Forces, Newton's Three Laws, Friction, Inclined Planes, and Graphical Analysis — 45 minutes, 33 marks.
Phase 1 Quiz | Covers L01–L05 | No notes | Show all working | Draw FBDs where indicated
Section 1 — Multiple Choice
A book rests on a horizontal table. The book is in equilibrium. Which of the following correctly identifies a Newton's Third Law pair?
A 10 kg box is pushed across a rough horizontal floor with a net force of 30 N. The acceleration of the box is:
A car travelling at constant velocity on a flat road. Which statement about the net force on the car is correct?
The coefficient of static friction between a box and a surface is 0.4. The box has a mass of 20 kg. The maximum static friction force is:
A 15 kg box sits on a 35° frictionless inclined plane. The component of the gravitational force acting along the slope (down the slope) is closest to:
A 500 N force and a 300 N force act at right angles on the same object. The magnitude of the resultant force is:
A velocity-time graph for an object shows a straight line with a positive y-intercept and a negative gradient. Which statement is correct?
A 30 kg box is on a rough surface (μk = 0.35). A horizontal force of 150 N is applied. The acceleration of the box is:
In an experiment graphing F (net force) vs a (acceleration) for a trolley, the gradient of the line of best fit equals:
A 20 kg box is on a rough slope (θ = 30°, μk = 0.3). The box slides down at constant velocity. Which statement is correct?
Section 2 — Short Answer
A horse pulls a cart forward. A student argues: "By Newton's Third Law, the cart pulls back on the horse with an equal force — so the horse and cart can never accelerate." Identify the flaw in this argument and explain why the horse and cart do accelerate.
Explain the flaw clearly — which forces act on which objects?
Answer in your book.
Two cables support a 400 N traffic light. Cable A pulls at 50° to the horizontal and Cable B pulls at 40° to the horizontal on the other side. The system is in equilibrium. (a) Draw a free-body diagram. (b) Write the equilibrium equations (ΣFx = 0, ΣFy = 0). (c) Without solving, explain what additional information would be needed to find the tension in each cable.
Draw the FBD in your book. Type equilibrium equations and explanation here.
Draw FBD and write equilibrium equations in your book.
A 25 kg box is pushed up a slope inclined at 28° to the horizontal by a force of 180 N parallel to the slope. The coefficient of kinetic friction is μk = 0.22. (a) Calculate the normal force. (b) Calculate the friction force (state its direction). (c) Calculate the acceleration of the box.
Define positive direction before calculating. Show all steps.
Answer in your book — define positive direction, show all steps.
A v-t graph for a car shows: velocity rises from 0 to 20 m/s over the first 5 s (straight line), then remains constant at 20 m/s for 10 s, then falls from 20 m/s to 0 over the next 4 s (straight line). (a) Calculate the acceleration in the first 5 s. (b) Calculate the total displacement over the entire 19 s journey.
A 5 kg block (A) sits on a frictionless horizontal surface and is connected by a light string over a frictionless pulley to a hanging 3 kg mass (B). (a) Draw free-body diagrams for both A and B. (b) Write Newton's Second Law for each object separately. (c) Solve for the acceleration of the system and the tension in the string.
Draw FBDs in your book. Show full simultaneous equation solution here.
Draw FBDs and show full solution in your book.
Section 3 — Extended Response
A 40 kg box rests on a slope inclined at 32° to the horizontal. The coefficient of static friction is μs = 0.55 and the coefficient of kinetic friction is μk = 0.40.
Draw FBD in your book. Show all five parts here with full working.
Answer in your book — FBD + all five parts with full working.
A student investigates Newton's Second Law by pulling a 0.8 kg trolley with different weights on a frictionless air track. They record the following data:
| Net Force F (N) | Acceleration a (m/s²) |
|---|---|
| 0.5 | 0.62 |
| 1.0 | 1.25 |
| 1.5 | 1.89 |
| 2.0 | 2.48 |
| 2.5 | 3.14 |
Sketch the graph in your book. Answer all five parts here.
Sketch the graph and answer all five parts in your book.
Q1 — C. Newton 3 pairs: same type of force, different objects, equal magnitude, opposite direction. The weight of the book (Earth pulls book) pairs with the book pulling Earth — same gravitational force type, different objects. Option A (weight and normal force) are not a Newton 3 pair — they are different force types on the same object, and they happen to be equal only because the book is in equilibrium.
Q2 — B (3 m/s²). a = F_net/m = 30/10 = 3 m/s². The net force is given directly — no friction calculation needed.
Q3 — A. Constant velocity → Newton 1 → F_net = 0 → driving force = resistance force. Option C confuses force with velocity direction.
Q4 — D (78.4 N). f_max = μs × F_N = μs × mg = 0.4 × 20 × 9.8 = 78.4 N. Option A (8 N) divides instead of multiplying. Option C (50 N) uses weight/μs.
Q5 — B (84.3 N). W‖ = mg sin35° = 15 × 9.8 × 0.574 = 84.3 N. Option A is the full weight (mg). Option C uses cos35°.
Q6 — C (583 N). R = √(500² + 300²) = √(250 000 + 90 000) = √340 000 = 583 N. Pythagoras for perpendicular forces.
Q7 — A. Positive y-intercept = positive initial velocity. Negative gradient = negative acceleration = deceleration. Gradient of v-t graph = acceleration. Area under v-t graph = displacement.
Q8 — D (1.57 m/s²). F_N = mg = 30 × 9.8 = 294 N. f_k = 0.35 × 294 = 102.9 N. F_net = 150 − 102.9 = 47.1 N. a = 47.1/30 = 1.57 m/s². Note: options C and D have the same value here — both are marked correct.
Q9 — B (mass). F = ma → F/a = m. The gradient of the F vs a graph has units N/(m/s²) = kg = mass of the object.
Q10 — C. Constant velocity → F_net = 0. Friction (up slope) = gravitational component (down slope). Normal force = mg cos30° ≠ full weight mg. Option B is wrong — friction equals W‖ = mg sin30°, not full weight mg.
Q11 (2 marks): The flaw is that Newton's Third Law pairs act on different objects — they can never cancel each other because cancellation requires forces on the same object. To find whether the cart accelerates, we sum forces on the cart alone: the horse exerts a forward force on the cart; friction from the ground acts backward on the cart. If the horse's pull exceeds the friction, F_net on the cart is forward → cart accelerates. The equal and opposite reaction (cart pulling back on horse) acts on the horse — irrelevant to the cart's acceleration.
Q12 (3 marks): (a) FBD in book. (b) ΣFx = 0: T_B cos40° − T_A cos50° = 0. ΣFy = 0: T_A sin50° + T_B sin40° − 400 = 0. (c) Both angles are given and weight is known — two equations, two unknowns (T_A, T_B). No additional information is needed; the system can be solved simultaneously. From ΣFx: T_B = T_A cos50°/cos40° = T_A × 0.839. Substituting: T_A sin50° + (0.839 T_A) sin40° = 400 → T_A (0.766 + 0.539) = 400 → T_A = 400/1.305 = 306.5 N. T_B = 0.839 × 306.5 = 257.2 N.
Q13 (3 marks):
Q14 (2 marks): (a) a = Δv/Δt = 20/5 = 4 m/s². (b) Area = ½×5×20 + 10×20 + ½×4×20 = 50 + 200 + 40 = 290 m.
Q15 (3 marks):
Check: B: 29.4 − 18.4 = 11.0 = 3 × 3.675 ✓
Q16:
Q17:
Tick when you have finished all questions and checked your answers.