A block is dragged across a level floor. The floor pushes up on it with a normal reaction N through the whole journey. How much work does N do? A large number of aspirants answer N × s. The answer is zero, and the reason is the only thing this topic asks you to hold onto.
Work done by a constant force is the dot product W = F·s = F s cos θ, where θ is the angle between the force vector and the displacement vector (NCERT Class 11 Physics Chapter 5, page 72). The cos θ is not decoration. It selects the component of F that lies along s, and discards everything perpendicular. For the block on the floor, N points vertically, s points horizontally, θ = 90°, cos 90° = 0. Zero work — not because N is small, but because it is sideways to the motion.
The sign of W follows the same factor, and it is worth memorising as three bands:
Work is a scalar with a sign. It has no direction, but "negative work" is a real, meaningful quantity, not an error in your algebra.
Two conditions are attached to this formula. F must be constant in magnitude and direction over the displacement — a varying force needs the integral form, covered separately in this unit. And when several forces act, apply W = F s cos θ to each force separately, each with its own θ; there is no single θ for the whole set.
Watch-out: displacement, not distance travelled. A body pushed 3 m forward and 3 m back has covered 6 m of path but has s = 0, so the constant applied force has done zero net work on it.
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
In the expression W = F s cos θ for the work done by a constant force, the angle θ is measured between:
Answer: D. D is correct. The definition of work in NCERT Class 11 Physics Chapter 5, page 72, is the dot product F·s, and the angle in a dot product is always the angle between the two vectors being multiplied — here force and displacement.
Why A is wrong: A is wrong because the horizontal is an arbitrary reference direction that plays no part in the definition; when the displacement happens to be horizontal the two agree, which is exactly why the error survives unnoticed.
Why B is wrong: B is wrong because the normal reaction is just one more force acting on the body. Each force gets its own θ measured against the displacement; no force is a reference axis for another.
Why C is wrong: C is wrong because it inverts the roles: θ is not an inclination of the displacement to any fixed axis, it is a relative angle between the two vectors in the product.
Work done by a constant force is:
Answer: D. D is correct. NCERT Class 11 Physics Chapter 5, page 72, defines work as the dot product of two vectors, and a dot product is a scalar. Its sign is inherited from cos θ, so it may be positive, negative or zero.
Why A is wrong: A is wrong because a dot product of two vectors is a scalar, not a vector. Work has magnitude and sign but no direction.
Why B is wrong: B is wrong for the same reason as A — the output of F·s carries no direction at all, so it cannot point along s either.
Why C is wrong: C is wrong because it confuses 'scalar' with 'non-negative'. When 90° < θ ≤ 180°, cos θ is negative and so is the work; a force opposing the motion does genuinely negative work.
The formula W = F s cos θ, as stated in NCERT Class 11 Physics Chapter 5, applies on the condition that the force is:
Answer: A. A is correct. The formula is stated for a constant force (NCERT Class 11 Physics Chapter 5, page 72); constancy means both magnitude and direction are fixed over the displacement, otherwise the integral form for a variable force is required.
Why B is wrong: B is wrong because a force of fixed magnitude that swings in direction has a changing θ along the path, so no single cos θ describes the whole displacement.
Why C is wrong: C is wrong because the formula applies to any one constant force among several. When multiple forces act, it is applied to each separately with that force's own angle.
Why D is wrong: D is wrong because it describes the special case θ = 0. The formula is general: the cos θ factor exists precisely so that non-aligned forces can be handled.
A constant force of 12 N acts on a body at 60° to its displacement of 5.0 m. The work done by this force is:
Answer: A. A is correct. W = F s cos θ = 12 × 5.0 × cos 60° = 12 × 5.0 × 0.50 = 30 J, i.e. 3.0 × 10¹ J (NCERT Class 11 Physics Chapter 5, page 72).
Why B is wrong: B is wrong because it is F × s = 12 × 5.0 with the cos θ factor dropped entirely — the component-selection step skipped.
Why C is wrong: C is wrong because it uses sin 60° ≈ 0.87 instead of cos 60°. The dot product takes the cosine of the angle between the vectors; the sine belongs to the cross product.
Why D is wrong: D is wrong because zero work requires θ = 90°. At 60° the force has a substantial component along the displacement and does positive work.
A block slides a distance of 4.0 m along a horizontal floor. The normal reaction exerted by the floor on the block is 50 N. The work done by the normal reaction on the block during this slide is:
Answer: C. C is correct. The normal reaction is vertical and the displacement is horizontal, so θ = 90° and cos 90° = 0, giving W = 0 whatever the magnitude of N (NCERT Class 11 Physics Chapter 5, page 72).
Why A is wrong: A is wrong because it computes N × s = 50 × 4.0 and treats the normal reaction as if it were aligned with the motion. It is perpendicular to it.
Why B is wrong: B is wrong because it applies a minus sign to the same N × s product, as though N opposed the motion. A perpendicular force neither helps nor opposes — it does no work at all.
Why D is wrong: D is wrong because it halves the N × s product, which corresponds to no angle in this situation; the perpendicularity makes the work exactly zero, not a fraction of Ns.
A constant frictional force of 8.0 N acts on a crate that is dragged 3.0 m across a floor. The work done by the frictional force on the crate is:
Answer: C. C is correct. Kinetic friction acts opposite to the displacement, so θ = 180° and cos 180° = −1, giving W = 8.0 × 3.0 × (−1) = −24 J, i.e. −2.4 × 10¹ J (NCERT Class 11 Physics Chapter 5, page 72).
Why A is wrong: A is wrong because it drops the sign, treating the magnitude of the product as the answer. The direction of friction relative to the displacement is what the cos θ factor is there to register.
Why B is wrong: B is wrong because friction here is antiparallel to the displacement, not perpendicular to it. Zero work needs θ = 90°.
Why D is wrong: D is wrong because it doubles the product for no reason available in the data; F s is 24 J in magnitude.
A student is asked for the work done by a 20 N horizontal force on a trolley and answers 'the work is 60 J, directed forwards'. The defect in that answer is that:
Answer: B. B is correct. Work is the dot product F·s, and a dot product of two vectors yields a scalar (NCERT Class 11 Physics Chapter 5, page 72). The number and its sign are the complete answer; 'directed forwards' is not a property work can have.
Why A is wrong: A is wrong because a force with a component along the displacement does positive work. The sign is set by cos θ, and there is nothing in the description forcing it negative.
Why C is wrong: C is wrong because the joule is the SI unit of work and is defined as one newton-metre; both labels denote the same unit, so this is not a defect.
Why D is wrong: D is wrong because it relocates the error rather than identifying it. Force is indeed a vector with direction, but the student attached the direction to the work, which is the scalar output.
A constant horizontal force of 15 N pushes a box 6.0 m east along a level floor, and then the same force, still pointing east, pushes the box 6.0 m east again in a second stage. Separately, a second box is pushed 6.0 m east by a 15 N eastward force and is then pushed back 6.0 m west by a 15 N westward force. Comparing the total work done by the applied force on each box over the full journey:
Answer: B. B is correct. For the first box, both stages have θ = 0°, so W = 15 × 6.0 + 15 × 6.0 = 180 J. For the second box the return force reverses with the motion, so the second stage also has θ = 0° and gives another +90 J: total 180 J, i.e. 1.8 × 10² J (NCERT Class 11 Physics Chapter 5, page 72).
Why A is wrong: A is wrong because it assumes the second box's return trip cancels the outward work. It would only cancel if the force still pointed east while the box moved west; the stem states the force reverses too, so θ = 0° again and the work adds.
Why C is wrong: C is wrong on the first box as well: it counts only one 6.0 m stage of a two-stage journey with an unchanged force.
Why D is wrong: D is wrong because the first box moves with the force throughout, which is the clearest possible case of positive work; assigning it zero inverts the situation.
Get a structured 30-day Mechanics plan and a complete formula booklet — delivered to your inbox instantly.
Given
• Displacement s = 8.0 m, horizontal• Rope tension T = 50 N, constant, at 60° above the horizontal• Normal reaction N = 1.6 × 10² N, vertical (= mg − T sin 60° = 200 − 43.3 = 157 N)• Mass m = 20 kg (not needed — see step 3)
Required
W_T, the work done by the tension, and W_N, the work done by the normal reaction.
Concept
Work done by a constant force is the dot product of that force with the displacement. Each force is treated separately, with its own angle to the displacement. Mass does not appear in the definition of work, so the 20 kg is surplus data here — a routine NEET inclusion.
Formula
W = F s cos θ
Substitution
• Tension: θ = 60° (the rope's stated inclination to the horizontal displacement), so W_T = 50 × 8.0 × cos 60°.• Normal reaction: vertical force, horizontal displacement, so θ = 90° and W_N = 1.3 × 10² × 8.0 × cos 90°.
Calculation
• cos 60° = 0.500 (exact). W_T = 50 × 8.0 × 0.500 = 200 J.• cos 90° = 0 (exact). W_N = 1.3 × 10² × 8.0 × 0 = 0 J.
The angles 60° and 90° are exact values, and so the cosines derived from them are exact; they do not contribute to the significant-figure count. The measured inputs 50 N and 8.0 m both carry two significant figures, so W_T is reported to two.
Final answer
W_T = 2.0 × 10² J. W_N = 0 J exactly.
Common trap
Reporting W_N = 1.3 × 10² × 8.0 = 1.0 × 10³ J, by treating the normal reaction as though it acted along the motion. The normal reaction is perpendicular to a horizontal displacement, so cos θ = 0 and its work is zero no matter how large N is. The same slip in reverse produces W_T = 50 × 8.0 = 4.0 × 10² J from dropping the cos 60°. Check the angle for every force before multiplying.
Similar NEET-style question
A sledge is pulled 12 m along level ground by a constant 40 N force directed at 30° above the horizontal, while a constant 15 N friction force opposes the motion. Find the work done by the pulling force and the work done by friction, separately.
Work W done by a constant force F on an object that undergoes a displacement s is the scalar product: W = F · s = |F| |s| cos θ, where θ is the angle between F and s. SI unit: joule (J) = N·m. Work is a scalar but has a sign: positive when 0 ≤ θ < 90°, zero when θ = 90°, negative when 90° < θ ≤ 180°.
-- NCERT Class 11 Physics, Ch. 5, p. 74The work done by a constant force F on an object that undergoes a displacement s is the dot product F.s. Equivalently, W = (magnitude of F) * (magnitude of s) * cos(angle between them). Work is a scalar but has a sign.
| Symbol | Quantity | SI Unit |
|---|---|---|
| W | Work (scalar, signed) | J |
| F | Constant force (vector) | N |
| s | Displacement (vector) | m |
| theta | Angle between F and s | rad/deg |
These are the exact patterns that cause wrong answers in NEET. Each trap includes when it triggers and how to avoid it.
Root cause: direction ignored
Work = F . s = F * s * cos(theta). For a block sliding horizontally, the normal force is perpendicular to displacement (theta = 90 deg, cos = 0), so the work done by N is ZERO. Always check the angle between force and displacement before computing work.
Distractor reports W = N * s for a block sliding on a level surface.
More in Work, Energy and Power: 11 exam traps and mistakes · 9 formulas · 7 question patterns from its other lessons.
No question in our NEET 2020–2025 set targets this topic directly.
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
Test yourself on this topic with real past-paper questions:
Practice this topic →