BPSC 71st CCE Prelims General Studies · Question 121 of 150

Correct — B, 200 Newtons. The decisive words in the stem are 'constant velocity', and everything else in the question is decoration. Constant velocity means the velocity is not changing, so the acceleration is zero; and by Newton's second law, F(net) = ma, zero acceleration forces the net external force on the box to be exactly zero. Resolve the forces on the box: vertically, its weight acts downwards and the floor's normal reaction acts upwards, and these two cancel because the box is not sinking or lifting off; horizontally, only two forces exist — your 200 N push forwards and the friction of the floor backwards. For the horizontal sum to vanish, friction must be 200 N, equal in magnitude and opposite in direction to the push. That is option (b), and no other option can be reconciled with steady motion. Notice how little the question actually gives you: no mass, no coefficient of friction, no material of the floor beyond 'wooden Box'. That absence is deliberate and it is the clue — the examiner has stripped out every quantity you would need to compute friction independently from f = (mu)N, because the answer is meant to come from the equilibrium condition alone. On terminology, one honest clarification, because it is what makes the question look harder than it is. The paper's phrase 'marginal friction force' is the Hindi seemant gharshan bal of the parallel Hindi text, the standard Indian-textbook term normally rendered in English as limiting friction. Strictly, limiting friction is the maximum value that static friction can reach at the instant sliding is about to begin, whereas the force acting on a body that is already sliding is kinetic or sliding friction, which for the same pair of surfaces is a little smaller. The box in the stem is already moving, so what balances your hand is kinetic friction — and because the box neither speeds up nor slows down, that kinetic friction is exactly 200 N. Whichever of the two names you attach to it, 200 N is the only value on the option list consistent with the motion described. A concrete feel for the number: if the wood-on-floor kinetic coefficient is a typical 0.2, then the normal reaction is N = f/(mu) = 200/0.2 = 1,000 N, which is a box of roughly 102 kg — a heavy crate, pushed steadily by one determined person.

• (a)100 Newtons — The intuitive answer, and the one this item is built to punish. It assumes half your push is 'used up' by friction and the other half is 'left over' to keep the box going. But an unopposed 100 N forward would be a net force, and a net force means acceleration: a 100 kg box would gain 1 m/s of speed every second and be moving at 10 m/s after ten seconds. That directly contradicts 'constant velocity'. This is Aristotle's physics — motion needs a surviving force — not Newton's, in which only a CHANGE of motion needs one.
• (c)300 Newtons — If friction were 300 N against a 200 N push, the net force would be 100 N backwards, the box would decelerate and would come to rest within seconds. A body cannot slide at unchanging speed while a net backward force acts on it. Candidates arrive here by treating friction as an active opposing agent that always 'wins' over the applied force, instead of a passive contact force whose value during steady sliding is pinned by the balance condition to exactly match the push.
• (d)400 Newtons — Twice the applied force — usually chosen by half-remembering f = (mu)N and treating some remembered number as a multiplier of the push, which the friction law never is: the normal reaction N, not the applied force, sets the friction. It also fails a second, simpler test. If this floor could really supply 400 N of resistance, then the limiting static friction would exceed your 200 N push and the box would never have started moving at all — yet the stem says it is moving.

Friction is the contact force that opposes relative sliding between two surfaces, and it comes in three regimes that examiners deliberately blur. Static friction is self-adjusting: push a stationary crate with 40 N and static friction returns exactly 40 N, push with 90 N and it returns 90 N, up to a ceiling called limiting friction, equal to (mu-s)N. Push harder than that ceiling and the crate breaks away; from then on the resistance is kinetic (sliding) friction, f = (mu-k)N, which stays roughly constant while sliding and is smaller than the limiting value — which is precisely why a heavy almirah is hard to start and noticeably easier to keep moving. The empirical rules are the Amontons-Coulomb laws, set out by Guillaume Amontons in 1699 and put on a quantitative footing by Charles-Augustin de Coulomb in 1785: friction is proportional to the normal reaction, is independent of the apparent area of contact, and is nearly independent of the speed of sliding. The apparent-area rule looks absurd until you know the microscopic reason — two surfaces touch only at the tips of microscopic asperities, so the REAL contact area is a tiny fraction of the visible one and grows in proportion to the load, not to how the block is laid out. Sitting underneath all of this is Newton's first law, from the Principia of 1687: rest and uniform motion in a straight line are the same dynamical state, and a body stays in it unless an unbalanced external force acts. A box sliding at steady speed is therefore in equilibrium exactly as a box standing still is — a condition physicists call dynamic equilibrium.

Reason to this answer in three moves and it takes about ten seconds. First, underline 'constant velocity'. Those two words convert a dynamics problem into a statics problem, because constant velocity means zero acceleration, and zero acceleration means zero net force. Second, draw the free body: weight down, normal reaction up (these balance), 200 N push forward, friction f backward. Third, set the horizontal sum to zero, which gives f = 200 N immediately. The single discriminating fact is the equivalence 'constant velocity = zero acceleration = zero net force'; every wrong option on the list corresponds to a net force and therefore to a changing speed, so each one contradicts the stem on its own terms. Option (a) would make the box accelerate, options (c) and (d) would make it decelerate. There is also a meta-clue worth internalising for the whole physics block of this paper: the stem supplies no mass, no coefficient and no surface data. When a one-mark numerical withholds every quantity a formula would need, the examiner is telling you the answer comes from a conservation or balance principle, not from substitution. The trap is psychological rather than mathematical. Everyday experience says that if you stop pushing a crate it stops, so pushing must be 'beating' friction — hence option (a). What everyday experience actually shows is that removing your push leaves friction unbalanced, which decelerates the crate. You have to exceed friction only to START the box moving, when you must beat the larger limiting value, or to ACCELERATE it. To move it at a steady speed you match friction exactly, never more.

• Newton's first law, stated in the Principia (1687), makes rest and uniform straight-line motion the same state: a body changes that state only under an unbalanced external force. So constant velocity implies zero acceleration implies zero net force — the entire content of this question.
• The Amontons-Coulomb laws of dry friction (Guillaume Amontons, 1699; Charles-Augustin de Coulomb, 1785): friction is proportional to the normal reaction N, is independent of the apparent area of contact, and is nearly independent of sliding speed. Static friction is self-adjusting between 0 and the limiting value (mu-s)N; kinetic friction has the fixed value (mu-k)N.
• For the same pair of surfaces (mu-k) is less than (mu-s), which is why the push needed to start a load is larger than the push needed to keep it moving. Representative dry values: wood on wood (mu-s) about 0.25-0.5 and (mu-k) about 0.2; rubber on dry concrete (mu-s) about 1.0; steel on steel about 0.57 dry but roughly 0.06 when lubricated.
• At a kinetic coefficient of 0.2, a friction force of 200 N implies a normal reaction of 1,000 N and hence a box of about 102 kg (taking g = 9.8 m/s squared) — the arithmetic the stem deliberately withholds, and a useful sanity check that the situation described is a physically ordinary one.
• Rolling friction is smaller than sliding friction by one to two orders of magnitude — the coefficient for a steel railway wheel on a steel rail is around 0.001 — which is why wheels, ball bearings and roller bearings exist, and why the same 200 N would move a far heavier load on a trolley than on a bare floor.
• Dynamic equilibrium is not a special case: a skydiver at terminal velocity (about 53 m/s, roughly 195 km/h, in the belly-to-earth posture), a car cruising at a fixed speed on a level road, and this wooden box are all bodies with zero net force and non-zero velocity.

• Reading 'constant velocity' as though it merely meant 'moving' — it is the whole question, because it fixes the acceleration, and hence the net force, at zero
• Believing that motion needs a surviving net force, so that the applied force must exceed friction; a net force is needed only to start or to change motion, never to sustain it
• Confusing limiting friction (the maximum static value, just before sliding starts) with kinetic friction (what acts during sliding, and slightly smaller) — the paper's 'marginal friction force' is the former term, but the box in the stem is already sliding

BPSC asks mechanics as a one-step substitution wrapped in slightly non-standard translated English — 'marginal friction force' for limiting friction here — with four round numbers as options and a single law doing all the work; the reliable strategy is to hunt for the phrase that fixes the acceleration. UPSC has almost stopped setting plug-in numericals since the 1990s and asks friction conceptually instead: in 2013 it asked WHY ball bearings help (the effective contact area is reduced), and in 2000 it set an assertion-reason on a man on a frictionless surface propelling himself by whistling, which had to be reasoned out from conservation of momentum.

• practice — not a real PYQ

  A box of mass 20 kg is pulled along a horizontal floor at a constant speed by a horizontal force of 50 N. Taking g = 10 m/s squared, the coefficient of kinetic friction between the box and the floor is

  • (a)0.10
  • (b)0.25
  • (c)0.40
  • (d)0.50

  Answer(b) 0.25 — constant speed means zero net force, so friction equals the applied 50 N; the normal reaction is N = mg = 20 x 10 = 200 N; hence mu = f/N = 50/200 = 0.25.

• practice — not a real PYQ

  Which one of the following statements about friction is NOT correct ?

  • (a)Limiting friction is greater than kinetic friction for the same pair of surfaces
  • (b)Kinetic friction is directly proportional to the normal reaction
  • (c)Friction is directly proportional to the apparent area of contact between the surfaces
  • (d)Rolling friction is much smaller than sliding friction

  Answer(c) Friction is directly proportional to the apparent area of contact between the surfaces — this is the one false statement: dry friction depends on the normal reaction and not on the apparent contact area, because the real contact happens only at microscopic asperities. The other three are standard results.