Newton's Laws, Friction and Accelerated Frames: questions with solutions
3 free questions from this chapter (the full set has 30). Marking: Multi-correct: +4 all correct, +1 per correct option if none wrong, −2 if any wrong · Numerical: +4 / 0 · Matching: +3 / −1.
Question 1Level 2 · ExamSingle correct
Two blocks are held at rest against a fixed vertical wall. Block B (4kg) touches the wall; block A (2kg) touches B on the side away from the wall. A horizontal force F, perpendicular to the wall, pushes A towards the wall; nothing else touches the blocks. The coefficient of static friction is 0.2 between A and B and 0.75 between B and the wall. Take g=10m/s2. The least value of F for which both blocks remain at rest is ______. Exactly one option is correct.
A80N
B100N
C180N
D120N
Show answer and solution
Answer: B
Horizontally, every contact carries the same normal force F.
A: its weight 20N must be held by friction from B, so fAB=20N (upward on A, downward on B).
B: the wall must hold B's weight plus the downward 20N from A, so fW=60N.
Both frictions are fixed by equilibrium (not by the limits), so the conditions are 20≤0.2F⇒F≥100N and 60≤0.75F⇒F≥80N.
The A-B interface is the critical one: Fmin=100N.
Trap: 80N checks only the wall contact with the total weight.
Difficulty: Hard
Question 2Level 3 · AdvancedSingle correct
A small bob hangs by a light inextensible string from the ceiling of a car on a level road. The car moves with constant acceleration 7.5m/s2 and the bob is at rest relative to the car. At t=0 the car's acceleration suddenly becomes zero (it continues at constant velocity). Take g=10m/s2. The ratio of the string tension just after t=0 to the tension just before t=0 is ______. Exactly one option is correct.
A1
B54
C2516
D259
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Answer: C
Before: in the car frame, effective gravity geff=g2+a2=12.5m/s2; the string lies along it, at θ to the vertical with tanθ=a/g=3/4. T1=m⋅12.5.
Just after: the car is now an inertial frame. The bob still sits at angle θ and is at rest relative to the car (velocity cannot jump), so there is no centripetal term: T2=mgcosθ=m⋅10⋅0.8=8m.
Ratio =8/12.5=16/25 (equivalently cos2θ).
Insight: the angle and relative velocity are continuous, but the pseudo force vanishes instantly; the bob then starts to swing with tangential acceleration gsinθ=6m/s2.
Difficulty: Very hard
Question 3Level 3 · AdvancedSingle correct
A smooth wedge with inclined face at 37∘ to the horizontal rests on a horizontal table. Take the x-axis horizontal; the inclined face rises in the +x direction. A small block is held at rest on the inclined face. At t=0 the block is released and, at the same instant, the wedge is given a constant acceleration 20m/s2 in the +x direction (by an external agent). Take g=10m/s2, sin37∘=0.6, cos37∘=0.8. The magnitude of the block's acceleration relative to the wedge just after t=0 is ______. Exactly one option is correct.
A10m/s2
B20m/s2
C22m/s2
D105m/s2
Show answer and solution
Answer: D
Work in the wedge frame: pseudo force ma along −x. The outward normal of the face is (−sinθ,cosθ).
Component of (gravity + pseudo force) pressing the block into the face: mgcosθ−masinθ=m(8−12)<0.
So the face cannot push hard enough to keep contact: contact is lost as soon as a>gcotθ=13.3m/s2. Here a=20, so the block leaves the surface immediately and is in free fall (ground acceleration g downward).
Relative to the wedge: arel=(0,−10)−(20,0), magnitude 400+100=105≈22.4m/s2.
Trap: assuming the block stays on the face gives gsinθ+acosθ=22m/s2; always check N≥0 first.
Difficulty: Very hard
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