For a one-dimensional variable force F(x), the work done as the particle moves from x_i to x_f is W = ∫_{x_i}^{x_f} F(x) dx — the area under the F-vs-x curve between the limits. Generalises to W = ∫ F · dr in higher dimensions.
-- NCERT Class 11 Physics, Ch. 5, p. 75Work Variable Force
8 MCQs1 revision card9-step worked example
Source: NCERT Work, Energy and PowerPYQ coverage: NEET 2026Official key: NTA-verifiedLast updated: 25 Sep 2026
Work Variable Force, explained for NEET
The habit that costs marks here is reaching for W = Fs cos θ when the force is changing. That formula is licensed only when F is constant in magnitude and direction over the whole displacement. The moment F depends on position, the product F × s has no single F to use, and multiplying by the initial, final, or largest value of F gives a wrong number rather than an approximate one.
NCERT Class 11 Physics Chapter 5, page 75, states the replacement: the work done by a variable force is the integral of F(x) with respect to x from the initial to the final position,
W = ∫ F(x) dx, from x_i to x_f.
Read it geometrically. Divide the path into slices so narrow that F is effectively constant across each one; on each slice the work is F(x)·Δx, a thin rectangle under the force-position graph. Summing the rectangles and shrinking their width gives the integral. So in one dimension W is the area under the F-versus-x curve between the two positions — and that is often the fastest route in the exam hall, because the area under a straight line or a rectangle needs no calculus at all.
Two consequences worth holding onto. First, area below the x-axis counts as negative work, so a force that reverses direction partway contributes work of both signs, and they cancel in the total. Second, the constant-force result is not a separate law: put F(x) = F, a constant, inside the integral and it comes straight back out as F(x_f − x_i).
The extension beyond one dimension is W = ∫ F·dr, the line integral along the actual path.
Watch out for the word "average." An average force applied over the displacement does reproduce the work — but only if you compute that average correctly from the area, not by averaging the endpoint values of a curved F(x).
Can you answer these Work Variable Force MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
The work done by a variable force in one dimension, as stated in NCERT Class 11 Physics Chapter 5, is expressed as:
Show answer and why every option is right or wrong
Answer: C. C is correct. Page 75 of NCERT Class 11 Physics Chapter 5 gives the work done by a variable force as the integral of F(x) dx from x_i to x_f.
Why A is wrong: A is wrong because a single product F × s presumes one fixed value of F, which is exactly what a variable force does not provide.
Why B is wrong: B is wrong because differentiating the force with respect to position yields a force gradient (N/m), not an energy; work accumulates force over displacement rather than differentiating it.
Why D is wrong: D is wrong because averaging only the two endpoint values of a curved F(x) is not the correct average, and NCERT states the result as an integral, not as an endpoint average.
MCQ 2Easy RecallPractice
In one dimension, the integral expression for the work done by a variable force corresponds graphically to:
Show answer and why every option is right or wrong
Answer: D. D is correct. As NCERT Class 11 Physics Chapter 5 sets out on page 75, the integral of F(x) dx is the area under the force-position curve between the limits.
Why A is wrong: A is wrong because the slope of F against x is the rate at which the force changes with position, not the accumulated work.
Why B is wrong: B is wrong because the area under a position-time graph has units of metre-seconds and is unrelated to work.
Why C is wrong: C is wrong because arc length along the curve mixes newtons and metres in a quantity with no physical meaning here; work is the enclosed area, not the path length of the graph.
MCQ 3Easy RecallPractice
The conditions of use listed in NCERT Class 11 Physics Chapter 5 give the variable-force result a natural generalisation beyond one dimension. That generalisation is:
Show answer and why every option is right or wrong
Answer: B. B is correct. NCERT Class 11 Physics Chapter 5, page 75, gives the higher-dimensional form as the line integral of the dot product F·dr along the path.
Why A is wrong: A is wrong because dropping the dot product discards the angle between force and displacement, so a force perpendicular to the motion would wrongly register work.
Why C is wrong: C is wrong because a cross product returns a vector (this is the form of a torque or moment), whereas work is a signed scalar.
Why D is wrong: D is wrong because force integrated over time is impulse, measured in newton-seconds, not work.
MCQ 4Direct ApplicationPractice
A force acting along the x-axis varies as F(x) = 3x, with F in newtons and x in metres. The work done by this force as the particle moves from x = 0 to x = 4.0 m is:
Show answer and why every option is right or wrong
Answer: C. C is correct. W = ∫₀^4.0 3x dx = (3/2)x² evaluated from 0 to 4.0 m = (1.5)(16) = 24 J, following the variable-force integral on page 75 of NCERT Class 11 Physics Chapter 5.
Why A is wrong: A is wrong because 12 J comes from evaluating the force at x = 4.0 m to get 12 N and then treating that as the answer, confusing a force in newtons with a work in joules.
Why B is wrong: B is wrong because 48 J is the constant-force product using the final force value, 12 N × 4.0 m — it treats the maximum force as acting over the whole displacement and so overstates the work by a factor of two.
Why D is wrong: D is wrong because 96 J multiplies by 2 where the integral divides by 2: 3 × 4.0² × 2 = 96 instead of (3/2) × 4.0² = 24 — the power-rule factor applied upside down.
MCQ 5Direct ApplicationPractice
A force directed along the x-axis has the constant value 5.0 N from x = 0 to x = 2.0 m, and then the constant value 2.0 N from x = 2.0 m to x = 6.0 m. The total work done by this force over the displacement from x = 0 to x = 6.0 m is:
Show answer and why every option is right or wrong
Answer: B. B is correct. The area under the force-position graph is the sum of two rectangles: (5.0 N)(2.0 m) + (2.0 N)(4.0 m) = 10 J + 8.0 J = 18 J, per the area interpretation on page 75 of NCERT Class 11 Physics Chapter 5.
Why A is wrong: A is wrong because 14 J uses 3.5 N — the plain mean of 5.0 N and 2.0 N — across 4.0 m or similar mismatched pairing; the two force values act over unequal distances, so their unweighted mean is not the correct average force.
Why C is wrong: C is wrong because 30 J is 5.0 N × 6.0 m, treating the larger force as acting over the entire displacement and ignoring the step in the graph.
Why D is wrong: D is wrong because 42 J adds the two force values and multiplies by the whole displacement, (5.0 + 2.0) × 6.0; each force acts only over its own stretch.
MCQ 6Direct ApplicationPractice
A force along the x-axis is described by F(x) = 4.0 N for 0 ≤ x ≤ 3.0 m and F(x) = −4.0 N for 3.0 m < x ≤ 6.0 m. The work done by this force as the particle travels from x = 0 to x = 6.0 m is:
Show answer and why every option is right or wrong
Answer: A. A is correct. The area above the axis, (4.0 N)(3.0 m) = +12 J, is cancelled by the equal area below it, −12 J, giving zero net work — the signed-area reading of the integral on page 75 of NCERT Class 11 Physics Chapter 5.
Why B is wrong: B is wrong because 24 J adds the magnitudes of the two areas, ignoring that area below the x-axis represents negative work.
Why C is wrong: C is wrong because −24 J likewise sums magnitudes, and then assigns the whole total the sign of the second stage only.
Why D is wrong: D is wrong because 12 J is the first stage's contribution alone, omitting the equal and opposite second stage.
MCQ 7Concept TrapPractice
A particle moves along the x-axis from x = 0 to x = 5.0 m under a force whose F-versus-x graph is a straight line falling from 10 N at x = 0 to 0 N at x = 5.0 m. A student computes the work as (10 N)(5.0 m) = 50 J. The reason this is incorrect is that:
Show answer and why every option is right or wrong
Answer: D. D is correct. With F varying linearly, the work is the area of the triangle, ½(5.0 m)(10 N) = 25 J, not the initial force times the full displacement — the area interpretation given on page 75 of NCERT Class 11 Physics Chapter 5.
Why A is wrong: A is wrong because the area under the F-versus-x graph gives the work directly; no acceleration or mass information is required.
Why B is wrong: B is wrong because the particle's displacement is a stated 5.0 m and is not adjusted; it is the force value, not the distance, that the student mishandled.
Why C is wrong: C is wrong because the force is non-zero over most of the interval and the enclosed area is finite; only the final instant has F = 0.
MCQ 8CalculationPractice
A force along the x-axis varies as F(x) = 6x², with F in newtons and x in metres. Over the displacement from x = 0 to x = 2.0 m, the constant force that would do the same amount of work is:
Show answer and why every option is right or wrong
Answer: A. A is correct. The work done by the varying force is W = ∫₀^2.0 6x² dx = [2x³]₀^2.0 = 2(8.0) − 0 = 16 J. A constant force F doing the same work over the same 2.0 m displacement satisfies F × 2.0 m = 16 J, so F = 8.0 N.
Why B is wrong: B is wrong because 24 N is the peak force value F(2.0 m) = 6(2.0)² = 24 N, and the peak force exceeds the equivalent constant force for a force that rises across the interval.
Why C is wrong: C is wrong because 4.0 N halves the correct equivalent force, as though the ½ factor that appears for a linear force law also applied to a quadratic one.
Why D is wrong: D is wrong because 12 N is the plain mean of the endpoint values, 0 N and 24 N. That endpoint average is correct only for a linear F(x); for F(x) = 6x² it overstates the true average, which is 16 J ÷ 2.0 m = 8.0 N.
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Work Variable Force: quick recall before you leave
How do you solve a Work Variable Force question? A worked example
- 1
Given.
F(x) = (4.0 N/m)x + 2.0 N; x_i = 1.0 m; x_f = 3.0 m.
- 2
Required.
The work W done by this force over the stated displacement.
- 3
Concept.
The force depends on position, so the constant-force product F·s does not apply. Work is the accumulation of F(x) dx over the displacement, equivalently the area under the F-versus-x graph between the two positions.
- 4
Formula.
W = ∫ F(x) dx from x_i to x_f (NCERT Class 11 Physics Chapter 5, page 75).
- 5
Substitution.
W = ∫₁.₀^3.0 [(4.0)x + 2.0] dx = [2.0x² + 2.0x] evaluated from x = 1.0 m to x = 3.0 m.
- 6
Calculation.
At x = 3.0 m: 2.0(9.0) + 2.0(3.0) = 18 + 6.0 = 24 J. At x = 1.0 m: 2.0(1.0) + 2.0(1.0) = 2.0 + 2.0 = 4.0 J. W = 24 J − 4.0 J = 20 J. The coefficients 4.0 N/m and 2.0 N are the given data at two significant figures; the exponent 2 and the integration constant ½ that produced the 2.0 coefficient are exact and contribute nothing to the significant-figure count.
Cross-check by area. The graph is a straight line from F(1.0 m) = 6.0 N to F(3.0 m) = 14 N over a 2.0 m base. For a linear F(x) the endpoint mean is the correct average: ½(6.0 + 14) = 10 N, and (10 N)(2.0 m) = 20 J. Agreement confirms the integration. - 7
Final answer.
W = 20 J (two significant figures).
- 8
Common trap.
Evaluating the force at one endpoint and multiplying by the displacement. Using F(3.0 m) = 14 N gives 28 J; using F(1.0 m) = 6.0 N gives 12 J. Both are wrong because neither value holds across the interval. Note also that the endpoint-mean shortcut used in the cross-check is legitimate only because F(x) is linear here — for F(x) = 6x² it fails, as MCQ 8 shows.
- 9
Similar NEET-style question.
A force along the x-axis varies as F(x) = (6.0 N/m²)x², with x in metres. Find the work done as the particle moves from x = 0 to x = 2.0 m, and state the constant force that would do the same work over that displacement. *(Answers: W = 16 J; equivalent constant force 8.0 N.)*
What to remember before solving Work Variable Force questions
Which Work Variable Force formulas do you need for NEET?
1 formula — click to collapse
Work done by a variable force
When a force varies along the path, the work done is the line integral of force over displacement. In one dimension this is the area under the F-vs-x curve between the start and end positions.
| Symbol | Quantity | SI Unit |
|---|---|---|
| W | Work done | J |
| F(x) | Force as a function of position | N |
| dx | Infinitesimal displacement | m |
| x_i, x_f | Initial and final positions | m |
Valid when
- Force may depend on position (e.g. spring force F = -k*x)
- Generalises to W = integral F.dr in higher dimensions
More in Work, Energy and Power: 12 exam traps and mistakes · 9 formulas · 7 question patterns from its other lessons.
Work Variable Force questions from past NEET papers
1 question from NEET 2026. Answers verified against NTA official keys. — click to collapse
NEET 2026
13
21
31/3
41/2
NTA Answer: Option 4(final)
Sources
NCERT refs: Class 11 Physics Chapter 5, p.75
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.
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