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Work

Work is done when a force produces displacement. It measures energy transferred by a force along the displacement direction.

Definition And Physical Meaning

If a force F acts on a body and the body has displacement s, work is the product of force component along displacement and displacement.

W = F . s W = Fs cosθ

W = work done by force.
F = magnitude of force.
s = magnitude of displacement.
θ = angle between force and displacement.
Only F cosθ contributes to work.

Nature, Unit And Dimensions

Property	Result
Nature	Scalar. Work may be positive, negative or zero, but it has no direction.
SI Unit	joule, J = newton metre
Dimensions	[M L² T^-2]
1 joule	Work done when 1 N force displaces a body by 1 m along force direction.

Force Along Displacement

Force Opposite Displacement

Force At Angle θ

Positive Work

Work is positive when the force component is in the direction of displacement, so cosθ > 0.

Condition

0° ≤ θ < 90°cosθ > 0

Force helps the motion and transfers energy to the body.

Examples

Pulling a trolley forward.
Gravity during falling motion.
Engine pulling a train forward.

Diagram

Negative Work

Work is negative when force has a component opposite to displacement, so cosθ < 0.

Condition

90° < θ ≤ 180°cosθ < 0

Force removes energy from the body or opposes motion.

Examples

Friction on a sliding block.
Brakes on a moving vehicle.
Gravity during upward motion.

Diagram

Zero Work

Work is zero if displacement is zero, or force is perpendicular to displacement, or the force itself is zero.

Condition

θ = 90°cos90° = 0

Perpendicular force changes direction of motion but not speed, so it does no work.

Examples

Centripetal force in uniform circular motion.
Person carrying load horizontally at constant height.
Gravity on a satellite in circular orbit.

Diagram

Work By Constant Force

For constant force, resolve force along displacement and multiply by displacement.

Derivation

Component of force along displacement is F cosθ. Work is this component multiplied by displacement s.

W = (F cosθ)sW = Fs cosθ

Parallel force: θ = 0°, W = Fs.
Anti-parallel force: θ = 180°, W = -Fs.
Inclined force: use only F cosθ.

Numerical Examples

A 20 N force moves a body 5 m in same direction.

Given F = 20 N, s = 5 m, θ = 0°. W = Fs cos0° = 20 x 5 = 100 J. Exam tip: same direction means maximum positive work.

A 50 N force acts at 60° and displacement is 4 m.

W = Fs cosθ = 50 x 4 x cos60° = 100 J. Exam tip: always include cosθ for inclined force.

Friction 10 N opposes displacement 8 m.

θ = 180°. W = 10 x 8 x cos180° = -80 J. Friction removes mechanical energy.

Work By Variable Force

Very important: when force changes with position, work is found by integration or by area under the force-displacement graph.

Integration Formula

If force varies with position x, divide displacement into tiny parts dx. Small work dW = F dx.

dW = F dxW = ∫ F dx

Graphically, W equals area under F-x curve.

Hooke's Law And Spring Work

For a spring, restoring force magnitude is proportional to extension or compression.

F = kxW = ∫₀^x kx dx = 1/2 kx²

This is the elastic potential energy stored in the spring.

Increasing Force

Decreasing Force

Spring Force

Graphical Interpretation

Area under F-x graph is work done. Rectangles, triangles and trapeziums appear repeatedly in CBSE, NEET and JEE.

Constant Force: Rectangle

Variable Force: Triangle

Trapezium Area

Solved Graph Problem 1

Force is constant 12 N from x = 0 to x = 5 m. Find work.

Area = rectangle = 12 x 5 = 60 J. Final answer: 60 J.

Solved Graph Problem 2

Force rises linearly from 0 to 30 N over 6 m. Find work.

Area = triangle = 1/2 x 6 x 30 = 90 J. Final answer: 90 J.

Important Conceptual Questions

Conceptual traps used in NEET, JEE Main and JEE Advanced.

Can work be negative?

Yes. If force opposes displacement, θ is greater than 90° and work is negative. Example: friction.

Can work be zero?

Yes. If displacement is zero or force is perpendicular to displacement, work is zero.

Is work scalar or vector?

Work is scalar because it is a dot product of force and displacement.

Does centripetal force do work?

In uniform circular motion, no. It is perpendicular to instantaneous displacement.

Does normal reaction do work?

Usually no for a body moving on a fixed surface, but it may do work if the point of contact moves, such as in a moving wedge or lift.

Does gravity always do positive work?

No. Gravity does positive work during downward motion, negative work during upward motion and zero work during horizontal displacement.

High-Quality Numericals

Solved numerical bank with question, diagram idea, given data, formula, calculation, final answer and exam tip.

CBSE: A 15 N force moves a block 6 m in its direction.

Diagram: force arrow parallel to displacement. Given F = 15 N, s = 6 m, θ = 0°. Formula W = Fs cosθ. Calculation W = 15 x 6 x 1 = 90 J. Final answer: 90 J. Exam tip: parallel force gives maximum work.

NEET: A 40 N force at 60° displaces a body by 5 m.

Diagram: inclined force with horizontal displacement. W = Fs cosθ = 40 x 5 x 1/2 = 100 J. Final answer: 100 J. Exam tip: resolve force along displacement.

JEE Main: Force F = 3x acts from x = 0 to 4 m.

Diagram: increasing straight F-x graph. Given F = 3x. W = ∫0 to 4 3x dx = 3x²/2 from 0 to 4 = 24 J. Exam tip: variable force needs integration.

JEE Advanced: Force F = 2x + 4 acts from x = 1 m to 5 m.

Diagram: straight line F-x graph. W = ∫1 to 5 (2x + 4) dx = [x² + 4x]1 to 5 = 45 - 5 = 40 J. Final answer: 40 J. Exam tip: use limits carefully.

IB: A student lifts a 3 kg mass through 2 m at constant speed.

Diagram: upward force and upward displacement. Given m = 3 kg, h = 2 m. W = mgh = 3 x 9.8 x 2 = 58.8 J. Exam tip: constant speed means applied force equals weight.

IGCSE: A 20 N force moves a trolley 4 m.

W = force x distance = 20 x 4 = 80 J. Final answer: 80 J. Exam tip: use joule as unit.

A-Level: A spring of k = 200 N/m is compressed by 0.10 m.

Diagram: compressed spring. W = 1/2 kx² = 1/2 x 200 x 0.01 = 1 J. Final answer: 1 J. Exam tip: x must be in metres.

JEE Main: A braking force of 500 N stops a car over 20 m.

Force is opposite displacement. W = -Fs = -500 x 20 = -10000 J. Final answer: -1.0 x 10^4 J. Exam tip: braking work is negative.

NEET Question Bank

50 high-quality NEET-style MCQs on work, signs, variable force and graph interpretation.

1. NEET Exam-style Question: Work done by 10 N over 3 m along force is: A 10 J B 20 J C 30 J D 40 J

Answer: C. W = Fs = 10 x 3 = 30 J.

2. Force and displacement are perpendicular. Work is: A positive B negative C zero D infinite

Answer: C. W = Fs cos90° = 0.

3. Friction on a sliding body usually does: A positive work B negative work C zero work D no force

Answer: B. Friction opposes displacement.

4. Work is a: A vector B scalar C tensor D unit vector

Answer: B. Work is dot product of force and displacement.

5. SI unit of work is: A watt B joule C newton D pascal

Answer: B. Work is measured in joules.

6. Dimensions of work are: A MLT^-2 B ML^2T^-2 C ML^-1T^-2 D MLT^-1

Answer: B. Work = force x displacement = ML T^-2 x L.

7. If θ = 60°, F = 20 N and s = 2 m, work is: A 10 J B 20 J C 30 J D 40 J

Answer: B. W = 20 x 2 x 1/2 = 20 J.

8. Work by gravity on a falling body is: A positive B negative C zero D undefined

Answer: A. Gravity and displacement are downward.

9. Work by gravity on upward moving stone is: A positive B negative C zero D maximum positive

Answer: B. Gravity is downward, displacement upward.

10. Centripetal force in uniform circular motion does: A positive work B negative work C zero work D variable positive work

Answer: C. Force is perpendicular to instantaneous displacement.

11. Area under F-x graph gives: A power B work C momentum D acceleration

Answer: B. W = ∫F dx.

12. F = 5x from 0 to 2 m. Work is: A 5 J B 10 J C 15 J D 20 J

Answer: B. W = ∫0^2 5x dx = 10 J.

13. Spring work for k = 100 N/m and x = 0.2 m is: A 1 J B 2 J C 4 J D 20 J

Answer: B. W = 1/2 kx² = 0.5 x 100 x 0.04 = 2 J.

14. For negative work, θ is: A less than 90° B equal 0° C between 90° and 180° D always 45°

Answer: C. cosθ is negative in second quadrant.

15. Zero work happens when: A force is zero B displacement is zero C force perpendicular displacement D all of these

Answer: D. Each condition gives W = 0.

16. A porter carries luggage horizontally. Work by weight is: A positive B negative C zero D equal mgh

Answer: C. Weight is vertical, displacement horizontal.

17. A rectangular F-x graph has height 8 N and base 5 m. Work is: A 13 J B 20 J C 40 J D 80 J

Answer: C. Area = 8 x 5 = 40 J.

18. A triangular F-x graph has base 4 m and height 10 N. Work is: A 10 J B 20 J C 40 J D 80 J

Answer: B. Area = 1/2 x 4 x 10 = 20 J.

19. Work done by normal reaction on a block sliding on fixed horizontal table is: A zero B positive C negative D mgx

Answer: A. Normal is perpendicular to horizontal displacement.

20. 1 joule equals: A N/m B N m C kg m/s D W/s

Answer: B. 1 J = 1 N m.

21. Work by force F = 6 N opposite 2 m displacement is: A 12 J B -12 J C 3 J D -3 J

Answer: B. θ = 180°, W = -Fs = -12 J.

22. If force is doubled and displacement unchanged, work becomes: A half B same C double D four times

Answer: C. W is proportional to F for same angle and displacement.

23. If displacement is doubled and force unchanged, work becomes: A half B same C double D four times

Answer: C. W is proportional to s.

24. A force acts but body does not move. Work is: A Fs B F/s C zero D infinite

Answer: C. Displacement is zero.

25. Work done by resultant force can change: A mass B kinetic energy C charge D temperature only

Answer: B. By work-energy theorem, net work changes kinetic energy.

26. For θ = 120°, sign of work is: A positive B negative C zero D depends only on F

Answer: B. cos120° is negative.

27. For θ = 30°, sign of work is: A positive B negative C zero D impossible

Answer: A. cos30° is positive.

28. Work done in stretching spring from 0 to x depends on: A x B x² C 1/x D independent of x

Answer: B. W = 1/2 kx².

29. F = constant 7 N from x = 2 m to x = 6 m. Work is: A 14 J B 21 J C 28 J D 42 J

Answer: C. Displacement interval = 4 m, W = 7 x 4 = 28 J.

30. F = x² from 0 to 3 m. Work is: A 3 J B 6 J C 9 J D 27 J

Answer: C. ∫0^3 x² dx = x³/3 = 9 J.

31. Work done by a force can be zero even if force is non-zero when: A s = 0 B θ = 90° C both possible D never

Answer: C. Both conditions make W zero.

32. A force of 25 N makes 0° with displacement 4 m. Work is: A 0 B 50 J C 100 J D -100 J

Answer: C. W = 25 x 4 = 100 J.

33. A force of 25 N makes 180° with displacement 4 m. Work is: A 0 B 50 J C 100 J D -100 J

Answer: D. W = 25 x 4 x (-1) = -100 J.

34. If F-x graph lies below x-axis, work is: A positive B negative C zero D always maximum

Answer: B. Signed area below axis is negative.

35. A trapezium F-x graph with parallel sides 10 N and 20 N, width 4 m has work: A 30 J B 40 J C 60 J D 80 J

Answer: C. Area = 1/2(10+20)4 = 60 J.

36. Work by air resistance on falling object is: A positive B negative C zero D always mgh

Answer: B. Air resistance opposes downward displacement.

37. Work by applied force in lifting load slowly is: A mgh B -mgh C zero D 2mgh

Answer: A. Applied force and displacement are upward.

38. Work by gravity in lifting load height h is: A mgh B -mgh C zero D Fh

Answer: B. Gravity is opposite upward displacement.

39. Unit watt is related to: A work B power C force D displacement

Answer: B. Watt is power, not work.

40. Dot product in work uses: A sinθ B tanθ C cosθ D secθ

Answer: C. Work = Fs cosθ.

41. A body moves in circle at constant speed. Work by net force in one round is: A positive B negative C zero D mgh

Answer: C. Speed and kinetic energy remain constant.

42. F = 10 - x from x = 0 to 4. Work is: A 24 J B 28 J C 32 J D 40 J

Answer: C. ∫0^4 (10-x)dx = 40 - 8 = 32 J.

43. Work done by force over closed path can be non-zero for: A conservative force only B non-conservative force C no force D gravity only

Answer: B. Non-conservative forces may do non-zero work in closed path.

44. Work by gravity over closed path is: A zero B positive C negative D infinite

Answer: A. Gravity is conservative.

45. F = 4x² from 0 to 2 m. Work is: A 8/3 J B 16/3 J C 32/3 J D 16 J

Answer: C. ∫0^2 4x²dx = 4 x 8/3 = 32/3 J.

46. A force has no component along displacement. Work is: A zero B Fs C -Fs D F/s

Answer: A. Component along displacement is zero.

47. If W is negative, force tends to: A speed up only B slow down C increase mass D remove displacement

Answer: B. Negative work often decreases kinetic energy.

48. Work done by tension in ideal circular pendulum at lowest instant is: A zero instantaneously B positive C negative D infinite

Answer: A. Tension is radial and displacement is tangential.

49. Work done by a varying force is exactly found from: A Fs always B area under F-x C area under v-t D slope of x-t

Answer: B. W = ∫F dx.

50. If k doubles and x same, spring work becomes: A half B same C double D four times

Answer: C. W = 1/2 kx².

JEE Main Question Bank

50 difficult JEE Main style questions on inclined force, variable force, graphs and work-energy concepts.

1. F = 12 N acts at 60° for displacement 10 m. Find work.

W = Fs cos60° = 12 x 10 x 1/2 = 60 J.

2. F = 2x + 3 from x = 0 to 5. Find work.

W = ∫(2x+3)dx = [x²+3x]0^5 = 25 + 15 = 40 J.

3. F = 6x² from 0 to 2. Find work.

W = ∫0^2 6x²dx = 6(8/3) = 16 J.

4. A force of 50 N at 37° pulls a block through 8 m. Find work. Use cos37° = 0.8.

W = 50 x 8 x 0.8 = 320 J.

5. A triangular F-x graph has base 8 m and height 20 N.

Work = area = 1/2 x 8 x 20 = 80 J.

6. F = 10 - 2x from 0 to 3. Find work.

W = [10x - x²]0^3 = 30 - 9 = 21 J.

7. Work by gravity for mass 2 kg lifted 5 m. g = 10.

Wg = -mgh = -2 x 10 x 5 = -100 J.

8. Work by applied force lifting same mass slowly by 5 m.

W = +mgh = 100 J.

9. Spring k = 400 N/m stretched 0.05 m. Work stored?

W = 1/2 kx² = 0.5 x 400 x 0.0025 = 0.5 J.

10. F-x graph rectangle 0-2 m height 5 N plus triangle 2-6 m height 5 N.

Area = 5x2 + 1/2 x4x5 = 10 + 10 = 20 J.

11. Force has components 3i + 4j N and displacement 5i m.

W = F.s = 3x5 + 4x0 = 15 J.

12. Force 4i - 2j N, displacement 3i + 6j m.

W = 4x3 + (-2)x6 = 12 - 12 = 0 J.

13. A block moves 10 m while friction 6 N acts. Work by friction?

W = -6 x 10 = -60 J.

14. F = kx, k = 50, x from 1 to 3. Work?

W = 1/2 k(3² - 1²) = 25 x 8 = 200 J.

15. F = 8/x from x = 1 to 2. Work?

W = ∫1^2 8/x dx = 8 ln2 J.

16. F = 5 N for 0-3 m and -2 N for 3-7 m. Net work?

W = 5x3 + (-2)x4 = 15 - 8 = 7 J.

17. Force at 120°: F = 20 N, s = 5 m. Work?

W = 20 x 5 x cos120° = -50 J.

18. Force at 53°: F = 100 N, s = 2 m, cos53° = 0.6.

W = 100 x 2 x 0.6 = 120 J.

19. F = 3x² + 2 from 0 to 2.

W = [x³ + 2x]0^2 = 8 + 4 = 12 J.

20. F = 9 N and displacement 4 m at 90°.

W = 0 because cos90° = 0.

21. A force does -200 J over 20 m opposite motion. Magnitude?

W = -Fs, so F = 200/20 = 10 N.

22. Work done by resultant force if kinetic energy changes from 10 J to 45 J.

Net work = ΔK = 35 J.

23. F = ax, work from 0 to L is 20 J. Find a.

W = aL²/2 = 20, so a = 40/L².

24. F = 15 N from x = 2 to x = 9.

W = 15(9 - 2) = 105 J.

25. Work done by conservative force in closed path?

Zero. It depends only on initial and final positions.

26. Work by non-conservative friction around closed path?

Negative and non-zero because friction dissipates energy.

27. F = 2x - 6 from x = 0 to 6.

W = [x² - 6x]0^6 = 36 - 36 = 0 J.

28. Area above x-axis 50 J and below x-axis 20 J. Net work?

Signed work = 50 - 20 = 30 J.

29. Spring work doubles if x changes by factor?

W ∝ x². To double W, x becomes sqrt2 times.

30. A force vector is perpendicular to displacement vector. Dot product?

Zero, so work is zero.

31. F = 4x + 1 from x = 1 to 4.

W = [2x²+x]1^4 = (32+4) - (2+1) = 33 J.

32. F = 6/x² from x = 1 to 3.

W = ∫6x^-2 dx = [-6/x]1^3 = -2 + 6 = 4 J.

33. Work by weight for horizontal displacement?

Zero because weight is vertical and displacement horizontal.

34. F = 20 N, s = 10 m, work = 100 J. Find angle.

cosθ = W/Fs = 100/200 = 1/2, so θ = 60°.

35. F = 30 N, θ = 180°, s = 3 m.

W = -90 J.

36. F = 30 N, θ = 0°, s = 3 m.

W = +90 J.

37. A force F = 5x acts for 0 to a, work is?

W = 5a²/2.

38. F = x + x² from 0 to 1.

W = 1/2 + 1/3 = 5/6 J.

39. F = 10 sin x from 0 to pi.

W = [-10 cos x]0^pi = 20 J.

40. Work by static friction in pure rolling on fixed ground?

At contact point displacement is zero; work by static friction at contact is zero in ideal pure rolling.

41. A man pushes wall with 100 N for 10 s. Wall displacement zero. Work?

Zero because displacement is zero.

42. A 5 kg body raised 3 m. Work by gravity g=10.

Wg = -mgh = -150 J.

43. Work by applied force in lowering load slowly height h?

Applied force upward, displacement downward, so W = -mgh.

44. Work by gravity in lowering load slowly height h?

Gravity and displacement are downward, so W = +mgh.

45. F-x graph is line from (0,10) to (4,20). Work?

Trapezium area = 1/2(10+20)4 = 60 J.

46. F = 7 N from 0-2 m, 0 from 2-5 m, -3 N from 5-7 m.

W = 14 + 0 - 6 = 8 J.

47. If net work is zero, kinetic energy change is?

Zero, by work-energy theorem.

48. Work of central force in small displacement perpendicular to radius?

Zero instantaneously because force is radial and displacement tangential.

49. F = 2i + 3j + 4k, s = i - j + 2k. Work?

W = 2(1)+3(-1)+4(2)=2-3+8=7 J.

50. F = x i + y j, displacement from (0,0) to (a,a) along y=x. Find line work.

On y=x, dx=dy, F.dr = x dx + x dx = 2x dx. Integral 0 to a = a².

JEE Advanced Question Bank

50 difficult JEE Advanced style questions with compact complete solutions.

1. F = ax² + bx acts from 0 to L. Find work.

W = ∫0^L (ax²+bx)dx = aL³/3 + bL²/2.

2. Force F = kx on a particle from x1 to x2.

W = ∫x1^x2 kx dx = k(x2² - x1²)/2.

3. F = -kx is restoring spring force from x to 0.

Work by spring = ∫x^0 (-kx)dx = kx²/2, positive as spring returns to natural length.

4. F = -kx from 0 to x.

Work by spring = ∫0^x -kx dx = -kx²/2. External work slowly stretching is +kx²/2.

5. A block moves under F = 4x - x² from 0 to 4.

W = ∫0^4 (4x-x²)dx = [2x² - x³/3]0^4 = 32 - 64/3 = 32/3 J.

6. A particle goes from x=0 to x=3 under F = 9 - x².

W = [9x - x³/3]0^3 = 27 - 9 = 18 J.

7. Signed graph areas: +12, -5, +8 J.

Net work = 12 - 5 + 8 = 15 J.

8. Work along x-axis for F = (2xy)i + x²j from x=0 to a, y=0.

Along x-axis y=0 and dy=0. F.dr = 0 dx + x² dy = 0. Work = 0.

9. Same force along path y=x from 0 to a.

F.dr = 2x² dx + x² dy, dy=dx. Integral 3x² dx from 0 to a = a³.

10. A non-conservative force gives different work on two paths. What conclusion?

Work is path-dependent; closed-loop work may be non-zero.

11. F = C/x² from R to 2R.

W = ∫R^2R Cx^-2 dx = [-C/x]R^2R = C/(2R).

12. F = C/r² radial outward, displacement radial from R to 3R.

W = ∫R^3R C/r² dr = C(1/R - 1/3R)=2C/3R.

13. Force perpendicular to velocity at every point.

Instantaneous power F.v = 0, so work done over path is zero if always perpendicular.

14. F = 10 N constant, displacement vector 3i + 4j along force i direction.

Force vector = 10i. Work = 10i.(3i+4j)=30 J.

15. F = 5i + 12j, displacement magnitude 13 along same direction as F.

Magnitude of F = 13 N. Work = Fs = 13 x 13 = 169 J.

16. F = 5i + 12j, displacement 13 m opposite force.

W = -Fs = -13 x 13 = -169 J.

17. F-x curve is semicircle radius R above axis. Work?

Area under curve = area of semicircle = 1/2 pi R² in force-displacement units.

18. F-x curve quarter circle radius R. Work?

Area = pi R²/4, with axes units included.

19. Spring is compressed x then released to x/2. Work by spring?

Work by spring = decrease in spring energy = 1/2kx² - 1/2k(x/2)² = 3kx²/8.

20. External work slowly compressing spring from x/2 to x.

ΔU = 1/2kx² - 1/2k(x/2)² = 3kx²/8.

21. Force F = A sin(πx/L), x from 0 to L.

W = A[-L/π cos(πx/L)]0^L = 2AL/π.

22. F = A cos(πx/2L), x from 0 to L.

W = A[2L/π sin(πx/2L)]0^L = 2AL/π.

23. A variable force has average value Favg over displacement s.

Work = Favg s if Favg is average with respect to displacement.

24. Force varies linearly from F1 to F2 over distance L.

Work = average force x distance = (F1+F2)L/2.

25. F = 2x for 0-3, then F = 6 for 3-5.

Work = triangle 1/2x3x6 + rectangle 6x2 = 9+12=21 J.

26. F = 6 for 0-2, then decreases to 0 at x=5.

Work = rectangle 12 + triangle 1/2x3x6 = 21 J.

27. A body moves on rough horizontal surface μ, distance s.

Friction = μmg. Work by friction = -μmgs.

28. Block pulled by force F at angle θ over rough surface distance s.

Work by applied force = Fs cosθ. Normal changes friction, but applied work uses force component along displacement.

29. If F = 3t and v = 2t, work from 0 to T?

Power = Fv = 6t². Work = ∫0^T 6t²dt = 2T³.

30. F = kt and displacement x = at². Work from 0 to T?

dx/dt = 2at. W = ∫F dx = ∫0^T kt(2at)dt = 2akT³/3.

31. Force F = αx i + βy j from (0,0) to (a,b) along straight line.

Parametrize x=at, y=bt. dr=(a i+b j)dt. W=∫0^1(αa²t+βb²t)dt=(αa²+βb²)/2.

32. Work by constant force is path independent or dependent?

For constant force, W = F.(r2-r1), so it depends only on displacement, not path.

33. Work by friction between two fixed points depends on path length?

Yes. Wf = -μmg times path length on horizontal rough surface.

34. F = 2x i + 3y j around rectangle 0,a,0,b.

This force is conservative with potential-like form; closed-loop work is zero.

35. F = -y i + x j around circle radius R anticlockwise.

Parametrize x=Rcosθ, y=Rsinθ. F.dr = R² dθ. Work over 0 to 2π = 2πR².

36. Particle under force F = -kx has potential energy relation?

Work by spring from 0 to x = -1/2kx², so potential energy increases by 1/2kx².

37. A graph crosses x-axis. How find work?

Take algebraic area: area above axis positive and below axis negative.

38. If net work is -K/2, final kinetic energy?

Kf = Ki + Wnet = K - K/2 = K/2.

39. Work done by force to stop mass m moving speed v?

Net work = ΔK = 0 - 1/2mv² = -1/2mv².

40. Minimum external work to lift mass m by h slowly?

mgh, assuming no losses.

41. Force F = ax³ from 0 to L.

W = aL^4/4.

42. Force F = a/x from x1 to x2.

W = a ln(x2/x1).

43. F = a√x from 0 to L.

W = a ∫0^L x^1/2 dx = 2aL^(3/2)/3.

44. Force at angle changes from 0 to 90 while magnitude constant. Work trend?

Work decreases as cosθ decreases from 1 to 0.

45. A block moves vertically upward while normal force from vertical wall acts horizontally.

Work by normal is zero because normal is perpendicular to vertical displacement.

46. Tension in string for circular motion work?

For ideal circular motion, tension is radial and displacement tangential, so work is zero.

47. Force F = 2x from -a to +a.

W = ∫-a^a 2x dx = [x²]_-a^a = 0.

48. Force F = x² from -a to +a.

W = ∫-a^a x² dx = 2a³/3.

49. Work done in stretching spring from x to 2x by external agent.

ΔU = 1/2k(4x² - x²)=3kx²/2.

50. A force is conservative if closed path work is:

Zero for every closed path. Equivalent: work is path independent.

IB / IGCSE / A-Level Questions

Separate international practice sets with answers and explanations.

IB Questions - 25

IB 1. Define work done by a force.

Work is energy transferred when a force acts through a displacement: W = Fs cosθ.

IB 2. State why work is scalar.

It is a dot product and has magnitude/sign but no direction.

IB 3. Calculate work: F=12 N, s=3 m, same direction.

W = 36 J.

IB 4. Explain zero work by centripetal force.

Force is perpendicular to instantaneous displacement.

IB 5. A box is lifted at constant speed. Work by applied force?

+mgh, since applied force and displacement are upward.

IB 6. Work by weight during lifting?

-mgh.

IB 7. What does area under F-x graph represent?

Work done.

IB 8. Force increases linearly to 20 N over 4 m. Work?

Triangle area = 40 J.

IB 9. Spring energy formula?

E = 1/2kx².

IB 10. If work is negative, what happens to kinetic energy?

It tends to decrease if this is net work.

IB 11. Work done by friction over 5 m if friction is 8 N?

-40 J.

IB 12. Why is displacement used, not distance?

Work by a force depends on displacement component along that force for the considered motion.

IB 13. Work by normal reaction on horizontal table?

Zero.

IB 14. Unit of work?

Joule.

IB 15. F=30 N at 60°, s=2 m. Work?

30 J.

IB 16. Give one example of positive work.

Gravity on a falling object.

IB 17. Give one example of negative work.

Friction on a sliding block.

IB 18. Give one example of zero work.

Carrying a bag horizontally at constant height, work by weight.

IB 19. Explain Hooke's law connection.

For spring F = kx, so work is area under F-x graph = 1/2kx².

IB 20. F=2x from 0 to 3. Work?

9 J.

IB 21. Work by gravity in horizontal motion?

Zero.

IB 22. Work and energy relation?

Work is energy transfer, measured in joules.

IB 23. Why can work be negative?

Because force can oppose displacement.

IB 24. Is W = Fs always valid?

Only when force is parallel and constant.

IB 25. What is W for θ=120°, F=10, s=2?

W = 10 x 2 x (-1/2) = -10 J.

IGCSE Questions - 25

IGCSE 1. Formula for work done?

Work = force x distance moved in direction of force.

IGCSE 2. Unit of work?

Joule, J.

IGCSE 3. 5 N force moves 4 m. Work?

20 J.

IGCSE 4. A wall does not move when pushed. Work?

Zero.

IGCSE 5. Work done lifting weight 20 N by 3 m?

60 J.

IGCSE 6. What is energy transferred by force called?

Work done.

IGCSE 7. Force opposite motion does what sign work?

Negative work.

IGCSE 8. Friction does what type of work?

Negative work.

IGCSE 9. Weight during horizontal motion does work?

Zero work.

IGCSE 10. 12 N over 0.5 m?

6 J.

IGCSE 11. Work done climbing height h against weight W?

Wh.

IGCSE 12. A force of 50 N moves 2 m.

100 J.

IGCSE 13. Distance in formula must be in:

Metres.

IGCSE 14. Force unit in work formula?

Newton.

IGCSE 15. Work is zero when force is perpendicular to motion?

Yes, because there is no force component along motion.

IGCSE 16. Work done by engine on train moving forward?

Positive.

IGCSE 17. Work by brakes?

Negative.

IGCSE 18. If force doubles, work doubles?

Yes, if distance and direction are unchanged.

IGCSE 19. If distance doubles, work doubles?

Yes, if force and direction are unchanged.

IGCSE 20. 100 J work over 5 m. Force?

F = W/s = 20 N.

IGCSE 21. 240 J work by 30 N. Distance?

s = 240/30 = 8 m.

IGCSE 22. Work done against gravity increases what energy?

Gravitational potential energy.

IGCSE 23. Area under force-distance graph gives?

Work done.

IGCSE 24. Constant force graph area shape?

Rectangle.

IGCSE 25. Linearly increasing force graph area shape?

Triangle.

A-Level Questions - 25

A-Level 1. Define work using scalar product.

W = F.s = Fs cosθ.

A-Level 2. F=4i+3j, s=2i+5j. Work?

W = 8 + 15 = 23 J.

A-Level 3. F=2x from 1 to 4.

W = [x²]1^4 = 15 J.

A-Level 4. Work by spring force from 0 to x.

-1/2kx² for work by spring during stretching.

A-Level 5. External work to stretch spring from 0 to x.

+1/2kx².

A-Level 6. F-x graph trapezium sides 4,10, width 3.

Area = 21 J.

A-Level 7. Force perpendicular to path always.

Work is zero.

A-Level 8. F=6/x from 1 to e.

W = 6 ln e = 6 J.

A-Level 9. Work by friction μ over distance s.

W = -μRs, where R is normal reaction.

A-Level 10. Work-energy theorem statement.

Net work equals change in kinetic energy.

A-Level 11. If Wnet = 0, speed?

Speed remains same if mass is constant.

A-Level 12. F=3x² from 0 to 2.

W = [x³]0^2 = 8 J.

A-Level 13. Work to raise 0.5 kg by 4 m, g=9.8.

W = 19.6 J.

A-Level 14. Work by gravity for same raise.

-19.6 J.

A-Level 15. Force varies linearly from 2 N to 8 N over 5 m.

W = average force x distance = 5 x 5 = 25 J.

A-Level 16. F = 10cos x from 0 to pi/2.

W = 10 J.

A-Level 17. F = 10sin x from 0 to pi/2.

W = 10 J.

A-Level 18. Dot product sign determines what?

Whether work is positive, negative or zero.

A-Level 19. Work done by conservative force around closed loop.

Zero.

A-Level 20. Work done by non-conservative force may depend on?

Path.

A-Level 21. Spring k=50, x=0.4.

W = 1/2 x 50 x 0.16 = 4 J.

A-Level 22. F=8N, θ=60, s=10.

W = 40 J.

A-Level 23. A vector force does no work if:

It is perpendicular to displacement or displacement is zero.

A-Level 24. Work by tension in horizontal pull.

Positive if tension has component along displacement.

A-Level 25. F = x² + x from 0 to 3.

W = 9 + 4.5 = 13.5 J.