These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. This is the algebra-based course, so no calculus is needed. Use g = 9.8 m/s². All coefficients of friction are invented for these questions. Round final answers to 2 significant figures unless told otherwise, and keep unrounded values in your calculator until the end.
Question 1 (multiple choice · foundation)
Take +x in the direction of the push. An 8.0 kg box rests on a level floor. For this box and floor, μ_s = 0.65 and μ_k = 0.50. A student pushes the box horizontally with 30 N, and it does not move. What is the size of the friction force on the box?
- (A) 30 N
- (B) 51 N
- (C) 0 N
- (D) 39 N
Answer and explanation
Answer: (A). The box stays at rest, so the net horizontal force is zero. Static friction must balance the push: 30 N in the −x direction. The maximum static friction is μ_s mg = 0.65 × 78.4 N = 51 N, which is more than 30 N, so the box can indeed stay at rest.
- (B) is μ_s F_n. That is the maximum static friction, reached only at the point of slipping.
- (C) assumes there is no friction when nothing moves. Without friction, the 30 N push would accelerate the box.
- (D) is μ_k F_n = 0.50 × 78.4 N. Kinetic friction applies only when the surfaces slide.
Question 2 (multiple choice · core)
Take +x to the right. A conveyor belt moves to the right at a steady 1.5 m/s. A parcel is placed on it at rest. For a short time the parcel slips on the belt before it reaches the belt’s speed. While it is slipping, what kind of friction does the belt exert on the parcel, and in which direction?
- (A) Kinetic friction, to the right
- (B) Kinetic friction, to the left
- (C) Static friction, to the right
- (D) No friction until the parcel starts moving
Answer and explanation
Answer: (A). The surfaces slide over each other, so the friction is kinetic. Relative to the belt, the parcel is moving left (the belt is running ahead of it). Kinetic friction opposes that relative motion, so it points right. This is the force that speeds the parcel up.
- (B) uses “friction opposes motion” with the wrong motion. What matters is the motion of the parcel relative to the belt, which is to the left.
- (C) names the right direction but the wrong kind: static friction only acts once the parcel stops slipping and moves with the belt.
- (D) confuses “the parcel is at rest relative to the ground” with “the surfaces are not sliding”. The belt slides under the parcel from the first instant.
Question 3 (multiple choice · core)
Take +y upward. A student presses a 0.50 kg block against a vertical wall with a horizontal force of 20 N. The block does not move. For the block and wall, μ_s = 0.60. What is the static friction force on the block?
- (A) 12 N, upward
- (B) 4.9 N, upward
- (C) 4.9 N, downward
- (D) 20 N, upward
Answer and explanation
Answer: (B). Here the normal force comes from the horizontal push, so F_n = 20 N and the maximum static friction is 0.60 × 20 N = 12 N. But the block is at rest, so along y the forces balance: static friction must equal the weight, mg = 0.50 × 9.8 = 4.9 N, pointing up. 4.9 N is below the 12 N maximum, so the block holds.
- (A) is μ_s F_n, the maximum. Static friction only needs to be as large as the weight.
- (C) has the right size but the wrong direction. Without friction the block would slide down, so friction must point up.
- (D) confuses the normal force with friction. The 20 N push is perpendicular to the wall.
Extension: the push could be reduced to mg ÷ μ_s = 4.9 ÷ 0.60 = 8.2 N before the block slips.
Question 4 (calculation · core)
Take +x in the direction of motion. A wooden block is flicked across a level tabletop with a speed of 4.0 m/s. It slides 2.5 m in a straight line and stops. Assuming the only horizontal force is kinetic friction, calculate μ_k for the block and tabletop.
Worked solution
- Kinematics (constant acceleration, final speed 0): v_x² = v_x0² + 2a_x Δx gives 0 = (4.0)² + 2a_x(2.5), so a_x = −16 ÷ 5.0 = −3.2 m/s².
- Forces: vertically F_n = mg. Horizontally the only force is kinetic friction, −μ_k mg. Newton’s second law: −μ_k mg = m a_x.
- The mass cancels: μ_k = −a_x ÷ g = 3.2 ÷ 9.8 = 0.33.
Suggested mark points (3): 1 for the deceleration of 3.2 m/s² (sign consistent with the axis); 1 for writing μ_k mg = m|a_x| with F_n = mg; 1 for μ_k = 0.33 with no units.
Common error: writing a = v² ÷ d and forgetting the factor of 2 gives μ_k = 0.65, double the correct value.
Question 5 (graph · core)
Take +x in the direction of the pull. A 3.0 kg block rests on a level bench. For the block and bench, μ_s = 0.40 and μ_k = 0.30. Starting at t = 0, a horizontal force-sensor pull increases steadily from 0 at a rate of 2.0 N every second.
(a) Find the time at which the block starts to slide. (b) Find the size of the friction force at t = 3.0 s and at t = 8.0 s. (c) Sketch a graph of friction force against time from 0 to 10 s. Label the key values. (d) Find the block’s acceleration at t = 8.0 s.
Worked solution and suggested Marlbridge rubric
(a) F_n = mg = 29.4 N. Maximum static friction = 0.40 × 29.4 = 11.76 N. The pull reaches 11.76 N at t = 11.76 ÷ 2.0 = 5.9 s.
(b) At 3.0 s the pull is 6.0 N and the block is still at rest, so friction = 6.0 N (static). At 8.0 s the block is sliding, so friction = μ_k F_n = 0.30 × 29.4 = 8.8 N (kinetic).
(c) A straight line from (0, 0) rising to a peak of 11.8 N at 5.9 s (static friction equals the pull). A sudden drop of 2.9 N at 5.9 s to 8.8 N. Then a horizontal line at 8.8 N to 10 s.
(d) At 8.0 s the pull is 16 N. a_x = (16 − 8.82) N ÷ 3.0 kg = +2.4 m/s².
| Point | What earns it |
|---|---|
| 1 | 5.9 s from μ_s mg ÷ (2.0 N/s) |
| 1 | 6.0 N at 3.0 s, justified by “at rest, so friction = pull” |
| 1 | 8.8 N at 8.0 s using μ_k |
| 1 | Sketch: rising straight line through the origin, then a drop at the slip time |
| 1 | Sketch: horizontal kinetic level below the peak, with values labelled |
| 1 | 2.4 m/s² with net force = pull − kinetic friction |
Question 6 (experimental design · stretch)
A student wants to find the coefficient of static friction between her trainer and a wooden plank. She has the plank, the trainer, a protractor and a set of 100 g masses that fit inside the trainer.
(a) Describe a procedure she could use. Say what she measures and how she reduces uncertainty. (b) Derive an expression for μ_s in terms of the angle measured. (c) Her five slip angles are 33°, 35°, 34°, 36° and 34°. Calculate μ_s. (d) She then adds masses inside the trainer. Predict how the slip angle changes, with a reason.
Worked solution and suggested Marlbridge rubric
(a) Place the trainer on the plank with the plank flat. Raise one end of the plank slowly until the trainer just starts to slide. Measure the angle between the plank and the horizontal with the protractor at that moment. Repeat at least five times, putting the trainer back in the same place, and take the mean.
(b) Just before slipping, along the slope mg sin θ = μ_s F_n, and perpendicular F_n = mg cos θ. So μ_s = sin θ ÷ cos θ = tan θ.
(c) Mean angle = (33 + 35 + 34 + 36 + 34) ÷ 5 = 34.4°. μ_s = tan 34.4° = 0.68. (Taking the mean of the five tan values also gives 0.68.)
(d) The slip angle should stay the same. Adding mass increases both the pull down the slope (mg sin θ) and the maximum friction (μ_s mg cos θ) by the same factor, so the mass cancels in μ_s = tan θ.
| Point | What earns it |
|---|---|
| 1 | Raises the plank slowly and records the angle at which slipping just begins |
| 1 | Repeats and averages (or another valid way to reduce uncertainty) |
| 1 | F_n = mg cos θ and mg sin θ = μ_s F_n at the threshold |
| 1 | μ_s = tan θ derived |
| 1 | μ_s = 0.68 |
| 1 | Angle unchanged because mass cancels from both forces |
Alternative method. Pulling the trainer along a level plank with a spring scale and reading the largest force just before it moves gives μ_s = F_max ÷ mg. This earns full credit for (a) and (b) if the reading is taken at the point of slipping and the weight is measured.
Question 7 (constructed response · stretch)
Take +x towards the front of a pickup truck. A 20 kg crate sits on the flat, level bed of the truck. The truck starts from rest and speeds up at 1.5 m/s², and the crate moves with the truck without slipping. A student says: “Friction always opposes motion. The crate moves forwards, so the friction on it points backwards.”
(a) Calculate the size and direction of the friction force on the crate. (b) Evaluate the student’s claim. (c) For the crate and truck bed, μ_s = 0.30. Find the largest acceleration the truck can have without the crate slipping.
Model answer and suggested Marlbridge rubric
(a) The crate has the same acceleration as the truck, +1.5 m/s². The only horizontal force on it is static friction from the truck bed. Newton’s second law: F_f,s = ma = 20 kg × 1.5 m/s² = 30 N in the +x direction (forwards).
(b) The claim is incorrect. Friction opposes sliding between the surfaces, not motion relative to the ground. Without friction the truck bed would move forwards from under the crate, so the crate would slide backwards relative to the bed. Static friction opposes that relative sliding, so it points forwards. A forward force is also needed, because the crate accelerates forwards. If friction pointed backwards, the crate would accelerate backwards, which it does not.
(c) The largest static friction is μ_s mg. Setting μ_s mg = m a_max gives a_max = μ_s g = 0.30 × 9.8 = 2.9 m/s². The mass cancels.
| Point | What earns it |
|---|---|
| 1 | 30 N forwards, from F = ma |
| 1 | States friction opposes relative sliding between the surfaces |
| 1 | Explains the crate would slide backwards relative to the bed without friction |
| 1 | Concludes the claim is wrong, linked to the forward acceleration |
| 1 | a_max = μ_s g = 2.9 m/s² |
An answer to (b) that says only “static friction can point either way” earns the conclusion point but not the two reasoning points.
How did you do?
- Q1 or Q3 wrong: re-read “Static friction” and Worked example 1 in the study guide. Static friction comes from a force balance.
- Q2 or Q7 wrong: revisit “Kinetic friction” (direction) and “Friction can point the way you move”.
- Q4 wrong: check “Finding the normal force” and your kinematics from Unit 1.
- Q5 incomplete: work through Figure 1.
- Q6 incomplete: see Worked example 3. Your reasoning needs the why: the mass cancels.
Then tick off the topic checklist.