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CLASS 11 PHYSICS - WORK ENERGY POWER
Kinetic Energy and Work Energy Theorem
Premium notes with complete derivation, theorem applications, SVG graphs, solved numericals and exam question banks.
CBSENEETJEE MainJEE AdvancedIBIGCSEA-Level
K
K = 1/2 mv²Energy possessed due to motion.
W
Wnet = ΔKNet work changes kinetic energy.
∫
W = ∫F dxVariable-force work is graph area.
k
W = 1/2 kx²Spring energy from Hooke's law.
01
Kinetic Energy
Kinetic energy is the energy possessed by a body because of its motion.
Definition And Formula
Kinetic energy measures the work needed to accelerate a body from rest to speed v, or the work the body can do before stopping.
K = 1/2 mv²- K = kinetic energy.
- m = mass of the body.
- v = speed of the body.
- It is scalar and always non-negative for ordinary motion.
Unit, Dimension And Meaning
| Property | Result |
|---|---|
| SI Unit | joule, J |
| Dimension | [M L2 T-2] |
| Nature | Scalar |
| Dependence | K ∝ m and K ∝ v² |
Moving Particle
Car Moving With v
Different Velocities
Energy Comparison
02
Derivation of Kinetic Energy
Derive K = 1/2mv² using Newton's law, kinematics and work formula.
Step 1
From Newton's second law:
F = maStep 2
From kinematics:
v² - u² = 2asSo, as = (v²-u²)/2
Step 3
Work done by constant force:
W = Fs = masStep 4
Substitute as:
W = 1/2 m(v²-u²)If the body starts from rest, u = 0, so K = W = 1/2mv².
03
Work-Energy Theorem
Very important: net work done by all forces equals change in kinetic energy.
Statement
Wnet = ΔKWnet = Kf - KiPositive net work increases speed. Negative net work decreases speed. Zero net work keeps kinetic energy unchanged.
Flow Diagram
Complete Derivation
For constant mass and motion along force: F = ma and Wnet = Fs = mas. From v² - u² = 2as, we get as = (v²-u²)/2. Therefore:
Wnet = 1/2mv² - 1/2mu² = ΔKFor variable force, the theorem remains valid because the total work is the integral of net force over displacement.
04
Applications of Work-Energy Theorem
Use the theorem whenever force, displacement and speed change are linked.
Car Acceleration
Concept: engine does positive work.
W = 1/2mv² - 1/2mu²Method: equate engine work to increase in kinetic energy.
Car Braking
Concept: brake force does negative work.
-Fs = 0 - 1/2mv²Method: stopping distance from initial kinetic energy.
Falling Object
Concept: gravity does positive work.
mgh = 1/2mv²Method: use energy change instead of time equations.
Particle Under Force
Concept: net force work changes speed.
∫F dx = ΔKSpring Systems
Concept: spring work converts to kinetic energy.
1/2kx² = 1/2mv²Rough Surface
Concept: friction reduces kinetic energy.
Wfriction = -μmgsInclined Plane
Concept: gravity component does work.
W = mg s sinθNeed help with Work-Energy Theorem?
Contact Kumar Sir for one-to-one online Physics classes.
Phone: +91-9958461445Email: kumarsirphysics@gmail.comWebsite: kumarphysicsclasses.com
05
Variable Force Problems
When force changes with position, use integration or area under the F-x graph.
Core Idea
W = ∫F dxFor a small displacement dx, work dW = F dx. Adding all small work elements gives total work.
Spring Force
F = kxW = ∫0x kx dx = 1/2kx²The F-x graph is a triangle, so area = 1/2 base x height = 1/2 x kx = 1/2kx².
Force vs Displacement
Spring Graph
Graph Problem
Force increases linearly from 0 to 40 N in 5 m. Find work.
Area of triangle = 1/2 x 5 x 40 = 100 J. This work equals increase in kinetic energy if this is net force.
06
Important Graphs
Beautiful graph summaries for WET, kinetic energy and spring force.
F vs x
Area under F-x gives work.
W vs x
For constant force, work increases linearly with x.
K vs v
Parabolic graph because K = 1/2mv².
K vs p
For fixed mass, K is proportional to p².
Variable Force
Use signed area under the graph.
Spring Force
Slope of spring graph is k.
07
High-Quality Numericals
Solved bank covering CBSE, NEET, JEE, IB, IGCSE and A-Level.
CBSE: Find KE of 2 kg body moving at 5 m/s.
Diagram: moving particle. Given m=2 kg, v=5 m/s. Formula K=1/2mv². Calculation K=1/2 x 2 x 25 = 25 J. Final answer: 25 J. Exam tip: velocity is squared.
NEET: A 1000 kg car speeds from 10 m/s to 20 m/s. Net work?
Wnet=ΔK=1/2(1000)(400-100)=150000 J. Final answer: 1.5 x 10^5 J. Exam tip: use final minus initial kinetic energy.
JEE Main: Force F=4x acts from x=0 to 3 m on 2 kg body initially at rest. Find speed.
W=∫0^3 4x dx=18 J. W=1/2mv², so 18=1/2 x 2 x v²=v². v=√18=3√2 m/s.
JEE Advanced: Spring k=200 N/m compressed 0.2 m launches 1 kg block. Find speed.
Spring energy=1/2kx²=1/2 x 200 x 0.04=4 J. 1/2mv²=4, v=√8=2√2 m/s.
IB: A 0.5 kg ball doubles speed from 4 to 8 m/s. Change in KE?
ΔK=1/2(0.5)(64-16)=0.25x48=12 J. Final answer: 12 J.
IGCSE: A 10 kg object moves at 3 m/s. KE?
K=1/2 x 10 x 9 = 45 J.
A-Level: Braking force 500 N stops 1000 kg car moving 20 m/s. Distance?
Initial KE=1/2 x 1000 x 400=200000 J. Work by brake=Fs. s=200000/500=400 m.
Graph: Force rises 0 to 30 N over 4 m. If mass 3 kg starts from rest, find speed.
Work=triangle area=1/2 x 4 x 30=60 J. 60=1/2 x 3 v², v²=40, v=2√10 m/s.
08
NEET Question Bank
50 high-quality NEET-style MCQs on kinetic energy, work-energy theorem, variable force and graphs.
1. NEET Exam-style Question: KE of mass m and speed v is: A mv B mv² C 1/2mv² D 2mv²
Answer: C. K = 1/2mv².
2. If speed doubles, KE becomes: A 2K B 4K C K/2 D K/4
Answer: B. K ∝ v².
3. If mass doubles at same speed, KE becomes: A 2K B 4K C K/2 D K/4
Answer: A. K ∝ m.
4. Unit of kinetic energy is: A watt B joule C newton D kg m/s
Answer: B. Energy is measured in joules.
5. Work-energy theorem states: A W=K B Wnet=ΔK C F=ma D P=W/t
Answer: B. Net work equals change in kinetic energy.
6. Net work is positive. KE: A increases B decreases C zero D unchanged
Answer: A. Positive net work means ΔK positive.
7. Net work is negative. Speed generally: A increases B decreases C infinite D unchanged
Answer: B. Negative work reduces kinetic energy.
8. A 2 kg body at 4 m/s has KE: A 8 J B 16 J C 32 J D 64 J
Answer: B. K=1/2 x2x16=16 J.
9. K=100 J, m=2 kg. Speed is: A 5 B 10 C 20 D 50 m/s
Answer: B. 100=1/2 x2 v², v=10.
10. Area under F-x graph gives: A KE directly always B work C velocity D mass
Answer: B. W=∫Fdx.
11. F=5x from 0 to 2 m. Work is: A 5 J B 10 J C 15 J D 20 J
Answer: B. ∫0^2 5x dx=10 J.
12. Spring energy is: A kx B 1/2kx² C k/x D 2kx
Answer: B.
13. K vs v graph is: A straight line B parabola C circle D hyperbola
Answer: B. K∝v².
14. K vs p for fixed mass is: A linear B parabolic C constant D zero
Answer: B. K=p²/2m.
15. KE can be negative? A yes B no C only springs D only gravity
Answer: B. K=1/2mv² is non-negative.
16. Body initially at rest, net work 50 J. Final KE: A 0 B 25 C 50 D 100 J
Answer: C. ΔK=Wnet.
17. Initial KE 80 J, net work -30 J. Final KE: A 30 B 50 C 80 D 110 J
Answer: B. Kf=80-30=50 J.
18. Work needed to stop a body of KE K is: A K B -K by stopping force C 2K D zero
Answer: B. Net work = -K.
19. Mass 4 kg speed 5 m/s. KE: A 25 B 50 C 100 D 200 J
Answer: B. 1/2 x4x25=50 J.
20. Work done by variable force equals: A slope B area C intercept D time
Answer: B.
21. If momentum doubles, KE at same mass becomes: A 2K B 4K C K/2 D K
Answer: B. K=p²/2m.
22. If mass doubles at same momentum, KE becomes: A 2K B 4K C K/2 D K/4
Answer: C. K=p²/2m.
23. Stopping distance for same brake force is proportional to: A v B v² C 1/v D m only
Answer: B for same mass and force, since Fs=1/2mv².
24. Work by gravity in falling height h equals: A mgh B -mgh C zero D 1/2kx²
Answer: A.
25. Falling from rest height h gives speed: A √gh B √2gh C 2gh D gh
Answer: B. mgh=1/2mv².
26. Spring k=100, x=0.2 m. Energy: A 1 J B 2 J C 4 J D 20 J
Answer: B.
27. Force graph triangle base 4, height 10. Work: A 10 B 20 C 40 D 80 J
Answer: B.
28. Force graph rectangle 5 N by 6 m. Work: A 11 B 15 C 30 D 60 J
Answer: C.
29. Net work zero implies KE: A increases B decreases C unchanged D negative
Answer: C.
30. KE dimension is: A MLT^-2 B ML²T^-2 C MLT^-1 D LT^-2
Answer: B.
31. 1/2mv² depends on: A velocity direction only B speed C displacement D acceleration only
Answer: B. KE depends on speed magnitude.
32. Car braking work is generally: A positive B negative C zero D infinite
Answer: B.
33. Engine work during acceleration is: A positive B negative C zero D none
Answer: A.
34. Variable force F=x² from 0 to 3. Work: A 3 B 6 C 9 D 27 J
Answer: C. ∫x²dx=9.
35. Kinetic energy at rest is: A zero B mv C mgh D infinite
Answer: A.
36. A 1 kg object has KE 18 J. Speed: A 3 B 4 C 6 D 9
Answer: C. 18=1/2v², v=6.
37. A 3 kg object has speed 2 m/s. KE: A 3 B 6 C 9 D 12 J
Answer: B.
38. Work-energy theorem is based on: A net work B only applied force C only gravity D only friction
Answer: A.
39. Spring graph F-x slope is: A x B k C 1/k D energy
Answer: B.
40. If x doubles in spring, energy becomes: A 2 times B 4 times C half D same
Answer: B.
41. KE lost by a body equals work done by: A accelerating force B retarding net force C mass D speed
Answer: B.
42. F=10-2x from 0 to 3. Work: A 18 B 21 C 24 D 30 J
Answer: B. ∫(10-2x)=30-9=21.
43. KE is independent of: A mass B speed C direction of velocity D square of speed
Answer: C.
44. Work done by net force if speed unchanged: A positive B negative C zero D mgh
Answer: C.
45. A body speeds up from 3 to 5 m/s, mass 2 kg. ΔK: A 8 B 16 C 25 D 34 J
Answer: B. 1/2x2(25-9)=16.
46. KE of 5 kg at 0 speed: A 0 B 5 C 10 D 25 J
Answer: A.
47. Work needed to increase KE from 20 to 70 J: A 20 B 50 C 70 D 90 J
Answer: B.
48. If Wnet = 100 J and Ki = 40 J, Kf: A 60 B 100 C 140 D 400 J
Answer: C.
49. Formula relating KE and momentum: A K=p²/2m B K=pm C K=p/m D K=2pm
Answer: A.
50. Work by spring while returning to natural length is: A positive B negative C zero D impossible
Answer: A, because spring force and displacement are in same direction during return.
09
JEE Main Question Bank
50 difficult JEE Main style questions.
1. A 2 kg body changes speed from 3 to 7 m/s. Net work?
W=1/2x2(49-9)=40 J.
2. F=3x+2 from 0 to 4 m, body initially KE 10 J. Final KE?
W=[1.5x²+2x]0^4=24+8=32 J. Kf=42 J.
3. Spring k=500, x=0.1, mass 2 kg. Speed?
Energy=2.5 J. 1/2x2v²=2.5, v=√2.5 m/s.
4. Braking force 200 N stops 4 kg body moving 20 m/s. Distance?
KE=800 J. s=800/200=4 m.
5. F-x trapezium sides 10 N, 30 N, width 5 m. Work?
W=1/2(10+30)5=100 J.
6. A 1 kg particle under F=4x reaches x=5 from rest. Speed?
W=∫0^5 4x dx=50 J. 1/2v²=50, v=10 m/s.
7. F=10-x from 0 to 6. Work?
W=[10x-x²/2]0^6=60-18=42 J.
8. KE is 72 J for mass 4 kg. Momentum?
K=p²/2m, p=√(2mK)=√576=24 kg m/s.
9. Momentum doubles. Work needed to change from p to 2p?
ΔK=(4p²-p²)/2m=3p²/2m.
10. A block slides down smooth incline length s angle θ from rest. Speed?
mg s sinθ=1/2mv², v=√(2gs sinθ).
11. Rough horizontal surface friction μ, initial speed v. Stopping distance?
μmgs=1/2mv², s=v²/(2μg).
12. F=6x² from 0 to 2 m. Work?
W=∫6x²dx=16 J.
13. Force graph rectangle 12 N for 3 m then triangle to zero over 2 m. Work?
W=36+12=48 J.
14. Ki=20 J, Kf=5 J. Net work?
Wnet=-15 J.
15. Work 64 J on 2 kg body initially at rest. Speed?
64=1/2x2v², v=8 m/s.
16. F=8/x from 1 to 2. Work?
W=8ln2 J.
17. Kinetic energy ratio for speeds 2v and 3v same mass?
K1:K2=4:9.
18. Same momentum, masses m and 4m. KE ratio?
K=p²/2m, ratio = 4:1.
19. A 3 kg body falls 5 m. Speed from rest g=10?
mgh=1/2mv², v=10 m/s.
20. Work done by net force if velocity changes direction but speed same?
Zero, because kinetic energy unchanged.
21. F=2x-4 from 0 to 4. Work?
[x²-4x]0^4=16-16=0.
22. F=x²+x from 0 to 3. Work?
9+4.5=13.5 J.
23. Spring compressed x launches mass m on smooth table. Speed?
1/2kx²=1/2mv², v=x√(k/m).
24. If spring compression doubles, launch speed?
v∝x, so speed doubles.
25. If spring constant doubles same compression, speed?
v∝√k, so speed becomes √2 times.
26. A 10 N retarding force acts for 6 m on 2 kg body moving 10 m/s. Final speed?
Ki=100 J, W=-60 J, Kf=40 J. 40=1/2x2v², v=√40=2√10.
27. F=5 N for 0-4 m, then -3 N for 4-9 m. Net work?
20-15=5 J.
28. Graph area above axis 80 J and below axis 30 J. ΔK?
Net work=50 J, so ΔK=50 J.
29. K vs v slope at speed v?
dK/dv=mv.
30. K vs p slope?
dK/dp=p/m=v.
31. Power relation with KE rate?
Net power = dK/dt.
32. F=kt, v=at. Work from 0 to T?
W=∫Fvdt=∫kat²dt=kaT³/3.
33. F=4x, mass 8 kg, from 0 to 4 rest. Speed?
W=32 J. 32=1/2x8v²=4v², v=2√2.
34. Body initially speed u, retarding force F constant. Stopping distance?
s=mu²/(2F).
35. Car speed triples. KE ratio final/initial?
9.
36. A force does work W on rest mass m. Final momentum?
W=p²/2m, so p=√(2mW).
37. A force does W on rest mass m. Final speed?
v=√(2W/m).
38. F=20cosθ along displacement? If θ=60, s=10.
W=20x10x1/2=100 J.
39. Work by gravity for vertical downward displacement h.
+mgh.
40. Work by gravity for vertical upward displacement h.
-mgh.
41. If K=1/2mv² and v=sqrt(2gh), K equals?
K=mgh.
42. Work by friction in rough incline length s.
Wf=-μN s=-μmg cosθ s.
43. Smooth incline speed after distance s.
v=√(u²+2gsinθ s) from WET.
44. Kinetic energy of 0.2 kg bullet at 100 m/s.
K=1000 J.
45. Work to speed 5 kg body from 2 to 6 m/s.
1/2x5(36-4)=80 J.
46. If Wnet=-Ki, final speed?
Zero.
47. F=3x²+2x from 0 to 2.
W=[x³+x²]0^2=8+4=12 J.
48. Force varies linearly from 4 N to 12 N over 5 m.
W=(4+12)/2 x 5=40 J.
49. If KE increases by 25 J, net work is?
25 J.
50. If KE decreases by 25 J, net work is?
-25 J.
10
JEE Advanced Question Bank
50 difficult JEE Advanced questions focusing on WET, variable force, integration, springs and graphs.
1. F=ax²+bx acts from 0 to L on rest mass m. Final speed?
W=aL³/3+bL²/2=1/2mv². v=√(2W/m).
2. F=-kx, particle moves from x=A to x=0. Work by spring?
W=1/2kA², positive.
3. Spring block starts from compression A, rough surface friction μmg over distance A. Speed at natural length?
1/2mv²=1/2kA²-μmgA.
4. Force F=C/x² from R to 2R. Change in KE?
ΔK=∫R^2R Cx^-2 dx=C/(2R).
5. F=A sin(πx/L) from 0 to L. Work?
W=2AL/π.
6. F=A cos(πx/2L) from 0 to L. Work?
W=2AL/π.
7. Body under F=αx³ from 0 to L. Final KE if initial KE K0?
Kf=K0+αL⁴/4.
8. F=β√x from 0 to L. Work?
W=2βL^(3/2)/3.
9. A graph has signed areas +20, -8, +12 J. ΔK?
ΔK=24 J.
10. A particle moves in circular path with tangential force F_t constant over arc length s.
Work=F_t s, so ΔK=F_t s.
11. Radial force only in circular motion. Change in KE?
Zero if force is always perpendicular to displacement.
12. Mass m moves from rest under F=kx from 0 to x. Momentum?
K=1/2kx²=p²/2m, so p=x√(mk).
13. External work to stretch spring from A to 2A.
ΔU=1/2k(4A²-A²)=3kA²/2.
14. Work by spring in same stretch A to 2A.
-3kA²/2.
15. Force F=2x-6 from 0 to 6. Net ΔK?
W=[x²-6x]0^6=0, so ΔK=0.
16. F=3x²-12 from 0 to 3. Work?
W=[x³-12x]0^3=27-36=-9 J.
17. If K vs x is given by K=K0+ax², force is?
F=dK/dx=2ax.
18. If W(x)=5x³, force is?
F=dW/dx=15x².
19. Kinetic energy changes with x as K=4x+7. Net force?
F=dK/dx=4 N.
20. Force F=10/x from 1 to e². Work?
W=10ln(e²)=20 J.
21. Particle moving under conservative spring force returns to original position. Net spring work?
Zero over complete closed path.
22. Friction over closed rough path length L.
W=-μmgL, non-zero negative.
23. Block descends rough incline length s angle θ. Final speed from rest.
1/2mv²=mg s sinθ-μmg cosθ s.
24. Minimum compression x of spring k to give speed v to mass m.
1/2kx²=1/2mv², x=v√(m/k).
25. Same with friction over compression distance x.
1/2kx²-μmgx=1/2mv².
26. Force F=F0(1-x/L), x 0 to L. Work?
W=F0L/2.
27. Force F=F0(x/L), x 0 to L. Work?
W=F0L/2.
28. Force F=F0(x/L)², x 0 to L. Work?
W=F0L/3.
29. Kinetic energy is K and momentum is p. Find mass.
K=p²/2m, so m=p²/(2K).
30. Kinetic energy is K and speed is v. Find momentum.
m=2K/v², p=mv=2K/v.
31. Work needed to change momentum p to 3p for mass m.
ΔK=(9p²-p²)/2m=4p²/m.
32. If force is perpendicular to velocity, show dK/dt.
dK/dt=P=F.v=0.
33. If power P is constant, KE after time t from rest.
K=Pt.
34. Constant power P, mass m starts from rest. Speed after t?
1/2mv²=Pt, so v=√(2Pt/m).
35. F=λv and motion over time with v=at. Work 0 to T?
P=Fv=λv²=λa²t². W=λa²T³/3.
36. F=αx+β/x from 1 to 2.
W=α(4-1)/2+βln2=3α/2+βln2.
37. A force graph is semicircle radius R above axis. Work?
Area=πR²/2.
38. A force graph quarter circle radius R. Work?
Area=πR²/4.
39. If K(x)=ax²+bx+c, net work from x1 to x2?
W=K(x2)-K(x1).
40. If force changes sign, how handle work?
Integrate algebraically; areas below x-axis are negative.
41. A smooth loop bottom speed condition from energy to reach top height 2R.
Energy condition ignoring contact at top: 1/2mv_b² ≥ mg(2R), so v_b ≥ 2√gR. For complete loop with contact, v_b ≥ √5gR.
42. Work by normal in smooth fixed track?
Zero because normal is perpendicular to instantaneous displacement.
43. Work by normal in moving surface can be?
Non-zero if point of contact has displacement component along normal in ground frame.
44. Speed at bottom after sliding down height h with friction work Wf.
1/2mv²=mgh+Wf. Since Wf negative, v=√(2(mgh+Wf)/m).
45. Work needed to increase speed from v to 2v.
ΔK=1/2m(4v²-v²)=3/2mv².
46. Work needed to increase speed from 2v to 3v.
ΔK=1/2m(9v²-4v²)=5/2mv².
47. Ratio of works for v→2v and 2v→3v.
(3/2):(5/2)=3:5.
48. F=x i + y j along y=x from 0 to a.
F.dr=x dx+x dx=2x dx. W=a².
49. F=-y i + x j around circle radius R once.
W=2πR², non-conservative field.
50. Work-energy theorem in rotating curved path with tangential force only.
Only tangential component changes kinetic energy; normal/radial component does no work.
11
IB / IGCSE / A-Level Questions
Separate international practice sets with answers and explanations.
IB Questions - 25
IB 1. Define kinetic energy.
Energy possessed by a body due to motion.
IB 2. State formula for KE.
K=1/2mv².
IB 3. State work-energy theorem.
Net work done equals change in kinetic energy.
IB 4. KE of 2 kg at 3 m/s.
9 J.
IB 5. What happens to KE if speed doubles?
It becomes four times.
IB 6. What does area under F-x graph show?
Work done.
IB 7. Spring energy formula.
1/2kx².
IB 8. Net work is -10 J. KE change?
-10 J.
IB 9. Net work is zero. KE?
Unchanged.
IB 10. Is KE scalar?
Yes.
IB 11. Unit of KE?
Joule.
IB 12. Dimension of KE?
ML²T^-2.
IB 13. Explain braking using WET.
Brake force does negative work, reducing KE to zero.
IB 14. Explain falling body using WET.
Work by gravity mgh becomes kinetic energy.
IB 15. KE of 0.5 kg at 4 m/s.
4 J.
IB 16. Work needed to stop KE 60 J body.
-60 J by retarding force.
IB 17. Variable force F=2x from 0 to 3.
Work=9 J.
IB 18. If p doubles, KE?
Four times for same mass.
IB 19. K vs v graph shape?
Parabola.
IB 20. K vs p graph shape?
Parabola.
IB 21. Why use net work?
Only total work of all forces equals change in KE.
IB 22. Work by friction sign?
Usually negative for sliding motion.
IB 23. Work by normal on fixed smooth track?
Zero.
IB 24. KE is 50 J, mass 4 kg. Speed?
5 m/s.
IB 25. Force 10 N over 5 m increases KE by?
50 J if it is the net force along displacement.
IGCSE Questions - 25
IGCSE 1. Formula for kinetic energy?
1/2mv².
IGCSE 2. Unit of energy?
Joule.
IGCSE 3. KE of 10 kg at 2 m/s.
20 J.
IGCSE 4. If speed doubles, KE?
Four times.
IGCSE 5. Work done changes what energy?
Kinetic energy if it is net work.
IGCSE 6. A car brakes. KE decreases because work is?
Negative.
IGCSE 7. A falling object gains KE due to?
Work done by gravity.
IGCSE 8. KE at rest?
Zero.
IGCSE 9. Mass unit?
kg.
IGCSE 10. Speed unit?
m/s.
IGCSE 11. KE of 1 kg at 10 m/s.
50 J.
IGCSE 12. Work done 30 J from rest gives KE?
30 J.
IGCSE 13. If KE is 100 J and speed 10 m/s, mass?
2 kg.
IGCSE 14. Force-distance graph area?
Work done.
IGCSE 15. Spring stores what energy?
Elastic potential energy.
IGCSE 16. Work by friction does what to KE?
Decreases it.
IGCSE 17. KE is scalar or vector?
Scalar.
IGCSE 18. Does direction of velocity affect KE?
No, speed magnitude matters.
IGCSE 19. Work needed to stop 40 J object?
-40 J by stopping force.
IGCSE 20. KE of 3 kg at 4 m/s.
24 J.
IGCSE 21. A 5 N net force acts over 4 m. KE change?
20 J.
IGCSE 22. What is Wnet if KE unchanged?
Zero.
IGCSE 23. Speed of mass 2 kg with KE 25 J.
5 m/s.
IGCSE 24. KE formula includes v or v²?
v².
IGCSE 25. Positive net work means speed usually?
Increases.
A-Level Questions - 25
A-Level 1. Express KE in terms of momentum.
K=p²/2m.
A-Level 2. Derive WET from F=ma.
W=Fs=mas and v²-u²=2as, so W=1/2m(v²-u²)=ΔK.
A-Level 3. F=4x from 0 to 5.
W=50 J.
A-Level 4. Spring k=80, x=0.5.
Energy=10 J.
A-Level 5. Work by friction μR over s.
-μRs.
A-Level 6. K=200 J, p=40. Mass?
m=p²/2K=1600/400=4 kg.
A-Level 7. Force varies linearly 2 N to 10 N over 6 m.
Work=36 J.
A-Level 8. Body falls 20 m from rest, g=9.8.
v=√(2gh)=√392 m/s.
A-Level 9. Net power equals?
Rate of change of kinetic energy, dK/dt.
A-Level 10. K vs p slope?
p/m.
A-Level 11. K vs v slope?
mv.
A-Level 12. F=A/x from x1 to x2.
W=A ln(x2/x1).
A-Level 13. KE change if speed from u to v.
1/2m(v²-u²).
A-Level 14. Stopping distance under resistance R.
s=mu²/(2R).
A-Level 15. F=2x² from 0 to 3.
W=18 J.
A-Level 16. Energy stored from x1 to x2 in spring.
1/2k(x2²-x1²).
A-Level 17. Work by spring from x1 to x2.
-1/2k(x2²-x1²).
A-Level 18. If net force perpendicular velocity, dK/dt?
Zero.
A-Level 19. Smooth incline: energy equation.
1/2mv²=1/2mu²+mg s sinθ.
A-Level 20. Rough incline: energy equation.
1/2mv²=1/2mu²+mg s sinθ-μmg cosθ s.
A-Level 21. K=1/2mv² proves KE cannot be?
Negative.
A-Level 22. A particle has KE 100 J and mass 8 kg. Speed?
5 m/s.
A-Level 23. Work by conservative force around loop.
Zero.
A-Level 24. Work by friction around loop.
Negative, generally non-zero.
A-Level 25. F=10 sin x from 0 to π.
W=20 J.
12
Assertion Reason
30 assertion-reason questions with hidden answers.
1. Assertion: KE is scalar. Reason: KE depends on speed squared.
Both true; reason supports scalar nature.
2. Assertion: KE can be negative. Reason: Velocity can be negative.
Assertion false, reason true in one-dimensional sign sense; KE uses v².
3. Assertion: Net work equals change in KE. Reason: This is work-energy theorem.
Both true.
4. Assertion: Positive net work increases KE. Reason: Wnet=ΔK.
Both true; reason explains.
5. Assertion: Negative net work decreases KE. Reason: Kinetic energy is vector.
Assertion true, reason false.
6. Assertion: Speed doubling makes KE four times. Reason: K∝v².
Both true.
7. Assertion: Momentum doubling makes KE four times for same mass. Reason: K=p²/2m.
Both true.
8. Assertion: Area under F-x graph gives work. Reason: W=∫Fdx.
Both true.
9. Assertion: Spring energy is 1/2kx². Reason: Force varies linearly with x.
Both true; reason explains after integration.
10. Assertion: Braking force does positive work. Reason: Braking force opposes displacement.
Assertion false, reason true.
11. Assertion: Gravity increases KE during free fall. Reason: Gravity does positive work.
Both true.
12. Assertion: Normal always changes KE. Reason: Normal always does work.
Both false in common fixed-track cases.
13. Assertion: If net work is zero, speed remains same. Reason: KE unchanged for constant mass.
Both true.
14. Assertion: WET applies only to constant force. Reason: Variable force cannot do work.
Both false. WET also applies to variable forces using integration.
15. Assertion: K vs v is parabolic. Reason: K=1/2mv².
Both true.
16. Assertion: K vs p is parabolic. Reason: K=p²/2m.
Both true for fixed mass.
17. Assertion: Work by friction is usually negative in sliding. Reason: It opposes relative motion.
Both true.
18. Assertion: KE dimension equals work dimension. Reason: Work changes KE.
Both true.
19. Assertion: At rest KE is zero. Reason: v=0.
Both true.
20. Assertion: If mass doubles at same speed, KE doubles. Reason: K∝m.
Both true.
21. Assertion: If mass doubles at same momentum, KE halves. Reason: K=p²/2m.
Both true.
22. Assertion: Work-energy theorem uses net work. Reason: Individual works may cancel.
Both true.
23. Assertion: Spring work by restoring force during stretching is negative. Reason: Spring force opposes displacement.
Both true.
24. Assertion: External work during slow spring stretch is positive. Reason: External force and displacement are same direction.
Both true.
25. Assertion: Graph area below x-axis gives negative work. Reason: Work is signed area.
Both true.
26. Assertion: A body can turn without change in KE. Reason: Force may be perpendicular to velocity.
Both true.
27. Assertion: Falling speed from height h is √2gh. Reason: mgh=1/2mv².
Both true for free fall from rest ignoring air resistance.
28. Assertion: WET is useful for stopping distance. Reason: Brake work equals loss of KE.
Both true.
29. Assertion: KE depends on frame. Reason: Speed depends on frame.
Both true.
30. Assertion: Work done by net force is path independent. Reason: Net force is always conservative.
Both generally false; work can be path dependent if non-conservative forces act.
13
Case Study Questions
Case studies on braking, falling body, roller coaster, spring system and variable force graph.
Case 1: Braking system. A 1000 kg car at 20 m/s is stopped by brakes.
Initial KE=200000 J. Work by brakes=-200000 J. If brake force is 4000 N, stopping distance is 50 m. Tip: braking work is negative.
Case 2: Falling body. A 2 kg object falls from 10 m.
Gravity work=mgh=200 J for g=10. KE gain=200 J. Speed=√(2gh)=10√2 m/s.
Case 3: Roller coaster descends height h without friction.
Loss in potential energy becomes kinetic energy: mgh=1/2mv². Speed depends on height, not mass.
Case 4: Spring system k=250 N/m compressed 0.2 m launches 0.5 kg cart.
Spring energy=1/2kx²=5 J. 1/2mv²=5, so v=√20=2√5 m/s.
Case 5: Variable force graph rises 0 to 50 N over 4 m then falls to 0 over 2 m.
Work=triangle1 1/2x4x50 + triangle2 1/2x2x50 = 100+50=150 J. KE increases by 150 J if this is net work.
14
Common Student Mistakes
Avoid these frequent mistakes in kinetic energy and work-energy theorem.
Confusing Work And Energy
Work is energy transfer. Kinetic energy is energy possessed due to motion.
Forgetting Net Work
WET uses total work by all forces, not just one applied force.
Wrong Sign Convention
Friction and braking usually do negative work. Gravity sign depends on displacement direction.
Incorrect Graph Area
Use signed area under F-x graph. Areas below axis are negative.
Incorrect Spring Work
Spring energy is 1/2kx², not kx². The factor 1/2 comes from triangular graph area.
Forgetting Velocity Squared
Kinetic energy depends on v². Doubling speed makes KE four times.
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