Forces and Interactions: Key Takeaways for GCSE Revision

1. Scalars vs. Vectors

  • Scalars: Magnitude only.
    Examples: Mass (3 kg), temperature (27°C), energy (200 J).
  • Vectors: Magnitude and direction.
    Examples: Force (250 N left), velocity (30 m/s east), displacement.
  • Tip: Velocity ≠ speed! Velocity includes direction (e.g., 180 m/s east).

2. Contact and Non-Contact Forces

  • Contact Forces: Require physical touch.
    Examples: Friction, tension, air resistance, normal contact force.
  • Non-Contact Forces: Act at a distance.
    Examples: Gravity, magnetism, electrostatic forces.
  • Key Rule: Weight (a non-contact force) is calculated using W=mgW=mg.
    Example: A 50 kg person on Earth: W=50×9.8=490 NW=50×9.8=490N. On the Moon: W=50×1.6=80 NW=50×1.6=80N.

3. Resultant Forces

  • Same Direction: Add forces.
    Example: Two people push a car with 300 N and 500 N → resultant force = 800 N.
  • Opposite Direction: Subtract forces.
    Example: Thrust = 500 N, drag = 300 N → resultant = 200 N.
  • Free Body Diagrams: Show all forces acting on an object (e.g., weight, normal contact force, friction).
    Tip: For stationary objects, resultant force = 0 (equilibrium).

4. Moments and Levers

  • Moment Formula: Moment=F×dMoment=F×d (force × perpendicular distance from pivot).
    Example: A 30 N force on a 0.2 m spanner: 30×0.2=6 Nm30×0.2=6Nm.
  • Principle of Moments: Sum of clockwise moments = sum of anticlockwise moments.
    Example: Balanced seesaw: 450 N×1 m=300 N×1.5 m450N×1m=300N×1.5m.
  • Levers: Increase force using a longer distance from the pivot.
    Tip: Use F1×d1=F2×d2F1​×d1​=F2​×d2​ to solve lever problems.

5. Work Done and Energy Transfer

  • Formula: W=F×sW=F×s (force × distance moved in the direction of the force).
    Example: Lifting a 20 N tin 0.5 m: 20×0.5=10 J20×0.5=10J.
  • No Work Done: If there’s no movement (e.g., holding a shopping bag) or force is perpendicular to motion.
  • Energy Transfer: Work done becomes kinetic, potential, or thermal energy.

6. Forces and Elasticity

  • Hooke’s Law: F=k×eF=k×e (force = spring constant × extension).
    Example: A spring (k = 300 N/m) stretches 8 cm: F=300×0.08=24 NF=300×0.08=24N.
  • Elastic Deformation: Returns to original shape (e.g., spring).
  • Inelastic Deformation: Permanent shape change (e.g., crushed can).
  • Elastic Potential Energy: Ee=12ke2Ee​=21​ke2.

7. Pressure in Fluids

  • Pressure Formula: P=FAP=AF​ (force ÷ area).
    Example: A 900 N man on shoes (0.06 m²): P=9000.06=15, ⁣000 PaP=0.06900​=15,000Pa.
  • Hydraulics: Transmit pressure equally. Small force on a small area creates a large force on a large area.
  • Pressure in Liquids: P=hρgP=hρg (height × density × gravity).
    Example: At 16 m depth: P=16×1000×9.8=156, ⁣800 PaP=16×1000×9.8=156,800Pa.

8. Floating and Upthrust

  • Upthrust: Resultant upward force due to pressure difference. Equals weight of displaced fluid.
  • Floating Condition: Upthrust = object’s weight.
    Example: Wood (density < water) floats; iron (density > water) sinks.

9. Atmospheric Pressure

  • Caused By: Air molecules colliding with surfaces.
  • Decreases With Altitude: Fewer molecules at higher altitudes.
    Example: Suction cups use atmospheric pressure: F=P×AF=P×A.

Key Tips for Exams

  1. Unit Conversions: Always convert cm → m, g → kg (e.g., 8 cm = 0.08 m).
  2. Free Body Diagrams: Label forces clearly (weight, normal, friction, applied).
  3. Moments: Use the perpendicular distance from the pivot.
  4. Elasticity: Hooke’s Law applies only up to the limit of proportionality.
  5. Work Done: Direction matters! No work if force is perpendicular to motion.

Practice Questions

  1. Calculate the moment of a 120 N force applied 0.5 m from a pivot.
  2. A 60 kg astronaut weighs 222 N on Mars. What is Mars’ gravitational field strength?
  3. Explain why a hydraulic jack can lift heavy loads with small input forces.

Answers:

  1. 120×0.5=60 Nm120×0.5=60Nm
  2. g=22260=3.7 N/kgg=60222​=3.7N/kg
  3. Pressure is transmitted equally; larger area at output creates larger force.

50 GCSE Questions on Forces and Interactions

Section A: Scalars, Vectors, and Forces

  1. Define scalar and vector quantities. Give two examples of each.
  2. Explain why velocity is a vector quantity, but speed is not.
  3. A car travels 15 km north, then 10 km east. What is its total distance and displacement?
  4. Identify the scalar quantity in this list: energy, force, displacement, velocity.
  5. A force of 250 N acts to the left. Represent this as a vector diagram.

Section B: Contact and Non-Contact Forces

  1. List three contact forces and three non-contact forces.
  2. Explain why gravity is a non-contact force.
  3. A balloon sticks to a wall after being rubbed. Which non-contact force is responsible?
  4. Calculate the weight of a 75 kg astronaut on Mars, where g=3.7 N/kgg=3.7N/kg.
  5. Why does a long spanner make it easier to tighten a nut?

Section C: Resultant Forces and Free Body Diagrams

  1. Two forces act on a box: 40 N left and 25 N right. Calculate the resultant force.
  2. Draw a free body diagram for a book resting on a table.
  3. A ball is falling at terminal velocity. Describe the forces acting on it.
  4. Resolve a 50 N force acting at 30° to the horizontal into its components.
  5. A stationary object has a resultant force of 0 N. What does this mean?

Section D: Moments and Levers

  1. Calculate the moment of a 120 N force applied 0.5 m from a pivot.
  2. State the principle of moments.
  3. A seesaw is balanced with a 400 N child 1.2 m from the pivot. Where should a 600 N adult sit?
  4. Explain why door handles are placed far from the hinges.
  5. A crowbar applies a force of 200 N at 1.5 m from a pivot. What force is exerted on a rock 0.3 m from the pivot?

Section E: Work Done and Energy Transfer

  1. Define work done and give its unit.
  2. Calculate the work done lifting a 25 N box through 4 m.
  3. Why is no work done when holding a heavy suitcase still?
  4. A spring stores 80 J of elastic potential energy when stretched 0.2 m. Calculate its spring constant.
  5. A crane does 12,000 J of work lifting a load. If the load weighs 1500 N, how high was it lifted?

Section F: Elasticity and Hooke’s Law

  1. State Hooke’s Law. What happens beyond the limit of proportionality?
  2. A spring extends by 5 cm under a 15 N force. Calculate its spring constant.
  3. Differentiate between elastic and inelastic deformation.
  4. A spring with k=200 N/mk=200N/m is compressed by 0.1 m. Calculate the energy stored.
  5. A rubber band stretches 10 cm under 8 N. Is this elastic or inelastic? Explain.

Section G: Pressure in Fluids

  1. Calculate the pressure exerted by a 900 N man standing on shoes with area 0.03 m20.03m2.
  2. A hydraulic jack has pistons of area 0.01 m20.01m2 and 0.1 m20.1m2. If 50 N is applied to the smaller piston, what force is exerted by the larger piston?
  3. Explain why pressure increases with depth in a fluid.
  4. Calculate the pressure at 20 m depth in water (ρ=1000 kg/m3ρ=1000kg/m3, g=9.8 N/kgg=9.8N/kg).
  5. Why do suction cups stick to surfaces?

Section H: Floating and Upthrust

  1. State the condition for an object to float.
  2. A block of wood displaces 300 N of water. If its weight is 250 N, will it float?
  3. Explain why a steel ship floats despite its high density.
  4. A 500 N object is fully submerged in water. Calculate the upthrust if it displaces 400 N of water.
  5. Why does a helium balloon rise in air?

Section I: Atmospheric Pressure

  1. Why does atmospheric pressure decrease with altitude?
  2. Calculate the force exerted by atmospheric pressure (100, ⁣000 Pa100,000Pa) on a window of area 2 m22m2.
  3. Explain how a barometer works.
  4. Why do mountaineers need oxygen at high altitudes?
  5. A suction cup has an area of 0.02 m20.02m2. What weight can it lift if atmospheric pressure is 100, ⁣000 Pa100,000Pa?

Section J: Mixed Applications

  1. A 50 kg mass is hung from a spring, causing it to stretch 12 cm. Calculate kk.
  2. A car engine exerts 800 N. If the frictional force is 600 N, calculate the resultant force.
  3. A diver experiences 250,000 Pa pressure at depth. If their mask has area 0.03 m20.03m2, calculate the force on the mask.
  4. A gear system has input radius 0.1 m and output radius 0.4 m. If 100 N is applied to the input, calculate the output force.
  5. A crane lifts a 2000 N load at 15 m radius. What counterbalance weight is needed at 5 m radius?

Detailed Answers

  1. Scalars have magnitude only (e.g., mass, energy). Vectors have magnitude and direction (e.g., force, velocity).
  2. Velocity includes direction (e.g., 30 m/s east); speed does not.
  3. Distance = 25 km. Displacement = 152+102=18 km152+102​=18km northeast.
  4. Energy is a scalar.
  5. ←[250 N]
  6. Contact: friction, tension, air resistance. Non-contact: gravity, magnetism, electrostatic.
  7. Gravity acts without physical contact (e.g., Earth pulling the Moon).
  8. Electrostatic force.
  9. W=75×3.7=277.5 NW=75×3.7=277.5N.
  10. Longer spanner increases distance from pivot, increasing moment (M=F×dM=F×d).
  11. Resultant = 40−25=15 N left40−25=15N left.
  12. Forces: weight (down), normal contact force (up).
  13. Weight (down) = air resistance (up). Resultant force = 0 N.
  14. Horizontal: 50cos⁡30°=43.3 N50cos30°=43.3N. Vertical: 50sin⁡30°=25 N50sin30°=25N.
  15. Forces are balanced; object is in equilibrium.
  16. Moment = 120×0.5=60 Nm120×0.5=60Nm.
  17. Total clockwise moments = total anticlockwise moments.
  18. 400×1.2=600×d⇒d=0.8 m400×1.2=600×d⇒d=0.8m.
  19. Longer distance from pivot increases moment, making it easier to turn.
  20. 200×1.5=F×0.3⇒F=1000 N200×1.5=F×0.3⇒F=1000N.
  21. Work = force × distance in direction of force. Unit: joule (J).
  22. W=25×4=100 JW=25×4=100J.
  23. No movement → no work done (s=0s=0).
  24. Ee=12ke2⇒80=12k(0.2)2⇒k=4000 N/mEe​=21​ke2⇒80=21​k(0.2)2⇒k=4000N/m.
  25. s=WF=12, ⁣0001500=8 ms=FW​=150012,000​=8m.
  26. F=keF=ke. Beyond limit, spring deforms permanently.
  27. k=150.05=300 N/mk=0.0515​=300N/m.
  28. Elastic: returns to shape. Inelastic: permanent deformation.
  29. Ee=12×200×(0.1)2=1 JEe​=21​×200×(0.1)2=1J.
  30. Elastic if returns to original length; inelastic if permanently stretched.
  31. P=9000.03=30, ⁣000 PaP=0.03900​=30,000Pa.
  32. 500.01=F0.1⇒F=500 N0.0150​=0.1F​⇒F=500N.
  33. Weight of fluid above increases with depth (P=hρgP=hρg).
  34. P=20×1000×9.8=196, ⁣000 PaP=20×1000×9.8=196,000Pa.
  35. Atmospheric pressure pushes the cup onto the surface.
  36. Upthrust ≥ weight.
  37. Yes (upthrust = 300 N > weight = 250 N).
  38. Hollow shape displaces enough water for upthrust = weight.
  39. Upthrust = 400 N. Resultant force = 500 – 400 = 100 N down.
  40. Helium’s density < air → upthrust > weight.
  41. Air molecules are fewer and spread out at higher altitudes.
  42. F=100, ⁣000×2=200, ⁣000 NF=100,000×2=200,000N.
  43. Measures atmospheric pressure using height of mercury column.
  44. Low atmospheric pressure → less oxygen per breath.
  45. F=100, ⁣000×0.02=2000 NF=100,000×0.02=2000N.
  46. k=50×9.80.12=4083 N/mk=0.1250×9.8​=4083N/m.
  47. Resultant = 800−600=200 N800−600=200N.
  48. F=250, ⁣000×0.03=7500 NF=250,000×0.03=7500N.
  49. Foutput=100×0.10.4=25 NFoutput​=0.4100×0.1​=25N.
  50. 2000×15=W×5⇒W=6000 N2000×15=W×5⇒W=6000N.