Cambridge IGCSE Physics · topic 1.5.1
Effects of forces
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Extra challenging questions
Work through the linked stages and explain what each result means.
Reading an accelerating lift's force sensor
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A 60 kg student stands on a scale in a lift. Use g = 10 N kg⁻¹. The graph gives the upward scale force during three separate constant-force phases. Initially the lift is stationary; ignore the short transitions between phases. Supplied data (Phase; Upward normal force / N): (1, 720), (2, 600), (3, 480). (a) Draw or describe the student's free-body diagram and find acceleration in each phase. Identify which forces form a Newton's-third-law pair. [3 marks] (b) Phase 1 lasts 2 s, phase 2 lasts 3 s and phase 3 lasts 2 s. Find total height gained and explain the 600 N reading while the lift moves. [3 marks]
Scale reading during lift phases
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Upward normal force / N
Horizontal axis (x): Phase
Read the exact data as a table
| Phase | Upward normal force / N |
|---|---|
| 1 | 720 |
| 2 | 600 |
| 3 | 480 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Draw or describe the student's free-body diagram and find acceleration in each phase. Identify which forces form a Newton's-third-law pair.
Reveal the worked solution
- The student experiences weight mg = 600 N downward and normal force N upward. There is no additional force called acceleration.
- Taking upward positive, N−600 = 60a gives accelerations +2.0, 0 and −2.0 m s⁻².
- Weight and normal act on the same body and are not a third-law pair. Each has a partner on a different body.
- Phase 1 ends at 4 m s⁻¹ and gains ½(2)(2²) = 4 m. Phase 2 gains (4)(3) = 12 m.
- Phase 3 slows from 4 to 0 m s⁻¹ in 2 s and gains 4 m, so total rise is 20 m.
- Zero resultant means constant velocity, which may be nonzero. Balanced forces do not require the lift to be stationary.
Answer (a) Accelerations: +2.0, 0, −2.0 m s⁻². The scale-on-student and student-on-scale forces form a third-law pair. (b) Height gained = 20 m; a 600 N reading is consistent with upward constant velocity.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Why a parachute opens before the new terminal speed
Optional enrichment: this challenge may extend beyond the usual syllabus depth and assumes confidence with algebra and graph interpretation. A 2.0 kg falling package has weight 20 N. Downward is positive. Before a parachute opens, drag follows the supplied measured speed–force data. After opening, drag at every speed is four times the previous value. Use the evidence to infer the drag model over this range. Supplied data (Speed / m s⁻¹; Upward drag / N): (0, 0), (2, 2), (4, 8), (6, 18), (8, 32). (a) Infer the drag relationship, find both terminal speeds, and determine the acceleration immediately after opening at 6 m s⁻¹. [3 marks] (b) Explain why the package does not instantaneously jump to its new terminal speed, and evaluate the model as it slows. [3 marks]
Drag before the parachute opens
Supplied question data. Read the labelled axes and use the observations to test the physical model.
Vertical axis (y): Upward drag / N
Horizontal axis (x): Speed / m s⁻¹
Read the exact data as a table
| Speed / m s⁻¹ | Upward drag / N |
|---|---|
| 0 | 0 |
| 2 | 2 |
| 4 | 8 |
| 6 | 18 |
| 8 | 32 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Begin with part (a): Infer the drag relationship, find both terminal speeds, and determine the acceleration immediately after opening at 6 m s⁻¹.
Reveal the worked solution
- Each nonzero data pair gives D/v² = 0.50 N s² m⁻², supporting quadratic drag over the measured range.
- Before opening, terminal motion requires 0.50v² = 20, giving √40 m s⁻¹. After opening D = 2.0v², giving √10 m s⁻¹.
- At 6 m s⁻¹ just after opening, drag is 72 N upward. Net force is 20−72 = −52 N, so a = −52/2 = −26 m s⁻².
- The package initially still moves downward; the upward acceleration changes its velocity over time rather than resetting it.
- As it slows, drag falls and the upward resultant becomes smaller, approaching zero at the new terminal speed.
- The fourfold-drag model assumes the canopy is fully inflated. Inflation time, air density and oscillation can invalidate it during the opening transient.
Answer (a) D = 0.50v²; terminal speeds ≈6.32 and 3.16 m s⁻¹; acceleration = −26 m s⁻² immediately after opening. (b) Finite upward resultant reduces downward speed continuously until drag balances weight.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.