Work and Energy – Case-based Questions with Answers
CBSE Class 9
Physics — Chapter 11: Work and Energy
20 Case-Based Questions & Answers — NCERT-aligned for CBSE Class 9 board exam practice
How to use these case-based questions
Each case presents a short scenario followed by 1–3 focussed questions and model answers. These are designed to test application of concepts — work, energy (kinetic, potential, elastic), conservation, work–energy theorem and power — in CBSE-style problems.
Case 1 — Lifting a suitcase
Priya lifts a 12 kg suitcase vertically from the floor to a height of 1.5 m at constant speed.
(a) Calculate the work done against gravity. (b) If the lift takes 4 s, what is the average power?
Answer: (a)
W = m g h = 12 × 9.8 × 1.5 = 176.4 J. (b) Average power P = W/t = 176.4 / 4 = 44.1 W.Case 2 — Towing a car
A tow truck applies a horizontal pull of 2000 N on a car causing it to move 8 m along a level road. The frictional resistive force is 500 N opposite to motion.
(a) Find the work done by the tow truck. (b) Work done by friction. (c) Net work and its meaning.
Answer: (a)
W_{truck} = 2000 × 8 = 16,000 J. (b) W_{fric} = −500 × 8 = −4,000 J. (c) W_{net} = 16,000 − 4,000 = 12,000 J which equals the increase in kinetic energy of the car (by work–energy theorem).Case 3 — Block on incline with friction
A 3 kg block is pulled up a 4 m long incline of angle 30° by a force parallel to the incline of 30 N. The coefficient of kinetic friction is 0.15.
Compute: (a) vertical rise, (b) work done by the applied force, (c) work done against friction, (d) net work (assume g = 9.8 m/s2).
Answer: (a) Vertical rise
h = 4 sin30° = 2 m. (b) W_{app} = 30 × 4 = 120 J. (c) Normal force N = mg cos30° = 3×9.8×0.866 ≈ 25.47 N; friction f_k = μN ≈ 0.15×25.47 ≈ 3.82 N; work by friction W_{fric} = −3.82 × 4 ≈ −15.28 J. (d) Change in potential ΔU = mgh = 3×9.8×2 = 58.8 J. Net work on block equals W_{net} = W_{app} + W_{fric} − ΔU = 120 − 15.28 − 58.8 ≈ 45.92 J, which is the net increase in kinetic energy.Case 4 — Spring launcher
A toy car mass 0.5 kg is held against a compressed spring (k = 200 N/m) by x = 0.1 m and released on a frictionless surface.
(a) Find the speed of the car as it leaves the spring. (b) State energy conversions.
Answer: Elastic PE
U = ½ k x² = 0.5×200×0.1² = 1 J. At release, KE = 1 J ⇒ ½ m v² = 1 ⇒ v = √(2 / 0.5) = 2 m/s. Energy conversion: elastic potential → kinetic.Case 5 — Pendulum at amusement park
A pendulum swing at rest is pulled aside so that bob rises by 0.8 m and released.
Find the speed of bob at lowest point (g = 9.8 m/s2) and explain energy change.
Answer:
½ m v² = m g h ⇒ v = √(2 g h) = √(2×9.8×0.8) ≈ √(15.68) ≈ 3.96 m/s. Energy: potential at top → kinetic at lowest point.Case 6 — Car braking to stop
A car of mass 1200 kg traveling at 20 m/s applies brakes and stops over distance 50 m. Assume constant braking force.
(a) Find change in kinetic energy. (b) Find braking force magnitude. (c) Where does the energy go?
Answer: (a) Initial KE = ½ m v² = 0.5×1200×20² = 240,000 J; final KE = 0 ⇒ ΔKE = −240,000 J. (b) Work by braking force = −240,000 J over 50 m ⇒ force ≈ 240,000 / 50 = 4,800 N opposite motion. (c) Energy converts to thermal energy in brakes and road, and some to sound.
Case 7 — Lifting with pulley and rope
A student uses a frictionless pulley to lift a 6 kg mass by pulling rope with constant force 70 N through distance 0.9 m.
Calculate the work done by the student and compare to increase in potential energy.
Answer: Work by student = 70 × 0.9 = 63 J. Vertical rise of mass = 0.9 m (frictionless ideal); ΔU = m g h = 6×9.8×0.9 = 52.92 J. The extra 10.08 J (63 − 52.92) goes into increasing kinetic energy or is due to assumptions — if mass lifted at constant speed, student’s force should equal weight (≈58.8 N), so given 70 N would accelerate mass; net work equals ΔKE + ΔU.
Case 8 — Rolling down hill with friction
A bicycle and rider of combined mass 75 kg roll down a small hill of height 10 m. Friction and air losses together dissipate 20% of the initial potential energy.
Compute: (a) initial potential energy, (b) mechanical energy available at bottom.
Answer: (a)
m g h = 75×9.8×10 = 7,350 J. (b) Loss = 20% ⇒ lost = 0.2×7,350 = 1,470 J; available = 7,350 − 1,470 = 5,880 J (sum of KE and residual PE).Case 9 — Satellite in low orbit (qualitative)
A satellite in circular orbit maintains constant speed and altitude (neglect atmospheric drag).
Explain whether gravitational force does work and whether mechanical energy is conserved.
Answer: Gravity acts perpendicular to instantaneous displacement over an infinitesimal time in circular motion (central force), so work by gravity over any small displacement is zero; hence gravity does no net work and kinetic energy stays constant. Mechanical energy (kinetic+potential) remains constant if no external non-conservative forces act.
Case 10 — Elevator with constant speed
An elevator of mass 800 kg is lifted upward at constant speed by a motor. It rises 12 m in 20 s. Neglect friction.
Find the work done by the motor and the power output.
Answer: Work = m g h = 800×9.8×12 = 94,080 J. Power = 94,080 / 20 = 4,704 W ≈ 4.70 kW.
Case 11 — Compressed gas cylinder heating (conceptual)
A technician squeezes a metal can, doing mechanical work on it rapidly which increases internal energy and temperature.
Explain energy transformation and why mechanical work increases thermal energy.
Answer: Mechanical work done on the can transfers energy to microscopic particles, increasing their internal (thermal) energy which raises temperature. Macroscopic ordered work becomes microscopic disordered energy (heat) due to inelastic collisions and friction at atomic scale.
Case 12 — Pushing a box in two ways
A boy pushes a box across the floor by applying force horizontally in one attempt and at 30° downward in another attempt; displacement is same in both cases.
Compare work done and frictional forces in both cases.
Answer: Horizontal push: work = F s (full horizontal component contributes). Downward-angled push: horizontal component is F cos30°, but the downward component increases normal reaction and hence friction; frictional force increases → more energy lost to heat. Net effect: though applied force may be same magnitude, effective work contributing to motion is lower for angled push due to larger frictional loss (and normal force change).
Case 13 — Spring and mass vertical oscillation
A 0.4 kg mass attached to vertical spring (k = 80 N/m) is pulled down and released, oscillating without damping.
If amplitude is 0.15 m, find total mechanical energy and maximum speed.
Answer: Total energy
E = ½ k A² = 0.5×80×0.15² = 0.9 J. At equilibrium KE max: ½ m v_{max}² = 0.9 ⇒ v_{max} = √(1.8 / 0.4) = √4.5 ≈ 2.12 m/s.Case 14 — Person carrying a tray
A waiter carries a tray of food horizontally at constant height across the restaurant.
Is work done by the waiter on the tray? Explain.
Answer: If the tray moves horizontally at constant height, the lifting force (vertical) is perpendicular to displacement (horizontal). So mechanical work done by vertical force against gravity is zero (no vertical displacement). However, the waiter expends metabolic energy to maintain posture — but it is not mechanical work on tray in physics terms.
Case 15 — Hydro turbine (energy chain)
Water drops through height to spin turbine blades generating electricity.
List the energy transformations and identify where work is done.
Answer: Transformations: gravitational potential energy of water → kinetic energy of flowing water → mechanical work on turbine (rotational KE) → electrical energy in generator. Work is done by water on turbine blades (fluid does pressure/force over displacement of blades).
Case 16 — Work done up a staircase vs elevator
A person of mass 65 kg climbs a 4 m high staircase or uses an elevator to the same height. Neglect losses in elevator.
Compare the mechanical work done in both cases and comment on energy source differences.
Answer: Mechanical work against gravity is same:
W = m g h = 65×9.8×4 ≈ 2,548 J. Climbing uses person's chemical energy; elevator uses electrical energy and possibly additional losses. Net mechanical energy change of person+earth system equal in both cases but sources differ.Case 17 — Power rating of motor
A 2 kW motor lifts goods continuously at rate of 0.5 m3 water equivalent mass 500 kg per hour to height 10 m. Is motor adequate?
Compute power required and compare with motor rating (use g = 9.8).
Answer: Mass per hour = 500 kg; work per hour = m g h = 500×9.8×10 = 49,000 J per hour. Convert to watts: 1 hour = 3600 s ⇒ power = 49,000 / 3600 ≈ 13.61 W. This is tiny; perhaps intended mass per hour bigger. If instead 500 kg per minute, power would be 49,000 / 60 ≈ 816.7 W. For 2 kW motor, it can deliver 2000 W so adequate for lifting 500 kg per minute; check units carefully. (Exam tip: always check time units.)
Case 18 — Energy in collisions (inelastic)
Two carts collide and stick together; kinetic energy is not conserved but momentum is.
Explain what happens to the 'lost' kinetic energy.
Answer: Lost kinetic energy is converted into internal energy: heat, deformation energy of carts, sound. Total energy conserved but mechanical kinetic energy is transformed into other forms.
Case 19 — Running up stairs vs running on flat
Two athletes run same horizontal distance; one runs up a flight of stairs gaining height, the other runs on flat ground.
Compare the work done against gravity and energy expenditure.
Answer: Work against gravity depends only on vertical gain. Athlete climbing stairs does additional work
m g h. Running on flat does not change gravitational PE. Total metabolic energy expended differs due to differing mechanical work and efficiency; stairs runner expends more energy overall.Case 20 — Energy audit of a small pump
A pump rated 500 W raises 100 kg of water per minute through 5 m. Observed electrical consumption is 600 W (including losses).
Compute: (a) useful hydraulic power, (b) pump efficiency.
Answer: Mass per second = 100 kg / 60 s ≈ 1.6667 kg/s. Useful power = m g h / s = 1.6667×9.8×5 ≈ 81.67 W. Efficiency = useful/electrical = 81.67 / 600 ≈ 0.136 ≈ 13.6% (very low — indicates large losses).
Note: Use g = 9.8 m·s−2 unless otherwise specified. All numerical answers are rounded sensibly for clarity. These cases adhere to NCERT topics: work, energy, power, conservation, and practical applications.