NEET UG Physics · Chapter 4

Work, Energy and Power

Work–energy theorem, potential energy, power, conservation of energy and collisions.

NCERT Class 11NEET: 1–2 Qs per paper 10 min read

1. Chapter at a Glance

Work, energy and power give a quick "before-and-after" way to solve mechanics problems. Instead of following forces step by step, we compare energies at two points. This chapter covers work done by constant and variable forces, kinetic and potential energy, the work–energy theorem, conservation of mechanical energy, power, and collisions. NEET asks one or two questions from it every year.

What NEET asks most
  • Work done from F·s or from the area under a force–displacement graph.
  • Kinetic energy–momentum relation.
  • Work–energy theorem and conservation of energy.
  • Elastic and inelastic collisions, coefficient of restitution.
  • Power of pumps, engines and motors.

2. Work

  • Work done by a constant force: W = F·s = Fs cosθ. Unit: joule (J).
  • W is positive when 0° ≤ θ < 90° (force helps motion), negative when 90° < θ ≤ 180° (friction, braking), and zero when θ = 90° (centripetal force, normal force on a level road, a coolie carrying a load on a level path).
  • Work done by a variable force: W = ∫F dx = area under the F–x graph.
  • Work done in stretching a spring from x₁ to x₂: ½k(x₂² − x₁²).
  • 1 kWh = 3.6 × 10⁶ J; 1 eV = 1.6 × 10⁻¹⁹ J; 1 erg = 10⁻⁷ J.

3. Kinetic Energy, Potential Energy and the Work–Energy Theorem

K = ½mv² = p²/2m    Wnet = ΔK
  • The work–energy theorem is valid for every kind of force: the net work done on a body equals its change in kinetic energy.
  • Potential energy is stored energy of position or configuration. Gravitational PE near the Earth: U = mgh. Spring PE: U = ½kx².
  • A force is conservative if its work depends only on the end points (gravity, spring force, electric force). For conservative forces F = −dU/dx. Friction is non-conservative.
  • Conservation of mechanical energy: if only conservative forces do work, K + U = constant.
top halfway bottom PEKEPEKE total energy = constant
For a freely falling body, potential energy changes into kinetic energy but their sum stays constant.

4. Power

Pavg = W/t    P = dW/dt = F·v

Unit: watt (W) = J s⁻¹; 1 horsepower (hp) = 746 W. A pump lifting mass m through height h in time t has power mgh/t. A vehicle moving at steady speed v against total opposing force F uses power Fv. Efficiency = (useful output/input) × 100%.

5. Collisions

In every collision, linear momentum is conserved (external forces are negligible during the short impact). Kinetic energy is conserved only in elastic collisions.

TypeMomentumKinetic energye
Perfectly elasticConservedConserved1
InelasticConservedPartly lost0 < e < 1
Perfectly inelasticConservedMaximum loss; bodies stick0

1D elastic collision with m₂ at rest: v₁ = (m₁ − m₂)u₁/(m₁ + m₂), v₂ = 2m₁u₁/(m₁ + m₂). Equal masses exchange velocities. The coefficient of restitution e = (speed of separation)/(speed of approach); a ball dropped from height h rebounds to e²h.

6. Formula Master Table

QuantityFormula
Work (constant force)W = Fs cosθ
Work (variable force)W = ∫F dx
Kinetic energyK = ½mv² = p²/2m
Potential energyU = mgh; Uspring = ½kx²
Work–energy theoremWnet = Kf − Ki
PowerP = W/t = Fv
Elastic collision (m₂ at rest)v₁ = (m₁ − m₂)u₁/(m₁ + m₂); v₂ = 2m₁u₁/(m₁ + m₂)
Rebound heighth′ = e²h

7. Master Summary Table

ConceptKey pointNEET Note
WorkScalar; can be +, − or 0Centripetal force does no work
Kinetic energyAlways ≥ 0p doubled ⇒ K four times
Potential energyDefined for conservative forcesF = −dU/dx
Energy conservationK + U constant without frictionWith friction, subtract work done by friction
PowerRate of doing workP = Fv at steady speed
CollisionsMomentum always conservedKE conserved only if elastic

8. Mind Map

Work, Energy & Power
Work
  • F·s
  • Area under F–x
  • Spring work
Energy
  • KE, PE
  • Work–energy theorem
  • Conservation
Power
  • W/t
  • F·v
  • Efficiency
Collisions
  • Elastic, inelastic
  • Restitution

9. Tips and Tricks

Shortcuts
  • If a question asks for speed at some point, try energy conservation before force equations.
  • For the same momentum, a lighter body has more kinetic energy (K = p²/2m).
  • In a perfectly inelastic collision with m₂ at rest, the fraction of KE lost = m₂/(m₁ + m₂).
  • Work to lift the hanging part of a chain (length L/n of total L, mass M) onto the table: MgL/(2n²).
Common NEET traps
  • Assuming kinetic energy is conserved in all collisions.
  • Forgetting the negative sign of work done by friction.
  • Using mgh when the height is comparable to the Earth's radius (use GMm/r instead).

10. Important One-liners

  • Work done by a centripetal force is always zero.
  • Work done by a conservative force around a closed path is zero.
  • If the momentum of a body is doubled, its kinetic energy becomes four times.
  • The work–energy theorem holds for all types of forces.
  • Momentum is conserved in both elastic and inelastic collisions.
  • Equal masses colliding elastically head-on exchange velocities.
  • 1 horsepower equals 746 watts.

11. Solved Numericals: Average to Advanced

Five average and five advanced numericals. Take g = 10 m s⁻². Click the button to see each answer and its working.

Average level

N1Average

A 50 N force acting at 60° to the horizontal pulls a trolley 10 m along a level floor. Find the work done by the force.

Show answer & solution

Concept: W = Fs cosθ.

  1. W = 50 × 10 × cos60°.
  2. = 500 × 0.5 = 250 J.

Answer: 250 J

N2Average

A 0.5 kg ball moves at 20 m s⁻¹. Find its kinetic energy and momentum, and check K = p²/2m.

Show answer & solution

Concept: K = ½mv², p = mv.

  1. K = ½ × 0.5 × 400 = 100 J.
  2. p = 0.5 × 20 = 10 kg m s⁻¹.
  3. Check: p²/2m = 100/1 = 100 J ✓.

Answer: 100 J; 10 kg m s⁻¹

N3Average

A motor raises a 200 kg load through 15 m in 20 s. Find its power output.

Show answer & solution

Concept: P = mgh/t.

  1. Work = 200 × 10 × 15 = 30 000 J.
  2. P = 30 000/20 = 1500 W = 1.5 kW (about 2 hp).

Answer: 1.5 kW

N4Average

A spring (k = 400 N m⁻¹) is already stretched by 5 cm. How much work is needed to stretch it further to 10 cm?

Show answer & solution

Concept: W = ½k(x₂² − x₁²).

  1. x₁ = 0.05 m, x₂ = 0.10 m.
  2. W = ½ × 400 × (0.01 − 0.0025) = 200 × 0.0075.
  3. W = 1.5 J.

Answer: 1.5 J

N5Average

A ball is dropped from a height of 10 m. Using energy conservation, find its speed just before it hits the ground.

Show answer & solution

Concept: loss of PE = gain of KE.

  1. mgh = ½mv² ⇒ v = √(2gh).
  2. v = √(200) ≈ 14.1 m s⁻¹.

Answer: about 14.1 m s⁻¹

Advanced level

N6Advanced

A 3 kg ball moving at 4 m s⁻¹ makes a head-on elastic collision with a 1 kg ball at rest. Find both velocities after the collision and verify that kinetic energy is conserved.

Show answer & solution

Concept: elastic-collision formulas with m₂ at rest.

  1. v₁ = (3 − 1)/(3 + 1) × 4 = 2 m s⁻¹ (moves on, slower).
  2. v₂ = 2 × 3/(3 + 1) × 4 = 6 m s⁻¹.
  3. KE before = ½ × 3 × 16 = 24 J; after = ½ × 3 × 4 + ½ × 1 × 36 = 6 + 18 = 24 J ✓.

Answer: 2 m s⁻¹ and 6 m s⁻¹

N7Advanced

A 20 g bullet moving at 400 m s⁻¹ embeds itself in a 4 kg block resting on a rough floor (μk = 0.2). How far does the block slide?

Show answer & solution

Concept: momentum conservation for the impact, then work–energy theorem for the sliding.

  1. Common velocity V = 0.02 × 400/4.02 ≈ 1.99 m s⁻¹.
  2. Friction deceleration = μg = 2 m s⁻².
  3. Distance = V²/(2μg) = 3.96/4 ≈ 0.99 m.

Answer: about 0.99 m

N8Advanced

A force F = (2x + 3) N acts on a body along the x-axis. Find the work done as the body moves from x = 1 m to x = 3 m.

Show answer & solution

Concept: W = ∫F dx for a variable force.

  1. W = ∫₁³ (2x + 3) dx = [x² + 3x]₁³.
  2. = (9 + 9) − (1 + 3) = 18 − 4.
  3. W = 14 J.

Answer: 14 J

N9Advanced

A uniform chain of mass 3 kg and length 1.5 m lies on a smooth table with one-third of its length hanging over the edge. How much work is needed to pull the hanging part back onto the table?

Show answer & solution

Concept: lift the centre of mass of the hanging part to table level.

  1. Hanging mass = 3/3 = 1 kg; hanging length = 0.5 m.
  2. Its centre of mass is 0.25 m below the table.
  3. W = 1 × 10 × 0.25 = 2.5 J (same as MgL/18).

Answer: 2.5 J

N10Advanced

A 1000 kg car climbs a road rising 1 m for every 20 m along the road, at a steady 10 m s⁻¹. Friction and air resistance total 500 N. Find the engine's power.

Show answer & solution

Concept: engine force = mg sinθ + resistance; P = Fv. Here sinθ = 1/20.

  1. mg sinθ = 1000 × 10 × 0.05 = 500 N.
  2. Total force = 500 + 500 = 1000 N.
  3. P = 1000 × 10 = 10 000 W = 10 kW.

Answer: 10 kW

12. NEET-pattern Questions with Detailed Answers

These questions are original and modelled on concepts repeatedly asked in NEET UG.

Q1Easy

The work done by the tension in a string on a stone moving in a horizontal circle is:

  1. Positive
  2. Negative
  3. Zero
  4. Depends on speed
Show solution

The tension is along the radius and the displacement is along the tangent; θ = 90° so W = 0.

Answer: (C)

Q2Easy

If the kinetic energy of a body becomes four times, its momentum becomes:

  1. Four times
  2. Twice
  3. Half
  4. Sixteen times
Show solution

p = √(2mK) ∝ √K, so p becomes √4 = 2 times.

Answer: (B)

Q3Medium

A ball dropped from 16 m rebounds to 4 m. The coefficient of restitution is:

  1. 0.25
  2. 0.5
  3. 0.75
  4. 1
Show solution

h′ = e²h ⇒ e² = 4/16 = 0.25 ⇒ e = 0.5.

Answer: (B)

Q4Medium

The potential energy of a particle is U = 3x² − 12x (SI units). The force on it at x = 1 m is:

  1. 6 N in +x
  2. 6 N in −x
  3. 12 N in +x
  4. 0
Show solution

F = −dU/dx = −(6x − 12) = 12 − 6x. At x = 1, F = +6 N, along +x. (Equilibrium is at x = 2 m.)

Answer: (A)

Q5Medium

A pump lifts 600 kg of water to a tank 20 m high in 1 minute. Its power is (g = 10 m s⁻²):

  1. 1 kW
  2. 2 kW
  3. 12 kW
  4. 120 kW
Show solution

P = mgh/t = 600 × 10 × 20/60 = 2000 W = 2 kW.

Answer: (B)

Q6Medium

In a perfectly inelastic collision:

  1. Both momentum and KE are conserved
  2. Momentum is conserved, KE is not
  3. KE is conserved, momentum is not
  4. Neither is conserved
Show solution

Momentum is conserved in every collision; in a perfectly inelastic one the bodies stick and the maximum KE is lost.

Answer: (B)

Q7Hard

A 2 kg body moving at 3 m s⁻¹ collides and sticks to a 1 kg body at rest. The loss of kinetic energy is:

  1. 3 J
  2. 6 J
  3. 9 J
  4. 0
Show solution

Common velocity = 6/3 = 2 m s⁻¹. KE before = 9 J, after = ½ × 3 × 4 = 6 J. Loss = 3 J (fraction m₂/(m₁ + m₂) = 1/3 of 9 J).

Answer: (A)

Q8Hard

A body is moved along a closed path by a conservative force. The total work done by that force is:

  1. Positive
  2. Negative
  3. Zero
  4. Equal to the path length × force
Show solution

For a conservative force, work depends only on the start and end points; on a closed path they coincide, so W = 0.

Answer: (C)

13. Quick Revision Summary

Remember before the exam
  • W = Fs cosθ; variable force: area under F–x graph.
  • K = p²/2m; Wnet = ΔK for all forces.
  • Conservative force: F = −dU/dx; closed-path work zero.
  • P = W/t = Fv; 1 hp = 746 W.
  • Momentum conserved in all collisions; e = 1 elastic, e = 0 sticking; h′ = e²h.
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