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.
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…
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NCERT Solutions for Class 11 Physics — Work Energy and Power. Step-by-step answers to all 6 textbook questions from this chapter, written for CBSE board preparation and free to use.
Work Energy and Power — All 6 Questions Solved
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 W = mgh = 1 × 10 × 1 = 10 J
Energy (E):
Energy is the capacity to do work. It is the ability of a system to perform work.
Types of mechanical energy:
1. Kinetic energy: Energy due to motion
2. Potential energy: Energy due to position or configuration
SI Unit: Joule (J)
Example: A ball thrown upward with velocity 10 m/s has kinetic energy KE = (1/2)mv² = (1/2)(1)(10)² = 50 J
Power (P):
Power is the rate of doing work. It is the amount of work done per unit time.
P = W/t = dW/dt
Alternatively: P = F · v (when force is in direction of velocity)
SI Unit: Watt (W) = J/s = kg⋅m²⋅s⁻³
Definition: One watt is the power when work of 1 joule is done in 1 second.
Example: If a person does 1000 J of work in 10 seconds, the power is P = 1000/10 = 100 W
Relationships:
- Work done by a force causes change in energy
- Power is the rate at which work is done or energy is transferred
- W = Pt (total work = power × time, when power is constant)
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 × cos 30° = 500 × (√3/2) = 250√3 J
Part 2: Work done against gravity
First, find the vertical displacement (height gained):
F_vertical = F sin θ = 50 sin 30° = 50 × (1/2) = 25 N
Vertical displacement (assuming the block stays on the surface):
If the force is sufficient to lift the block, vertical displacement h can be found.
But the question states "block moves 10 m horizontally", implying it stays on the surface.
If the block stays on the horizontal surface:
Vertical displacement h = 0
Work done against gravity:
W_gravity = mgh = 20 × 10 × 0 = 0 J
Note: The vertical component of the applied force (25 N) acts upward, and the normal force from the ground adjusts accordingly. The net vertical displacement is zero.
Part 3: Net work
Assuming the block moves on a horizontal frictionless surface with no vertical motion:
Forces perpendicular to displacement:
- Weight W = mg = 20 × 10 = 200 N (downward)
- Vertical component of applied force = 25 N (upward)
- Normal force N = W - 25 = 200 - 25 = 175 N (upward)
No work is done by these perpendicular forces (θ = 90°).
Net work = Work by applied force + Work against friction + Work against gravity
If frictionless surface and no vertical displacement:
Net work = W_applied = 250√3 J ≈ 433 J
Alternative interpretation (if there is friction):
Without friction coefficient given, we assume frictionless surface.
Answers:
Work done by applied force = 250√3 J ≈ 433 J
Work done against gravity = 0 J (on horizontal surface with no vertical displacement)
Net work = 250√3 J ≈ 433 J (if frictionless)
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 + 5H = 5H J
By conservation of energy:
E₁ = E₂
100 = 5H
H = 20 m
Part 2: Velocity when ball returns to initial position
At the return point (same height as initial position):
Height h₃ = 0 (same as initial)
Velocity v₃ = ? (to find)
KE₃ = (1/2)mv₃²
PE₃ = 0
Total energy E₃ = (1/2)mv₃²
By conservation of energy:
E₁ = E₃
100 = (1/2)(0.5)v₃²
100 = 0.25v₃²
v₃² = 400
v₃ = 20 m/s
The velocity is 20 m/s in the downward direction (negative direction if upward is positive).
Alternative using kinematic approach:
At maximum height: v = 0, h = H
Using v² = u² - 2gH
0 = 20² - 2(10)H
20H = 400
H = 20 m ✓
For return velocity:
Using v² = 2gh (falling from rest at height H)
v² = 2(10)(20) = 400
v = 20 m/s (downward) ✓
Part 3: Energy analysis at different heights
At any height h:
KE + PE = constant = 100 J
At h = 10 m:
PE = 0.5 × 10 × 10 = 50 J
KE = 100 - 50 = 50 J
v = √(2 × KE/m) = √(2 × 50/0.5) = √200 = 10√2 m/s ≈ 14.14 m/s
Answers:
Maximum height = 20 m
Velocity when returning to initial position = 20 m/s (downward, same magnitude as initial but opposite direction)
This confirms the principle that velocity at a given height is the same whether the ball is going up or coming down (ignoring air resistance)
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 delivered by the engine
The work done by the engine equals the kinetic energy gained (assuming no friction or on a horizontal surface):
W = ΔKE = 675,000 J
Average power:
P_avg = W/t = 675,000/10 = 67,500 W = 67.5 kW
Alternative derivation:
Acceleration: a = (v - u)/t = (30 - 0)/10 = 3 m/s²
Average velocity: v_avg = (u + v)/2 = (0 + 30)/2 = 15 m/s
Distance: s = v_avg × t = 15 × 10 = 150 m
Or using: s = ut + (1/2)at² = 0 + (1/2)(3)(10)² = 150 m ✓
Force: F = ma = 1500 × 3 = 4500 N
Work: W = F × s = 4500 × 150 = 675,000 J ✓
Power: P = W/t = 675,000/10 = 67,500 W ✓
Part (c): Work done by all forces
By work-energy theorem:
Net work = Change in kinetic energy
W_net = ΔKE = 675,000 J
This is the net work done by all forces (driving force minus friction, if any).
If the road is horizontal and frictionless:
W_driving = 675,000 J
If there is friction force f:
W_net = W_driving - W_friction = 675,000 J
Answers:
(a) Kinetic energy gained = 675,000 J = 6.75 × 10⁵ J
(b) Average power delivered = 67,500 W = 67.5 kW
(c) Work done by all forces (net work) = 675,000 J = 6.75 × 10⁵ J
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 = Work/Time
P = W/t = 20,000/60 = 1000/3 = 333.33 W ≈ 333 W or 1/3 kW
Alternative approach using rate:
Mass lifted per second = 100/60 = 5/3 kg/s
Power = (dm/dt) × g × h = (5/3) × 10 × 20 = (5/3) × 200 = 1000/3 W
Step 5: Alternative units
In kilowatts: P = 333.33/1000 ≈ 0.33 kW or 1/3 kW
In horsepower: P = 333.33/746 ≈ 0.45 hp (1 hp = 746 W)
Answer: The power of the pump is 333.33 W or approximately 333 W or 1/3 kW or 0.33 kW
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 perpendicular to motion).
Work done by friction: W_f = ΔKE = -250 J
The negative sign indicates friction opposes motion.
Step 3: Express friction force
Friction force: f = μ_k × N = μ_k × mg
Work done by friction:
W_f = -f × s = -μ_k × mg × s
-250 = -μ_k × 5 × 10 × 25
-250 = -μ_k × 1250
μ_k = 250/1250 = 0.2
Method 2: Using kinematic equation
Using v² = u² - 2as (deceleration due to friction)
0² = 10² - 2 × a × 25
0 = 100 - 50a
a = 2 m/s²
Friction provides this deceleration:
f = ma = 5 × 2 = 10 N
Normal force:
N = mg = 5 × 10 = 50 N
Coefficient of friction:
μ_k = f/N = 10/50 = 0.2
Method 3: Using energy directly
Initial kinetic energy = Work done against friction
(1/2)mu² = μ_k × mg × s
(1/2) × 10² = μ_k × 10 × 25
50 = 250μ_k
μ_k = 50/250 = 0.2
Verification:
Friction force: f = 0.2 × 50 = 10 N
Distance to stop: s = u²/(2a) = 100/(2 × 2) = 25 m ✓
Energy dissipated: f × s = 10 × 25 = 250 J = Initial KE ✓
Answer: The coefficient of kinetic friction is μ_k = 0.2
Work Energy and Power — Practice MCQs with Answers
Attempt these 20 multiple-choice questions after working through the solutions above. Each carries the correct option and the reasoning behind it, so a wrong answer tells you which idea to revisit.
MCQ 1. When is the work done by a force on an object considered zero?
- A. When the force is parallel to the displacement
- B. When the force is in the same direction as displacement
- C. When the force is perpendicular to the displacement
- D. When the object moves in a circular path
Answer: C. When the force is perpendicular to the displacement — Work done is calculated as W = F.d cosθ. If the force is perpendicular to the displacement, the angle θ is 90°, and cos 90° = 0, resulting in zero work done.
MCQ 2. What is the SI unit of power?
- A. Joule
- B. Watt
- C. Newton
- D. Kilogram-meter/second
Answer: B. Watt — The SI unit of power, which is the rate at which work is done or energy is transferred, is the Watt (W).
MCQ 3. A body of mass 'm' is moving with velocity 'v'. If its velocity is doubled, how does its kinetic energy change?
- A. It doubles
- B. It becomes half
- C. It quadruples
- D. It remains the same
Answer: C. It quadruples — Kinetic energy (KE) is given by the formula KE = 1/2 mv². If velocity 'v' is doubled to '2v', the new KE will be 1/2 m(2v)² = 1/2 m(4v²) = 4 * (1/2 mv²), so it quadruples.
MCQ 4. What type of energy is stored in a stretched or compressed spring?
- A. Kinetic energy
- B. Gravitational potential energy
- C. Elastic potential energy
- D. Thermal energy
Answer: C. Elastic potential energy — A stretched or compressed spring stores energy due to its deformation, which is known as elastic potential energy. This energy can be converted into kinetic energy when the spring is released.
MCQ 5. A body is projected vertically upwards. What is its kinetic energy at the peak of its trajectory?
- A. Maximum
- B. Half of its initial KE
- C. Zero
- D. Equal to its potential energy
Answer: C. Zero — At the peak of its trajectory, the body's vertical velocity momentarily becomes zero before it starts falling down. Since kinetic energy depends on velocity (KE = 1/2 mv²), its kinetic energy at this point is zero.
MCQ 6. Calculate the work done by a variable force F(x) = (2x + 3) N on an object as it moves from x = 0 m to x = 2 m.
- A. 6 J
- B. 8 J
- C. 10 J
- D. 12 J
Answer: C. 10 J — Work done by a variable force is calculated by integrating the force over the displacement: W = ∫ F(x) dx = ∫(2x + 3) dx from 0 to 2. This evaluates to [x² + 3x] from 0 to 2 = (2² + 3*2) - (0² + 3*0) = (4 + 6) - 0 = 10 J.
MCQ 7. A 2 kg object falls from a height of 10 m. What is its kinetic energy just before hitting the ground? (Assume g = 10 m/s² and negligible air resistance)
- A. 100 J
- B. 150 J
- C. 200 J
- D. 250 J
Answer: C. 200 J — By the principle of conservation of mechanical energy, the initial potential energy at 10m is converted entirely into kinetic energy just before hitting the ground. PE = mgh = 2 kg * 10 m/s² * 10 m = 200 J. Therefore, KE = 200 J.
MCQ 8. Which of the following is NOT a unit of energy?
- A. Joule
- B. Erg
- C. Electron Volt (eV)
- D. Watt
Answer: D. Watt — Joule, erg, and electron volt are all units of energy. Watt is the SI unit of power, which is energy per unit time, not energy itself.
MCQ 9. A constant force of 5 N acts on a 10 kg mass for 4 seconds, starting from rest. What is the average power delivered by the force during this time?
- A. 2.5 W
- B. 5 W
- C. 10 W
- D. 20 W
Answer: B. 5 W — First, find acceleration: a = F/m = 5 N / 10 kg = 0.5 m/s². Displacement: s = ut + 1/2 at² = 0 + 1/2 * 0.5 * (4s)² = 4 m. Work done: W = F * s = 5 N * 4 m = 20 J. Average power: P_avg = W/t = 20 J / 4 s = 5 W.
MCQ 10. A block of mass 'm' is pulled up an inclined plane (angle θ) by a force P parallel to the plane. The coefficient of kinetic friction is μ. If the block moves with constant velocity 'v', what is the power delivered by the force P?
- A. P v
- B. v (mg sinθ)
- C. v (mg cosθ)
- D. v (mg sinθ + μmg cosθ)
Answer: D. v (mg sinθ + μmg cosθ) — For constant velocity, the pulling force P must balance the gravitational component along the incline (mg sinθ) and the frictional force (μN = μmg cosθ). So, P = mg sinθ + μmg cosθ. Power delivered is P_force * v = (mg sinθ + μmg cosθ)v.
MCQ 11. The work done by the centripetal force on an object moving uniformly in a circular path is always:
- A. Positive
- B. Negative
- C. Zero
- D. Depends on the radius of the path
Answer: C. Zero — The centripetal force acts towards the center of the circle, perpendicular to the object's instantaneous displacement (which is tangential). Since work W = F.d cosθ and θ = 90°, the work done is always zero.
MCQ 12. If the momentum of a body increases by 50%, its kinetic energy increases by:
- A. 50%
- B. 100%
- C. 125%
- D. 225%
Answer: C. 125% — Kinetic energy (KE) and momentum (p) are related by KE = p²/(2m). If p' = 1.5p, then KE' = (1.5p)²/(2m) = 2.25 * (p²/(2m)) = 2.25 KE. The increase in KE is KE' - KE = 2.25 KE - KE = 1.25 KE, which is a 125% increase.
MCQ 13. A spring of spring constant 'k' is stretched by 'x'. The work done is W. If it is stretched further by 'x' (so total stretch is 2x), the additional work done is:
- A. W
- B. 2W
- C. 3W
- D. 4W
Answer: C. 3W — Work done to stretch a spring by 'x' is W₁ = 1/2 kx². Work done to stretch it by '2x' is W₂ = 1/2 k(2x)² = 1/2 k(4x²) = 4 * (1/2 kx²) = 4W₁. The additional work done is W₂ - W₁ = 4W₁ - W₁ = 3W₁.
MCQ 14. A light body and a heavy body have the same kinetic energy. Which one has greater momentum?
- A. Light body
- B. Heavy body
- C. Both have equal momentum
- D. Cannot be determined without mass values
Answer: B. Heavy body — The relationship between kinetic energy (KE) and momentum (p) is KE = p²/(2m), so p = √(2mKE). If KE is the same for both, momentum is directly proportional to the square root of mass (p ∝ √m). Thus, the heavier body will have greater momentum.
MCQ 15. The work-energy theorem states that the net work done on an object equals the change in its:
- A. Potential energy
- B. Total mechanical energy
- C. Kinetic energy
- D. Momentum
Answer: C. Kinetic energy — The Work-Energy Theorem is a fundamental principle in physics that states the net work done on an object is equal to the change in its kinetic energy (W_net = ΔKE).
MCQ 16. A body is dropped from a height H. What is the ratio of its kinetic energy to potential energy when it is at height H/2?
- A. 1:2
- B. 2:1
- C. 1:1
- D. 3:1
Answer: C. 1:1 — At height H/2, the potential energy (PE) is mg(H/2). The total mechanical energy (E) remains constant at mgH. So, KE = E - PE = mgH - mg(H/2) = mg(H/2). Therefore, the ratio KE:PE is (mgH/2) : (mgH/2), which is 1:1.
MCQ 17. A pump motor is used to deliver water. If the velocity with which the water leaves the pipe is 'v', what is the dependence of the instantaneous power of the motor (to impart kinetic energy) on 'v'?
- A. v
- B. v²
- C. v³
- D. v⁴
Answer: C. v³ — The mass flow rate (mass per unit time) is dm/dt = ρAv, where A is the cross-sectional area of the pipe and ρ is the density. The power required to impart kinetic energy is P = d(KE)/dt = 1/2 (dm/dt) v². Substituting dm/dt, P = 1/2 (ρAv) v² = 1/2 ρAv³, so power is proportional to v³.
MCQ 18. In an elastic collision, which of the following quantities are conserved?
- A. Only momentum
- B. Only kinetic energy
- C. Both kinetic energy and momentum
- D. Neither kinetic energy nor momentum
Answer: C. Both kinetic energy and momentum — In an elastic collision, both the total momentum and the total kinetic energy of the system are conserved. Internal forces during the collision are conservative.
MCQ 19. A 10 kg object is raised by 2 m. The work done by the gravitational force is: (g = 10 m/s²)
- A. 200 J
- B. -200 J
- C. 100 J
- D. -100 J
Answer: B. -200 J — Work done by gravity is W_g = -mgh when an object is lifted against gravity. Here, W_g = -(10 kg * 10 m/s² * 2 m) = -200 J. The negative sign indicates that gravity does work in the opposite direction to the displacement.
MCQ 20. A body is moving such that its kinetic energy is given by KE = At², where A is a constant and t is time. What is the instantaneous power delivered to the body?
- A. A
- B. At
- C. 2At
- D. 2A
Answer: C. 2At — Instantaneous power is the rate of change of kinetic energy with respect to time, P = d(KE)/dt. Given KE = At², P = d(At²)/dt = 2At.
How to Use These Solutions
Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.
Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.