μsN was computed as 50 N and marked. But 50 N is the ceiling on friction, not its value. The applied force was only 40 N, so the block never moved and friction only had to rise to 40 N.
Static friction is self-adjusting: f = F whenever F < μsN. The value μsN is reached only at the instant sliding begins.
Before using μN, prove the body is actually sliding. If F < μsN, the answer is just F.
For a block pressed against a vertical wall, the normal reaction is the horizontal push (40 N), not the weight. Friction acts vertically and must hold up mg = 20 N, which it exactly can, since μN = 20 N.
N = 40 N ⇒ fmax = 0.5 × 40 = 20 N. Weight to support = 20 N. Friction takes the value required: 20 N upward.
N is always perpendicular to the contact surface. Vertical wall ⇒ N is horizontal ⇒ N ≠ mg.
The tension was found correctly (37.5 N) and then submitted as the answer. The question asked for the force on the pulley, which is pulled down by two string segments.
Fpulley = 2T = 75 N. Equivalently 4m₁m₂g/(m₁+m₂) in one step.
Underline what the question asks for before writing the final line. Tension and pulley load are different quantities.
The standard Atwood tension formula was applied with g = 10, ignoring that the whole system sits inside a lift accelerating upward at 2 m s⁻².
geff = g + a₀ = 12 m s⁻². T = 2m₁m₂geff/(m₁+m₂) = 2(3)(1)(12)/4 = 18 N.
Any system inside an accelerating container: write geff = g ± a₀ as the first line, before any formula.
The lift was descending, but m(g + a) was used — the formula for upward acceleration. This is a direction reversal, not a formula gap: the right formula was applied with the sign flipped.
Descending ⇒ N = m(g − a) = 60(10 − 2) = 480 N = 48 kg.
Ask one physical question, not an algebraic one: does the floor push me harder or softer? Down ⇒ softer ⇒ subtract.
Zero is the tension at the top of a just-completing vertical circle. The question asked for the lowest point. The right physics was recalled and attached to the wrong station.
vbot² = 5gr, and at the bottom gravity acts away from the centre: T = mv²/r + mg = 6mg = 60 N.
Memorise the full triple for the just-completing case — bottom 6mg, horizontal 3mg, top 0 — then read which point is asked.
F = 6t was evaluated at t = 3 s to give 18 N and then treated as a constant force. A force that varies with time cannot be used in v = u + at.
J = ∫₀³ 6t dt = 27 N s ⇒ v = 27/3 = 9 m s⁻¹. (For a force linear in t, the average is half the final value.)
If t appears inside F, the tool is an integral or an area — never constant-acceleration kinematics.
The area under the force–time graph was found correctly as 20 N s, then submitted as the speed. The impulse is a momentum, not a velocity.
Δp = 20 kg m s⁻¹, m = 2 kg ⇒ v = 20/2 = 10 m s⁻¹.
Impulse–momentum is two operations: find the area, then divide by the mass. Check the units of your final number.
Rows A and B were paired correctly; rows C and D were interchanged. This is the exact procedure fault already on record — and option (B) was the decoy built into this paper to test for it.
Region II ⇒ friction constant, body accelerates (R). Slope of Region I ⇒ equals one, dimensionless (S).
After fixing the two rows you are sure of, do not infer the last two by elimination. Verify C and D independently, each against its own statement.
Both statements are true, and both concern static friction — which made (A) feel right. But self-adjustment describes the range of static friction; it says nothing about how the ceiling compares with kinetic friction. The two facts are independent.
μs > μk is an experimental fact about the size of the limit. Self-adjustment is about behaviour below the limit. R does not force A.
Two true statements on the same topic are not automatically an explanation. Step 3 is a separate test: does R force A to be true?
Both are the same one-line idea — momentum arriving or leaving at a steady rate. Q12 is sand landing on a belt (F = v·dm/dt = 10 N); Q40 is a water jet hitting a wall (F = ρAv² = 10 N). Leaving both blank suggests the form F = v·(dm/dt) is not yet automatic.
Learn the one relation and its two dressings: pick-up rate × speed, and ρAv² for a jet (doubled only if it rebounds).
The standard result is hanging fraction = μ/(1 + μ). The coupling — that a longer hanging part leaves a shorter part on the table to make friction — is what makes it look harder than it is.
Memorise ℓ/L = μ/(1 + μ). Here 0.25/1.25 = 20%.
This is a direct rearrangement of T = mv²/r, comparable in difficulty to Q24 (conical pendulum), which was answered correctly. The skip looks like time pressure or hesitation rather than a knowledge gap.
v = √(Tr/m) = √(50/0.5) = 10 m s⁻¹. Attempt questions of this shape — they are single-step.