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Work, Energy, Power and Collision

NEET > Physics > Work Energy and Power

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Overview content

Chapter Snapshot - Work, Energy, Power and Collision

Chapter 6 bridges kinematics and dynamics by introducing scalar energy quantities. Work (W = Fs cosθ) links force to displacement; the work-energy theorem ties net work to change in kinetic energy. Potential energy (gravitational mgh, elastic ½kx²) completes mechanical energy conservation. Power measures the rate of doing work (P = F·v). Collisions — elastic, inelastic, perfectly inelastic — are analysed via momentum conservation and the coefficient of restitution. Vertical circular motion unifies all concepts: minimum speed conditions at the top (√gr) and bottom (√5gr) of a loop govern tension and energy analysis.

✓ Use This To Plan Your First 2–3 Hours
Expected Questions (Typical)
Q
3-4
NEET consistently places 3–4 questions from this chapter every year, often blending work-energy theorem with collision scenarios.
Time Required (Practical)
⏱
10-12 hrs
Allow ~3 hrs for work and energy formulas, ~3 hrs for collisions, ~2 hrs for power, and ~2 hrs for vertical circular motion with practice problems.
Difficulty Level
⚡
Moderate-Hard
Formula application is straightforward but NEET questions combine multiple concepts (e.g., energy conservation plus collision type plus vertical circle) in a single problem.
Most Asked Style: Numerical problems requiring application of work-energy theorem, elastic collision velocity formulas, and minimum speed in vertical circular motion — often with a twist on which bodies are stationary or moving in opposite directions.Biggest Trap: Assuming kinetic energy is conserved in all collisions. Only perfectly elastic collisions conserve KE; perfectly inelastic (bodies stick) and plain inelastic do not. Also confusing W_spring = −½kx² (by spring on block) vs U_spring = +½kx² (stored in spring).Fast Win: Memorise the five special-case elastic collision outcomes (equal masses exchange velocities; heavy hits stationary light → light moves at 2u; light hits heavy stationary → light bounces back). These appear almost every year.Revision-Friendly: Draw energy pie charts (KE ↔ PE) for conservation problems and a standard elastic collision table. Both fit on one page and recover 80% of marks in this chapter.

Subtopics - Work, Energy, Power and Collision (NEET)

From W = Fs cosθ to collision formulas and vertical loops — master scalar energy methods

Revision tip: Always start with energy conservation; switch to momentum conservation only when forces are impulsive (collision). For vertical circle, check both tension and velocity conditions at top and bottom simultaneously.
NCERT LinesMCQsQuick Test

1) Work and Work-Energy Theorem

Defines work as the dot product of force and displacement (W = F·s cosθ), covers positive, negative and zero work, variable force work via integration, and proves the work-energy theorem W_net = ΔKE = ½mv² − ½mu².

W = Fs cosθW_net = ΔKEConservative forcesPath independenceF·d graph area
›
Constant Force WorkW = Fs cosθ; positive (θ < 90°), negative (θ > 90°), zero (θ = 90° or s = 0 or F = 0).
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Variable Force WorkW = ∫F·ds; area under F-x graph equals work done.
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Work-Energy TheoremW_net = ΔKE = ½mv² − ½mu²; valid for all force types, including non-conservative.
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Conservative vs Non-conservative ForcesConservative (gravity, spring): path-independent work, closed-loop work = 0. Non-conservative (friction): path-dependent, closed-loop work ≠ 0.

2) Energy Types and Conservation

Covers kinetic energy KE = ½mv² = P²/2m, gravitational PE = mgh, elastic PE = ½kx², and the law of conservation of mechanical energy. Relates momentum and kinetic energy (P = √(2mE)).

KE = ½mv²PE = mghU_spring = ½kx²E = KE + PE = constP = √(2mE)
›
Kinetic EnergyKE = ½mv²; always positive scalar; KE = P²/2m links momentum and energy.
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Gravitational Potential EnergyU = mgh (near Earth surface); zero reference chosen at ground.
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Elastic (Spring) Potential EnergyU = ½kx²; work done by spring on block = −½kx² (spring force opposes extension).
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Law of Conservation of Mechanical EnergyKE + PE = constant in conservative fields; loss in KE = gain in PE and vice versa. With friction: ΔE = W_friction.
›
Potential Energy Curves and EquilibriumF = −dU/dx; stable equilibrium at PE minimum (d²U/dx² > 0); unstable at maximum.

3) Power and Collisions

Power P = dW/dt = F·v (instantaneous) or P_avg = W/t. Collision types: elastic (e = 1, KE conserved), perfectly inelastic (e = 0, bodies merge), inelastic (0 < e < 1). Elastic 1D formulas: v1 = (m1−m2)u1/(m1+m2) + 2m2u2/(m1+m2); key special cases; loss in KE for perfectly inelastic = ½·m1m2/(m1+m2)·(u1−u2)².

P = F·ve = v_sep/v_appElastic: KE conservedPerfectly inelastic: e = 0ΔKE loss formula
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PowerP_avg = W/t; P_inst = F·v; units: Watt, hp (1 hp = 746 W); kWh is energy unit.
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Coefficient of Restitutione = (v2 − v1)/(u1 − u2); elastic e=1; perfectly inelastic e=0; inelastic 0 < e < 1.
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Perfectly Elastic Head-on CollisionKE conserved; velocities: v1=(m1−m2)u1/(m1+m2)+2m2u2/(m1+m2); v2=2m1u1/(m1+m2)+(m2−m1)u2/(m1+m2). Equal masses: exchange velocities.
›
Perfectly Inelastic CollisionBodies merge; v_comb=(m1u1+m2u2)/(m1+m2); ΔKE = ½·m1m2/(m1+m2)·(u1−u2)².
›
Rebounding Ball (e)After nth bounce: height hn = e²ⁿh0; total distance H = h0·(1+e²)/(1−e²); total time T = √(2h0/g)·(1+e)/(1−e).

4) Vertical Circular Motion

Motion in a vertical loop: velocity and tension vary with position. Critical conditions: minimum speed at top v_C = √(gl) (tension = 0); minimum speed at bottom v_A = √(5gl) to complete loop; T_bottom − T_top = 6mg. Energy conservation connects all positions.

v_top_min = √(gl)v_bottom_min = √(5gl)T_bottom − T_top = 6mgKE + PE = constCritical state
›
Velocity at Any Pointv = √(u² − 2gl(1 − cosθ)) where u is speed at lowest point and l is string length.
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Tension at Any PointT = (m/l)[u² − gl(2 − 3cosθ)]; at bottom T_A = mu²/l + mg; at top T_C = mu²/l − 5mg.
›
Critical ConditionFor just completing the circle, T_C = 0 → u_min = √(5gl) at bottom; v_min = √(gl) at top.
›
Tension DifferenceT_bottom − T_top = 6mg (independent of speed and radius at critical condition).
›
Block on Frictionless HemisphereBlock loses contact at height h = 2r/3 (angle cosθ = 2/3) from ground.

Work, Energy, Power and Collision Download Notes & Weightage Plan

For each topic in the Work, Energy, Power and Collision chapter below, you get (2) the exact resources to download and how to use them, and (3) a simple importance & time plan so NEET students know what to do first and what to revise last.

2 Downloads

Work and Work-Energy Theorem

Core concept linking force and displacement via scalar product; work-energy theorem is the fundamental tool for energy-based problem solving.

W = Fs cosθW_net = ΔKEConservative forcesArea under F-x graph

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)W = Fs cosθ (constant F); W = ∫F·ds (variable F) = area under F-x graph. Net work done = change in KE: W_net = ½mv² − ½mu². Conservative forces (gravity, spring): W independent of path, closed-loop W = 0. Non-conservative forces (friction): path-dependent, closed-loop W ≠ 0. Frame-dependent: work changes with frame.
Download NotesPrintable PDF
★
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Draw the F-x graph for spring and shade the triangular area. Verify work-energy theorem on 2-3 standard problems (inclined plane + friction). Practise closed-loop W = 0 proofs for gravity and spring.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1Usually 1 direct question on W = Fs cosθ or work-energy theorem, often combined with an inclined plane or circular path.
Time Required2-3 hrsCover theory in 1 hr; solve 10 past NEET questions in 2 hrs.
DifficultyModerateFormula application is easy; distinguishing conservative vs non-conservative contexts is where students lose marks.
  • Scoring Focus: Work-energy theorem applied to stopping distance (x = mv²/2F) and block-spring problems: these are direct 1-mark scorers.
  • High-risk Area: Mixing up W_by_spring (negative: −½kx²) and U_spring (positive: +½kx²). Also forgetting that centripetal force does zero work.
  • Best Practice Style: Mixed numerical + conceptual MCQs; 60% computation, 40% identification of sign and conditions.
Priority rule: Solve first: stopping distance, block on incline, spring compression. These three patterns cover ~70% of work-energy MCQs in NEET.

Energy Types and Conservation

KE, gravitational PE, elastic PE and their interconversion under conservation of mechanical energy. Equilibrium analysis using PE curves.

KE = ½mv² = P²/2mPE = mghU = ½kx²KE + PE = const

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)KE = ½mv² = P²/2m; momentum P = √(2mE). PE_gravity = mgh; PE_spring = ½kx². Conservation: KE + PE = E (constant) when only conservative forces act. With friction: E_final = E_initial + W_friction. Stable equilibrium: PE minimum, d²U/dx² > 0. Unstable: PE maximum. Body loses contact with hemisphere surface at h = 2r/3.
Download NotesPrintable PDF
★
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Practise energy bar charts (KE, PE, total E) at three positions: start, midpoint, end. Memorise P = √(2mE) and apply to comparison problems. Sketch PE curves and identify equilibrium points.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1One question typically tests conservation of energy in projectile, spring-block or roller-coaster context.
Time Required2-3 hrs1 hr theory + 2 hrs problem sets on spring-mass systems and PE curves.
DifficultyModeratePE concepts are manageable; PE curve equilibrium analysis (stable/unstable/neutral) can trip students in assertion-reason format.
  • Scoring Focus: Solving for velocity after a spring releases a block (½kx² = ½mv²) or height reached after collision using conservation — direct 1-mark problems.
  • High-risk Area: Sign error in work done by spring: W_by_spring = −½kx² (spring does negative work on block when block compresses or stretches it from mean). KE can never be negative.
  • Best Practice Style: Mostly numerical; assertion-reason for PE curves. Equal weight on spring-block and gravitational scenarios.
Priority rule: Master spring-block energy conservation and KE-momentum relations before moving to PE curves.

Power and Collisions

Power as rate of work; elastic and inelastic collision analysis via momentum conservation plus coefficient of restitution framework.

P = F·ve = v_sep/v_appv1,v2 elastic formulasΔKE = ½·μ·(u1−u2)²

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)P_avg = W/t; P_inst = F·v; 1 hp = 746 W; kWh is energy. Collision: total momentum conserved always; KE only in elastic. e = (v2−v1)/(u1−u2): elastic e=1, perfectly inelastic e=0. Elastic 1D: v1=(m1−m2)u1/(m1+m2)+2m2u2/(m1+m2); v2=2m1u1/(m1+m2)+(m2−m1)u2/(m1+m2). Equal masses: exchange velocities (v1=u2, v2=u1). Heavy hits light stationary: v1≈u1, v2=2u1. Light hits heavy stationary: v1≈−u1. Perfectly inelastic: v_comb=(m1u1+m2u2)/(m1+m2); ΔKE_loss = ½·m1m2/(m1+m2)·(u1−u2)². Bullet-block: v_block = mu/(m+M), then ½(m+M)v² = (m+M)gh gives h.
Download NotesPrintable PDF
★
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Write the three sets of elastic collision special cases from memory. Derive ΔKE_loss formula for perfectly inelastic once. Practice e = 0.5 inelastic problems. For power: solve 3 automobile-power problems using constant power equation v = √(2Pt/m).

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1-2Collisions get 1–2 questions most years; power occasionally appears as a standalone or embedded in a collision scenario.
Time Required3-4 hrs1 hr power + 3 hrs collision types with NEET past questions.
DifficultyHardElastic collision velocity derivation and special-case memory require sustained practice; traps exist in sign conventions and direction assumptions.
  • Scoring Focus: Equal-mass elastic collision (exchange velocities) and bullet-block perfectly inelastic collision appear almost every NEET year — learn these cold.
  • High-risk Area: Forgetting that KE is not conserved in 'inelastic' (most real-world) collisions. Also sign errors when one body moves opposite to the other before collision (negative initial velocity convention).
  • Best Practice Style: Mostly numerical with 4 options close in value, testing precision in v1/v2 formula. Conceptual MCQs on e=1 vs e=0 conditions.
Priority rule: Perfect elastic 1D special cases → perfectly inelastic ΔKE formula → power P=F·v. This order maximises marks per hour.

Vertical Circular Motion

Non-uniform circular motion under gravity; critical speed conditions, tension calculations, and energy distribution at every point of the loop.

v_top_min = √(gl)v_bottom_min = √(5gl)T_A − T_C = 6mgh_critical = 2r/3v = √(u²−2gl(1−cosθ))

1) Download Packs For This Topic (And How To Use Them)

Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.

↓
Topic Notes (Condensed)For circular loop of radius l with body at lowest point A: v = √(u²−2gl(1−cosθ)). Critical condition (T_C = 0 at top): u_min = √(5gl); v at top = √(gl); T at bottom = 6mg; T at horizontal = 3mg. Difference: T_A − T_C = 6mg always. If √(2gl) < u < √(5gl): string slackens between horizontal and top (parabolic trajectory). If u < √(2gl): oscillates. Block on hemisphere loses contact at h = 2r/3 (cosθ = 2/3).
Download NotesPrintable PDF
★
NCERT Key Lines (One-Liners)These are the lines NEET converts into "statement is correct/incorrect" questions.
NCERT LinesFlashcards
Q
Practice Set (MCQs + PYQs)Do 30–50 questions, then mark errors as "memory miss" or "confusion between options."
MCQ SetPYQs
How to revise: Draw the vertical loop, label positions A (bottom), B (side), C (top). Fill in Table 3.5 values from memory: velocities √(5gl), √(3gl), √(gl), tensions 6mg, 3mg, 0. Then derive T_A − T_C = 6mg from formulas.

2) Importance, Weightage & Time Allocation (Practical)

Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.

Expected Questions1One question on minimum speed or tension difference in vertical circle appears in most NEET exams.
Time Required2 hrs30 min theory + table memorisation; 1.5 hrs problem sets including NEET and AIIMS past questions.
DifficultyModerate-HardThe five-case table (√5gl, between √2gl-√5gl, √2gl, below √2gl) and tension difference formula require careful memorisation; application is straightforward once memorised.
  • Scoring Focus: T_A − T_C = 6mg is a direct formula question scorer. Minimum speed to complete loop (√(5gl) at bottom) appears as a 1-mark direct question.
  • High-risk Area: Confusing the speed at top (√gl) with speed at bottom (√5gl) — NEET distractors specifically exploit this. Also missing that the hemisphere detachment height is 2r/3, not r/2.
  • Best Practice Style: Direct formula application numericals (60%) and conceptual condition-based MCQs (40% — which regime does the particle fall into for a given u?).
Priority rule: Memorise: v_bottom_min = √(5gl), v_top_min = √(gl), T_difference = 6mg. All three appear in NEET at least once every 2 years.

Work, Energy, Power and Collision Chapter NEET Traps & Common Mistakes (Topic-Wise)

Each subtopic below is of the Work, Energy, Power and Collision chapter and shows what NEET students usually do wrong in NEET examination, a short example of the mistake, and how NEET frames the question to trick you with close options are given below.

! Avoid Easy Negatives
Work Done by Spring
WorkSpringSign Convention

Mistake Snapshot (What Students Do Wrong)

  • Confusing stored energy with work by spring: U_spring = +½kx² (energy stored in spring), but work done BY the spring on the block = −½kx² (spring force opposes displacement from natural length).
  • Double-counting work: Students add both external agent's work AND spring's work to get ΔKE, but they are equal and opposite — W_external = +½kx² = −W_spring.
2–3 Line Example (Typical Error)

A spring (k = 200 N/m) is compressed by 0.1 m. A student claims work done by spring is +1 J. Correct: W_spring on block = −½ × 200 × 0.01 = −1 J (negative; spring pushes block in direction of release, which is opposite to compression).

How NEET Frames The Trap

Work done by spring vs potential energy stored in spring

NEET-Style Trap Question Format

Q. A spring of spring constant k is compressed by x from its natural length. The work done BY the spring on the block during this compression is:
A. +½kx²   B. −½kx²   C. kx²   D. 0  
Trick: Answer is −½kx² (B). The spring exerts a restoring force opposing the compression direction, so work done by the spring is negative. U = +½kx² is the elastic PE stored, not the work done by the spring.

Quick rule: W_by_spring = −½kx² (always negative during compression or extension); U_spring = +½kx² (always positive, stored energy).
Elastic Collision Special Cases
CollisionElasticSpecial Cases

Mistake Snapshot (What Students Do Wrong)

  • Forgetting the heavy-hits-light formula: When a very heavy body (m1 >> m2) hits a stationary light body: v1 ≈ u1 (heavy barely slows down) and v2 = 2u1 (light flies off at twice the speed). Students often write v2 = u1.
  • Equal mass direction confusion: If m1 = m2 and both are moving in the same direction, they exchange velocities. But if u2 = 0, then v1 = 0, v2 = u1 — the first body stops completely.
2–3 Line Example (Typical Error)

A truck (mass 1000 kg, velocity 10 m/s) elastically hits a stationary ball (mass 1 kg). Students answer v_ball = 10 m/s. Correct: v_ball = 2 × 10 = 20 m/s (light body gets almost double the incoming speed).

How NEET Frames The Trap

Special cases of elastic 1D collisions — NEET loves heavy-vs-light scenarios

NEET-Style Trap Question Format

Q. A ball of mass m moving with speed u makes a perfectly elastic head-on collision with a stationary ball of mass 3m. The velocity of the lighter ball after collision is:
A. u/2 (forward)   B. −u/2 (backward)   C. 2u/3 (backward)   D. u/3 (forward)  
Trick: Use v1 = (m − 3m)/(m + 3m)·u = (−2m/4m)·u = −u/2. Answer is B (−u/2, backwards). When light hits heavy stationary body, the light ball bounces back.

Quick rule: Light ball hits heavy stationary: v_light = −[(m2−m1)/(m1+m2)]u ≈ −u (bounces back). Heavy hits stationary light: v_light ≈ 2u (flies forward at nearly double speed).
Minimum Speed in Vertical Circle
Vertical CircleCritical SpeedTension

Mistake Snapshot (What Students Do Wrong)

  • Swapping top and bottom speed: v_min at top of loop = √(gl), NOT √(5gl). v_min at bottom = √(5gl) to just complete the circle. Students regularly interchange these.
  • Missing T_bottom − T_top = 6mg: This result is derived from energy conservation + tension formulas and is independent of speed. However, students recompute T at each point instead of using this elegant result.
2–3 Line Example (Typical Error)

NEET 2018 asked: a body slides from height h to just complete a vertical circle of diameter D. Students chose h = D (wrong). Answer: h = 5D/4 from energy conservation (½mv² = mgh gives v² = √(5gl) at bottom, h = 5l/2 = 5D/4).

How NEET Frames The Trap

Minimum speed and tension at top vs bottom of a vertical loop

NEET-Style Trap Question Format

Q. A stone of mass m is tied to a string of length l and whirled in a vertical circle. The minimum velocity at the lowest point so that the string does not go slack at the highest point is:
A. √(gl)   B. √(2gl)   C. √(3gl)   D. √(5gl)  
Trick: Answer is D — √(5gl). The critical condition is T = 0 at top, giving v_top = √(gl). From energy conservation: v_bottom² = v_top² + 4gl = gl + 4gl = 5gl, so v_bottom = √(5gl).

Quick rule: √(gl) = minimum speed at TOP; √(5gl) = minimum speed at BOTTOM; difference in tension T_bottom − T_top = 6mg.
KE Conservation in Collisions
CollisionKinetic EnergyConservation

Mistake Snapshot (What Students Do Wrong)

  • Assuming all collisions conserve KE: Only perfectly elastic collisions (e = 1) conserve kinetic energy. ALL collisions conserve momentum.
  • Confusing total energy with kinetic energy: Total energy (including heat, deformation energy) is always conserved. KE is only conserved in elastic collisions.
2–3 Line Example (Typical Error)

A bullet embeds in a wooden block — KE is NOT conserved (perfectly inelastic, e = 0). Students incorrectly use KE_before = KE_after. Use momentum conservation to find v_combined, then KE_after = ½(M+m)v_combined².

How NEET Frames The Trap

Which quantity is conserved in elastic vs inelastic vs perfectly inelastic collisions?

NEET-Style Trap Question Format

Q. In a perfectly inelastic collision between two bodies of equal mass m, one of which is at rest, the fraction of kinetic energy lost is:
A. 0   B. 1/4   C. 1/2   D. 3/4  
Trick: Answer is C (1/2). By momentum conservation: mv = 2mv_f → v_f = v/2. KE_initial = ½mv²; KE_final = ½(2m)(v/2)² = mv²/4; ΔKE = ½mv²(1 − 1/2) = mv²/4. Fraction lost = (mv²/4)/(mv²/2) = 1/2.

Quick rule: Elastic: KE conserved, e = 1. Perfectly inelastic: maximum KE lost (consistent with momentum conservation), e = 0. All: total energy and momentum conserved.

Topics

Work Done

Energy

Power

Collision

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