not.
Q4. In a snake boat race, 100 oarsmen row together. Of these, 95 row to propel the boat forward, but 5 row in the opposite direction. If each applies a horizontal force of 200 N, calculate the net force on the boat. Ignore drag and air friction.
Force in the forward direction:
95 × 200 = 19,000 NForce in the opposite direction:
5 × 200 = 1,000 NNet force:
19,000 − 1,000 = 18,000 NAnswer: 18,000 N forward.
Q5. When a net force acts on an object, the object accelerates: (i) opposite to the force, with acceleration proportional to force; (ii) opposite to the force, with acceleration proportional to mass; (iii) in the direction of force, with acceleration inversely proportional to force; or (iv) in the direction of force, with acceleration proportional to force.
Newton’s Second Law states that acceleration is in the direction of the net force and its magnitude is proportional to the net force for a fixed mass.
Answer: Option (iv).
Q6. The position-time graphs for four objects A, B, C and D moving along a straight line are shown below. For which object does a net force act?
Answer: Object C (option iii). Its position-time graph is curved, so its slope—and therefore velocity—changes with time. A changing velocity means acceleration, which requires a non-zero net force. The straight-line graphs represent constant velocity, including rest for the horizontal line.
Q7. A sailor jumps out of a small boat towards the shore. Will the boat move? If so, in which direction and why?
Answer: Yes. As the sailor jumps forward, they push the boat backward. By Newton’s Third Law, the boat exerts an equal and opposite force on the sailor, and the boat moves backward, away from the shore.
Q8. During a high-jump event, a landing mat or sand bed is placed for the athlete to fall upon. Explain the reason.
The mat or sand bed increases the time taken for the athlete to come to rest. For a similar change in momentum, a longer stopping time reduces the average impact force.
Favg = Δp / ΔtTherefore, the athlete is less likely to suffer an injury.
Q9. A hand cart loaded with vegetables collides with an identical but empty hand cart. During the collision, which is correct? (i) the loaded cart exerts a larger force; (ii) the empty cart exerts a larger force; (iii) neither cart exerts a force on the other; (iv) both carts exert equal-magnitude forces on each other.
Answer: Option (iv). By Newton’s Third Law, the carts exert forces of equal magnitude and opposite direction on each other. Their accelerations can differ because their masses differ.
Q10. The acceleration-mass graph below shows the acceleration produced by a force on objects of different masses. Plot the force-mass graph for this case.
For a constant net force, Newton’s Second Law gives F = ma. As mass increases, acceleration decreases so that the product ma remains constant. From the graph, for example, m = 1 kg and a = 10 m/s² give F = 10 N.
Answer: The force-mass graph is a horizontal straight line at F = 10 N, because the force stays constant for all masses shown.
Q11. The velocity-time graph of an object of mass 10 kg is shown below. Calculate the force acting on the object.
Mass, m = 10 kg. From the graph, initial velocity u = 10 m/s, final velocity v = 30 m/s, and time t = 8 s.
a = (v − u)/t = (30 − 10)/8 = 2.5 m/s²Using Newton’s Second Law:
F = ma = 10 × 2.5 = 25 NAnswer: 25 N, in the direction of acceleration.
Q12. A bullet of mass 50 g moving at 100 m/s enters a heavy stationary wooden block and stops after penetrating 50 cm. Estimate the stopping force, assuming constant acceleration.
Convert units: m = 50 g = 0.05 kg; initial speed u = 100 m/s; final speed v = 0; distance s = 50 cm = 0.5 m.
Using v² = u² + 2as:
0 = 100² + 2a(0.5) ⇒ a = −10,000 m/s²Then:
F = ma = 0.05 × (−10,000) = −500 NAnswer: The stopping force has magnitude 500 N and acts opposite to the bullet’s motion.
Q13. A footballer kicks a football to a speed of 108 km/h. The force imparted is estimated to be 800 N, and the football’s mass is 0.4 kg. Calculate the time of contact between the foot and the ball.
Assuming the ball is initially at rest, convert the final speed:
108 km/h = 108 × (5/18) = 30 m/sChange in momentum:
Δp = m(v − u) = 0.4 × (30 − 0) = 12 kg·m/sUsing F = Δp/Δt:
Δt = Δp/F = 12/800 = 0.015 sAnswer: 0.015 s, or 15 milliseconds.
Q14. An object of mass 2 kg moves at 10 m/s and enters a rough patch. Friction is 7 N, and an additional constant force of 3 N also opposes the motion. How far does the object travel before stopping?
Total opposing force:
F = 7 + 3 = 10 NTaking the direction of initial motion as positive:
a = −F/m = −10/2 = −5 m/s²Use v² = u² + 2as, with v = 0 and u = 10 m/s:
0 = 10² + 2(−5)s ⇒ s = 10 mAnswer: 10 m.
Q15. A tractor pulls a harrow of mass m₁ with net force F and acceleration a₁. The same tractor pulls a trolley of mass m₂ with force F and acceleration a₂. If it pulls the trolley with the harrow placed on it using the same force F, obtain the resulting acceleration in terms of a₁ and a₂. Ignore friction.
For the harrow:
F = m₁a₁ ⇒ m₁ = F/a₁For the trolley:
F = m₂a₂ ⇒ m₂ = F/a₂When both are pulled together, total mass is m₁ + m₂, so:
a = F/(m₁ + m₂) = F/(F/a₁ + F/a₂)Cancel F and simplify:
a = 1/(1/a₁ + 1/a₂) = (a₁a₂)/(a₁ + a₂)Answer: a = (a₁a₂)/(a₁ + a₂).
Q16. When a bar magnet is brought close to a magnetic compass, the magnet and compass needle exert equal and opposite forces on each other. However, the compass needle moves, whereas the bar magnet appears not to move. Explain why.
Newton’s Third Law says the forces on the magnet and needle are equal in magnitude and opposite in direction. But their masses are very different. From a = F/m, the much lighter compass needle has a much larger acceleration. The magnet’s acceleration is extremely small, so its motion is usually not noticeable.
Final Revision: Five Ideas to Remember
- Force is a push or pull and is measured in newtons (N).
- Motion changes according to the net force.
- Zero net force means zero acceleration—not necessarily zero velocity.
- Newton’s Second Law is Fnet = ma.
- Action and reaction are equal and opposite forces acting on different objects.
