Work and Energy – Long Answer Type Questions
CBSE Class 9
Physics — Chapter 11: Work and Energy
30 Long Answer Type Questions & Answers — NCERT-aligned for CBSE Class 9 board exam preparation
CBSE Board Examinations — How to use this set
This collection covers conceptual explanations, derivations, worked examples and real-life applications from NCERT Class 9 Chapter 11: Work and Energy. Use for practice, assignments or classroom handouts.
Q1
Explain the concept of work in physics and derive the expression for work done by a constant force acting at an angle to the displacement.
Answer:
In physics, work is done when a force acting on an object causes a displacement of its point of application. For a constant force
where
𝐅 acting on an object that undergoes displacement 𝐬, the work done is the scalar (dot) product:
W = 𝐅 · 𝐬 = F s cos θ
θ is the angle between the force and the displacement. Only the component of the force along the displacement contributes to work. Special cases: θ = 0° ⇒ W = F s; θ = 90° ⇒ W = 0; θ = 180° ⇒ W = −F s (negative work).
Q2
Discuss the different cases when work done can be positive, negative or zero and give two practical examples for each case.
Answer:
- Positive work (force component along displacement, cos θ > 0): e.g., pushing a trolley forward; lifting a bag upward while it rises.
- Negative work (force component opposite displacement, cos θ < 0): e.g., friction opposing sliding motion; a parachute slowing a descending jumper.
- Zero work (force ⟂ displacement, cos θ = 0 or no displacement): e.g., centripetal force on an object in uniform circular motion; holding a box stationary.
Q3
Define kinetic energy. Derive the expression for kinetic energy of a body starting from the work–energy theorem for a constant net force.
Answer:
Kinetic energy (KE) is the energy possessed by a body due to its motion. Start with Newton's second law for constant net force
F = m a. Work done by net force when speed changes from u to v while moving through displacement s:
W = F s = m a sUsing kinematics:
v² = u² + 2 a s ⇒ a s = (v² − u²)/2, so
W = m · (v² − u²)/2 = ½ m v² − ½ m u²Therefore net work equals change in the quantity
½ m v², identifying KE = ½ m v². If initial speed is zero, KE = ½ m v².
Q4
What is potential energy? Explain gravitational potential energy near Earth's surface and derive its formula.
Answer:
Potential energy (PE) is stored energy due to position or configuration. Gravitational potential energy near Earth's surface: lifting a mass
m slowly by height h requires work against gravity equal to m g h, which is stored as potential energy. Thus,
U = m g hwhere
g ≈ 9.8 m·s−2. The value depends on the reference level where U = 0 is chosen.
Q5
Explain the work–energy theorem and illustrate how it applies to a car accelerating on a straight road, including role of friction and engine's energy input.
Answer:
The work–energy theorem states
Wnet = ΔKE. For a car accelerating, the engine supplies propulsive force doing positive work on the car (via wheels), increasing its kinetic energy. Resistive forces (rolling friction, air drag) do negative work, converting mechanical energy into heat. The net work (engine work minus resistive work) equals the increase in car's kinetic energy. Energy input comes from fuel (chemical energy) converted by the engine into useful mechanical work and heat; only part becomes kinetic energy due to inefficiencies.
Q6
Derive the expression for elastic potential energy stored in a spring and describe an experiment to determine the spring constant.
Answer:
For Hooke's law spring, restoring force is
F = −k x. Work done to stretch from 0 to x by external agent:
W = ∫₀ˣ k x' dx' = ½ k x²So elastic potential energy
U = ½ k x². Experiment: hang masses from a spring, measure extension x for different masses (F = m g) and plot F vs x; slope = k. Or find period T = 2π √(m/k) and compute k from measured T.
Q7
Explain conservation of mechanical energy. Use a frictionless roller-coaster hill as example to show conversion between potential and kinetic energy.
Answer:
In absence of non-conservative forces, total mechanical energy
E = KE + PE is conserved. For a frictionless roller-coaster: starting at height h₁ with KE ≈ 0, total energy = m g h₁. At lower height h₂, potential decreases and kinetic increases so that
m g h₁ = ½ m v² + m g h₂ ⇒ ½ m v² = m g (h₁ − h₂)Thus potential ↔ kinetic exchanges while
E remains constant.
Q8
A block of mass 2 kg is pulled along a rough horizontal surface by a force of 15 N at 30° above the horizontal. The block moves 4 m. The coefficient of kinetic friction is 0.2. Calculate the work done by the applied force, frictional force, normal force and gravity; also compute the net work and change in kinetic energy.
Answer:
Horizontal component:
Vertical component:
Normal reaction:
Friction:
Work by applied force:
Work by friction:
Work by normal and gravity: each ≈ 0 J (perpendicular to displacement).
Net work:
Fx = 15 cos 30° = 15 × (√3/2) ≈ 12.99 N.Vertical component:
Fy = 15 sin 30° = 7.5 N (upward). Weight: mg = 2 × 9.8 = 19.6 N.Normal reaction:
N = mg − Fy = 19.6 − 7.5 = 12.1 N.Friction:
fk = μ N = 0.2 × 12.1 = 2.42 N (opposes motion).Work by applied force:
Wapp = Fx × s = 12.99 × 4 ≈ 51.96 J.Work by friction:
Wfric = − fk × s = −2.42 × 4 ≈ −9.68 J.Work by normal and gravity: each ≈ 0 J (perpendicular to displacement).
Net work:
Wnet = 51.96 − 9.68 ≈ 42.28 J ⇒ ΔKE ≈ 42.28 J (by work–energy theorem).
Q9
Explain why friction is called a non-conservative force and discuss its effect on energy conservation in mechanical systems.
Answer:
Friction is non-conservative because work done by friction depends on the path (distance) traversed, not solely on endpoints. It dissipates mechanical energy as thermal energy (internal energy of surfaces), so mechanical energy decreases in the system; however, total energy (mechanical + internal heat) is still conserved globally (First Law of Thermodynamics). Because friction converts ordered mechanical energy into disordered thermal energy, system's useful mechanical energy is reduced.
Q10
Describe an experiment to verify the work–energy theorem using a dynamic cart, a pulley and a hanging mass. Explain the measurements and expected results.
Answer:
Setup: low-friction cart on track attached by a string over pulley to hanging mass. Release so hanging mass falls and accelerates cart. Measure initial and final speeds of cart (light gates or ticker-tape) and displacement. Compute ΔKE = ½ m (v_f² − v_i²). Compute work by net force ≈ (tension − friction) × displacement or equate gravitational potential energy lost by hanging mass (m g Δh) minus losses. Expected: within experimental error, net work ≈ ΔKE, validating theorem; account for friction/air resistance to improve accuracy.
Q11
How is the concept of power related to work and energy? Derive the relation between instantaneous power, force and velocity and give practical examples.
Answer:
Power is the rate of doing work:
P = dW/dt. Since dW = 𝐅 · d𝐫 and d𝐫/dt = 𝐯, instantaneous power delivered by force 𝐅 is
P = 𝐅 · 𝐯If force and velocity are collinear,
P = F v. Examples: an electric motor rated in watts (power) determines how fast it can lift a load or accelerate a vehicle.
Q12
A 1000 kg car moves up a slope of height 10 m at constant speed. The engine must overcome gravity and frictional losses amounting to 2000 J/s. If the car takes 20 s to climb, calculate total work done against gravity and total energy supplied. Also find the average power output required.
Answer:
Work against gravity:
Frictional losses: 2000 J/s for 20 s ⇒
Total energy supplied:
Average power:
Wg = m g h = 1000 × 9.8 × 10 = 98,000 J.Frictional losses: 2000 J/s for 20 s ⇒
Wloss = 2000 × 20 = 40,000 J.Total energy supplied:
Win = 98,000 + 40,000 = 138,000 J.Average power:
P = Win / t = 138,000 / 20 = 6,900 W ≈ 6.9 kW.
Q13
Explain path independence of work done by gravity and derive the relation between work done by gravity and change in potential energy.
Answer:
Gravity is conservative; work between two points depends only on vertical positions. For mass
m moved from height h₁ to h₂:
Wgrav = m g (h₁ − h₂)Change in potential energy:
ΔU = m g (h₂ − h₁), so Wgrav = −ΔU. This allows defining gravitational potential energy.
Q14
A stone is thrown vertically upward with speed 20 m·s−1. Neglecting air resistance, find the maximum height reached and the work done by gravity during ascent. Take g = 9.8 m·s−2.
Answer:
Using energy conservation: initial KE = ½ m v² converts to potential at top:
m g h = ½ m v² ⇒
h = v² / (2 g) = 20² / (2 × 9.8) = 400 / 19.6 ≈ 20.41 mWork done by gravity during ascent = −change in kinetic energy = −½ m v² = −200 m J (i.e., for m = 1 kg, W = −200 J). Alternatively gravity's work = m g (h_final − h_initial) = m g h = +m g h (negative relative to motion direction), conventionally negative when opposing motion.
Q15
Discuss energy transformations and losses when a cyclist brakes to a stop from a high speed. What happens to the kinetic energy?
Answer:
When braking, kinetic energy is removed by braking force doing negative work. Most kinetic energy is converted into thermal energy (heat) in brake pads and rims/rotors; some energy is radiated as sound and tiny amounts may heat the road. Mechanical energy is transformed into internal energy of materials and surroundings; total energy conserved but less available as useful mechanical work afterwards.
Q16
Explain how energy conservation applies to a simple pendulum and derive an expression relating speed at lowest point to initial height.
Answer:
For an ideal pendulum (no damping), mechanical energy is conserved. If released from rest at height
h above lowest point, initial energy = m g h. At lowest point all converts to kinetic:
½ m v² = m g h ⇒ v = √(2 g h)So speed at bottom depends on vertical drop
h, not on path or mass.
Q17
A mass m is attached to a spring (spring constant k) and stretched from equilibrium by distance x. Show total mechanical energy of the system and explain how energy changes as it oscillates.
Answer:
Total mechanical energy for undamped mass–spring:
E = ½ m v² + ½ k x²At amplitude (x = A),
v = 0 and E = ½ k A². At equilibrium (x = 0), E = ½ m vmax². Energy oscillates between KE and elastic PE; total remains constant.
Q18
Describe with equations how a hydraulic lift (like an elevator) uses work and energy principles to raise a car from ground to a platform, considering input work, output work and losses.
Answer:
Useful output work to lift car mass
m by height h:
Wout = m g hPump must supply input work
Win where
Win = Wout + WlossLosses
Wloss arise from pump inefficiency, viscous dissipation, friction. If pump supplies power Pin for time t, then Win = Pin t and efficiency η = Wout / Win.Q19
A 0.5 kg ball is dropped from a height of 10 m. Neglect air resistance. Calculate its speed just before hitting the ground and the work done by gravity. Use g = 9.8 m·s−2.
Answer:
Initial potential energy:
m g h = 0.5 × 9.8 × 10 = 49 J. At impact KE = 49 J:
½ m v² = 49 ⇒ v² = 98 / 0.5 = 196 ⇒ v = 14 m·s⁻¹Work done by gravity = +49 J (gravity does positive work bringing ball down).
Q20
Explain how energy transformations in a hydroelectric power plant follow the work and energy principles, starting from stored water to electrical energy output.
Answer:
Water stored at height has gravitational potential energy. As it flows down, potential converts to kinetic energy, which does work on turbine blades (mechanical energy). The turbine drives a generator converting mechanical energy into electrical energy (via electromagnetic induction). Losses (friction, turbulence, heat) reduce net electrical output. Thus sequential conversions: gravitational PE → kinetic energy → mechanical work → electrical energy, in accordance with energy conservation (accounting for losses).
Q21
Derive expression for work done by a variable force F(x) acting along x from x = a to x = b. Provide a simple example where this integral is used.
Answer:
For variable force
F(x), infinitesimal work dW = F(x) dx. Total work:
W = ∫ₐᵇ F(x) dxExample: stretching spring from
x = a to x = b where F = k x gives
W = ∫ₐᵇ k x dx = ½ k (b² − a²)
Q22
A 2 kg block slides down a frictionless incline of height 5 m. Determine its speed at the bottom using energy methods. Also compute the work done by gravity and by normal force.
Answer:
Initial PE:
m g h = 2 × 9.8 × 5 = 98 J. At bottom all → KE:
½ m v² = 98 ⇒ v² = 196 / 2 = 98 ⇒ v ≈ 9.90 m·s⁻¹Work by gravity = +98 J. Work by normal force = 0 J (normal ⟂ displacement along incline).
Q23
Explain the role of reference level in defining gravitational potential energy. How does choice of reference affect numerical value but not physical predictions?
Answer:
Gravitational potential energy
U = m g h is defined relative to chosen zero level. Changing reference shifts all U by constant, possibly making values negative, but differences ΔU remain unchanged. Physical predictions (forces, motion, energy changes) depend on differences, therefore independent of reference choice.
Q24
A block of mass 3 kg is attached to a spring (k = 300 N·m−1). The block is pulled 0.2 m from equilibrium and released from rest. Find the maximum speed of the block during oscillation.
Answer:
Total energy at amplitude
A = 0.2 m:
E = ½ k A² = 0.5 × 300 × 0.2² = 6 JAt equilibrium all KE:
½ m vmax² = 6 ⇒ vmax² = 12 / 3 = 4 ⇒ vmax = 2 m·s⁻¹.
Q25
Discuss how the First Law of Thermodynamics (energy conservation) is consistent with mechanical energy losses due to friction.
Answer:
First Law: total energy conserved. Friction converts mechanical energy into internal (thermal) energy of bodies and surroundings. Mechanical energy appears 'lost' from mechanical viewpoint but is accounted as increased internal energy (heat). Thus energy conservation holds; only the form of energy changes and entropy increases.
Q26
A man lifts a 10 kg suitcase from the floor to a height of 1.5 m in 3 s. Calculate the work done against gravity and the power expended. State assumptions and ignore other losses.
Answer:
Work against gravity:
Time = 3 s ⇒ average power
Assumptions: constant speed (no ΔKE), ideal mechanical lifting (no internal metabolic inefficiencies).
W = m g h = 10 × 9.8 × 1.5 = 147 J.Time = 3 s ⇒ average power
P = W/t = 147 / 3 = 49 W.Assumptions: constant speed (no ΔKE), ideal mechanical lifting (no internal metabolic inefficiencies).
Q27
Explain why centripetal force does no work on a body moving in uniform circular motion. What happens to the kinetic energy of the body?
Answer:
Centripetal force acts radially while instantaneous displacement is tangential → angle = 90°. Hence infinitesimal work
dW = F dr cos 90° = 0. No work means KE unchanged; speed remains constant (only direction changes).
Q28
A 1500 W electric motor lifts a 200 kg lift by 8 m. Assuming 80% efficiency, calculate the time taken to raise the lift. (Use g = 9.8 m·s−2)
Answer:
Useful work:
With 80% efficiency, input energy required:
Time:
Wout = m g h = 200 × 9.8 × 8 = 15,680 J.With 80% efficiency, input energy required:
Win = Wout / 0.8 = 19,600 J.Time:
t = Win / P = 19,600 / 1500 ≈ 13.07 s (≈13.1 s).
Q29
Discuss the energy transformations and the role of work in the operation of a household electric fan.
Answer:
Electrical energy from supply → motor converts it into mechanical rotational work on blades. Blades do work on air, imparting kinetic energy to air (airflow). Some electrical energy dissipates as heat (motor windings) and sound. Work per unit time is electrical power consumed; efficiency = mechanical power delivered to air / electrical power consumed.
Q30
Summarize the key differences between work, energy and power, and provide an example that highlights all three in a single scenario.
Answer:
- Work (J) — energy transferred when force causes displacement (e.g., W = F s cos θ).
- Energy (J) — capacity to do work (kinetic, potential, thermal, etc.).
- Power (W) — rate of doing work:
P = dW/dt.
m by height h using a motor: work done = m g h; energy used = chemical/electrical energy consumed; power = how fast the lift performs the work (e.g., shorter time ⇒ higher power).