Work Energy and Power — Class 11 Physics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 11 Physics chapter "Work Energy and Power" — 6 important questions with detailed answers for CBSE board exam preparation.

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TL;DR: Free step-by-step NCERT solutions for Class 11 Physics chapter "Work Energy and Power" — 6 important questions with detailed answers for CBSE board ex…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Jul 23, 2026

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Key Questions Covered:

Define work, energy, and power. Give SI units and one example of each.
A force of 50 N is applied on a block at an angle of 30° above the horizontal…
A ball of mass 0.5 kg is thrown vertically upward with an initial velocity of…
A car of mass 1500 kg accelerates from rest to 30 m/s in 10 seconds on a hori…
A pump lifts 100 litres of water per minute to a height of 20 m. If the densi…
A block of mass 5 kg sliding on a horizontal surface with initial velocity 10…

Solutions Summary:

Question	Status
Define work, energy, and power. Give SI units and one exa…	✓ Solved
A force of 50 N is applied on a block at an angle of 30° …	✓ Solved
A ball of mass 0.5 kg is thrown vertically upward with an…	✓ Solved
A car of mass 1500 kg accelerates from rest to 30 m/s in …	✓ Solved
A pump lifts 100 litres of water per minute to a height o…	✓ Solved
A block of mass 5 kg sliding on a horizontal surface with…	✓ Solved

Showing 6 of 6 questions

Q1: Define work, energy, and power. Give SI units and one example of each.

Definitions: Work (W): Work is defined as the product of force and displacement in the direction of the force. W = F · d · cos θ Where: F = magnitude of force d = magnitude of displacement θ = angle between force and displacement SI Unit: Joule (J) = N⋅m = kg⋅m²⋅s⁻² Definition: One joule is the work done when a force of 1 newton moves an object through a displacement of 1 metre in the direction of the force. Example: Lifting a 1 kg mass vertically upward by 1 m against gravity requires work...

Q2: A force of 50 N is applied on a block at an angle of 30° above the horizontal. The block moves 10 m horizontally. Calculate the work done by this force, the work done against gravity, and the net work if the block has mass 20 kg and g = 10 m/s².

Given: Force F = 50 N Angle with horizontal θ = 30° Displacement s = 10 m (horizontal) Mass m = 20 kg g = 10 m/s² Part 1: Work done by the applied force The work done depends on the component of force in the direction of displacement. Force component along displacement: F_parallel = F cos θ = 50 cos 30° = 50 × (√3/2) = 25√3 N ≈ 43.3 N Work done by applied force: W_applied = F_parallel × s = 25√3 × 10 = 250√3 J ≈ 433 J Alternative using dot product: W = F⃗ · d⃗ = F × d × cos θ = 50 × 10 × co...

Q3: A ball of mass 0.5 kg is thrown vertically upward with an initial velocity of 20 m/s. Using energy conservation, find the maximum height reached. Also calculate the velocity when the ball returns to the point from which it was thrown. (g = 10 m/s²)

Given: Mass m = 0.5 kg Initial velocity u = 20 m/s (upward) g = 10 m/s² Part 1: Maximum height using energy conservation At the point of throwing (initial position): Height h₁ = 0 (reference level) Velocity v₁ = 20 m/s KE₁ = (1/2)mv₁² = (1/2)(0.5)(20)² = 0.25 × 400 = 100 J PE₁ = 0 (at reference level) Total energy E₁ = KE₁ + PE₁ = 100 + 0 = 100 J At maximum height: Height h₂ = H (maximum height) Velocity v₂ = 0 (momentarily at rest) KE₂ = 0 PE₂ = mgh = 0.5 × 10 × H = 5H J Total energy E₂ = 0 ...

Q4: A car of mass 1500 kg accelerates from rest to 30 m/s in 10 seconds on a horizontal road. Calculate (a) the kinetic energy gained (b) the average power delivered by the engine (c) the work done by all forces.

Given: Mass m = 1500 kg Initial velocity u = 0 m/s (from rest) Final velocity v = 30 m/s Time t = 10 s Part (a): Kinetic energy gained Initial kinetic energy: KE_initial = (1/2)mu² = (1/2)(1500)(0)² = 0 J Final kinetic energy: KE_final = (1/2)mv² = (1/2)(1500)(30)² = 750 × 900 = 675,000 J = 6.75 × 10⁵ J Kinetic energy gained: ΔKE = KE_final - KE_initial = 675,000 - 0 = 675,000 J Alternative: ΔKE = (1/2)m(v² - u²) = (1/2)(1500)(30² - 0²) = 750 × 900 = 675,000 J Part (b): Average power deliv...

Q5: A pump lifts 100 litres of water per minute to a height of 20 m. If the density of water is 1000 kg/m³, calculate the power of the pump. (g = 10 m/s²)

Given: Volume of water pumped = 100 litres per minute Height h = 20 m Density of water ρ = 1000 kg/m³ g = 10 m/s² Step 1: Convert volume to SI units and find mass Volume = 100 litres = 100 × 10⁻³ m³ = 0.1 m³ per minute Mass of water: m = ρ × V = 1000 × 0.1 = 100 kg per minute Step 2: Convert time to seconds Time t = 1 minute = 60 seconds Step 3: Calculate work done against gravity Work done to lift the water: W = mgh = 100 × 10 × 20 = 20,000 J per minute Step 4: Calculate power Power = ...

Q6: A block of mass 5 kg sliding on a horizontal surface with initial velocity 10 m/s comes to rest after moving 25 m. Using energy methods, find the coefficient of kinetic friction between the block and the surface. (g = 10 m/s²)

Given: Mass m = 5 kg Initial velocity u = 10 m/s Final velocity v = 0 m/s (comes to rest) Distance s = 25 m g = 10 m/s² Find: Coefficient of kinetic friction μ_k Method 1: Using Work-Energy Theorem Step 1: Calculate change in kinetic energy Initial KE: KE_i = (1/2)mu² = (1/2)(5)(10)² = 2.5 × 100 = 250 J Final KE: KE_f = (1/2)mv² = 0 J Change in KE: ΔKE = KE_f - KE_i = 0 - 250 = -250 J Step 2: Apply work-energy theorem The only force doing work is friction (gravity and normal force are pe...

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