Work, Power and Energy | ISC Class 11 Physics Notes

This note covers work by constant and variable forces, kinetic and potential energy, the work-energy theorem, conservative and non-conservative forces, spring energy, conservation of energy, mass-energy equivalence, power, and elastic and inelastic collisions in one and two dimensions.

What is work, and how does the scalar product describe it?

Work is the product of the component of a force along a displacement and the magnitude of that displacement. A force is a push or pull; displacement is the directed change from an initial position to a final position.

A vector has magnitude and direction; a scalar has no direction. The scalar product, or dot product, multiplies one vector's magnitude by the component of the other along it. It therefore gives a scalar.

Let F⃗ be a constant force and s⃗ the displacement. Their magnitudes are F and s, and θ is the angle between them. An arrow identifies a vector; cos θ is the cosine of the angle. Work is denoted by W.

W = Fs cos θ

The equivalent vector expression is W = F⃗ · s⃗. The dot denotes the scalar product. This is also (F cos θ)s: only the force component along the displacement contributes. Interchanging the two vectors leaves their scalar product unchanged.

How are work and its dimensions expressed?

The SI unit of work is the joule, symbol J. SI means International System of Units. One joule is the work done by a force of one newton through one metre along the force: 1 J = 1 N m.

Here N denotes newton and m after a number denotes metre. Kilogram, kg, measures mass, and second, s, measures time. 1 N = 1 kg m s⁻². A quantity's dimensions express its dependence on base quantities.

Using M for mass dimension, L for length and T for time, force has dimensions [MLT⁻²], displacement [L], and work [ML²T⁻²]. Square brackets indicate dimensions. Context distinguishes the displacement symbol s from the unit second.

When is work positive, negative or zero?

The sign of work follows from the angle between force and displacement. Positive work transfers energy to the body's motion through that force; negative work removes energy from its motion. The change in motion depends on the work of all forces together. Angles below are measured in degrees, denoted °.

Work is also zero when force or displacement is zero. A person pushing an unmoving rigid wall does no work on the wall. Feeling tired does not establish that mechanical work has been done on the object.

For a block moving horizontally on a smooth table, gravity does no work because its force is vertical. If the Moon's orbit is assumed perfectly circular, Earth's gravitational force does no work: force is radial and instantaneous displacement is tangential.

How does friction affect a skidding cycle?

Worked example 1. A cycle skids to rest over 10 m. The road exerts a constant force of 200 N opposite its motion. Find the work done on the cycle and on the stationary road.

Formula: W = Fs cos θ.

Substitute: W = 200 × 10 × cos 180°.

Answer: Work on the cycle is −2000 J. The cycle exerts an equal opposite force on the road, but the road's displacement is zero, so work on the road is 0 J.

Note: Equal and opposite interaction forces need not do equal and opposite work. Their points of application can undergo different displacements.

How is work calculated when force varies with position?

A variable force changes as the motion proceeds. For motion along a straight coordinate axis, let x denote position, measured in metres, and F(x) the force component along that axis. Multiplying the entire displacement by an arbitrary force value generally gives an incorrect answer.

Divide the displacement into small intervals Δx, where Δ denotes a change. Within a sufficiently small interval, the force is approximately constant and the small work ΔW is approximately F(x)Δx. Adding these contributions gives an approximation to the total work.

In the limit of vanishing interval size, write dW = F(x) dx. The letter d indicates an infinitesimal change, while ∫ denotes integration, the limiting sum of these contributions. Subscripts i and f mean initial and final.

W = ∫ F(x) dx

The integration limits are xᵢ and x𝒻. More generally, dW = F⃗ · ds⃗, so it is the force component along each displacement that matters. A force-position graph gives signed area: for increasing x, areas below the horizontal axis contribute negative work.

How do rectangular and trapezoidal areas give work?

Worked example 2. A woman pushes a trunk with 100 N over the first 10 m. Over the next 10 m, her force decreases linearly to 50 N. A constant frictional force of 50 N opposes motion. Find the work of each force.

Formula: Applied work equals the area under its force-position graph; frictional work equals opposing force times displacement with a negative sign.

Substitute: Applied work = 100 × 10 + ½(100 + 50) × 10. Frictional work = −50 × 20.

Answer: The woman does 1750 J of work; friction does −1000 J.

The applied-force line is horizontal at 100 N until 10 m, then slopes down to 50 N at 20 m. The friction line lies at −50 N below the displacement axis.

See Fig. 5.4 in your NCERT textbook

What is kinetic energy, and how does it depend on speed?

Energy is the capacity to do work. Kinetic energy is energy associated with motion. For a body of mass m and speed v in classical mechanics, kinetic energy K is half the mass multiplied by the square of the speed.

Mass measures inertia, resistance to changes in velocity, and has SI unit kg and dimension [M]. Speed is the magnitude of velocity, the rate of change of position; both have unit m s⁻¹ and dimensions [LT⁻¹]. The symbol m in a formula denotes mass, rather than the metre unit.

K = ½mv²

The SI unit of kinetic energy is the joule. Its dimensions are [ML²T⁻²]. Kinetic energy is scalar and cannot be negative for a body of positive mass. Its value depends on speed in the chosen reference frame, the coordinate system relative to which motion is measured.

For a fixed mass, kinetic energy varies as the square of speed. A percentage reduction in kinetic energy is therefore different from the percentage reduction in speed. Recover speed by taking a square root, rather than applying the energy percentage directly.

How is speed found from the remaining kinetic energy?

Worked example 3. A bullet of mass 50.0 g travels at 200 m s⁻¹ through plywood 2.00 cm thick. It emerges with 10% of its initial kinetic energy. Find its emergent speed. Here g means gram and cm means centimetre.

Formula: Kᵢ = ½mvᵢ²; K𝒻 = 0.10Kᵢ; v𝒻 = √(2K𝒻/m). The symbol √ means square root.

Substitute: m = 0.0500 kg; Kᵢ = ½ × 0.0500 × 200² = 1000 J; K𝒻 = 100 J.

Answer: v𝒻 = √(200/0.0500) = 63.2 m s⁻¹. The speed decreases by approximately 68%, rather than 90%.

How does the work-energy theorem connect force with motion?

Definition: The work-energy theorem states that the change in a particle's kinetic energy equals the work done by the net force acting on it. The net force is the vector sum of all forces on the particle.

Wₙₑₜ = K𝒻 − Kᵢ

The subscript net identifies total work. The theorem applies to constant and variable forces. Forces that increase kinetic energy and forces that reduce it must all be included. Work by one selected force need not equal the full kinetic-energy change.

Derivation: Work-energy theorem for a variable force

Consider one-dimensional motion of constant mass m. Let t be time and a = dv/dt the acceleration, with unit m s⁻² and dimensions [LT⁻²]. A derivative such as dv/dt is an instantaneous rate of change. Here F is the net force component.

1. Differentiate K = ½mv² with respect to time: dK/dt = mv(dv/dt).
2. Use Newton's second law, F = ma, to obtain dK/dt = Fv.
3. Since v = dx/dt, write dK/dt = F(dx/dt), hence dK = F dx.
4. Integrate between the initial and final positions: K𝒻 − Kᵢ = ∫ F dx.

Wₙₑₜ = ΔK, where ΔK means K𝒻 − Kᵢ.

This method gives an energy change without necessarily giving the complete motion. It does not, in general, preserve explicit information about time or direction available from the vector equation of motion.

How can work be found for an unknown resistive force?

Worked example 4. A raindrop of mass 1.00 g falls from rest through 1.00 km and reaches 50.0 m s⁻¹. Take constant gravitational acceleration g = 10 m s⁻². Find gravitational and resistive work. Here km denotes kilometre.

Formula: Wg = mgh; ΔK = ½mv²; Wr = ΔK − Wg. Height fallen is h, Wg denotes gravitational work, and Wr denotes work by air resistance, a force opposing the fall.

Substitute: m = 10⁻³ kg and h = 10³ m. Wg = 10⁻³ × 10 × 10³; ΔK = ½ × 10⁻³ × 50.0².

Answer: Wg = 10.0 J; ΔK = 1.25 J; Wr = 1.25 − 10.0 = −8.75 J. The resistive force removes kinetic energy relative to the increase gravity alone would produce.

What are potential energy and conservative forces?

Potential energy, denoted U, is stored energy associated with position or configuration. Configuration means the relative arrangement of a system's parts. Its SI unit is the joule and its dimensions are [ML²T⁻²]. It is associated with conservative interactions.

A conservative force does work that depends on the initial and final positions, rather than the path between them. Equivalently, its work around a closed path is zero. A closed path returns the body to its starting position.

W = Uᵢ − U𝒻

Here W is work by the conservative force, while Uᵢ and U𝒻 are the initial and final potential energies. Positive work by that force decreases potential energy. In one dimension, F = −dU/dx: force is the negative spatial rate of change of potential energy.

When can gravitational potential energy be written as mgh?

U = mgh

Here h is height above the chosen zero level, and g is gravitational acceleration. This expression treats g as constant near Earth's surface, where height is very small compared with Earth's radius. It is not a formula for unrestricted distances from Earth.

Choose the zero level conveniently and retain it throughout the calculation. Raising a body without changing its kinetic energy requires work against gravity, stored as increased potential energy. During free fall without air resistance, this stored energy becomes kinetic energy.

A non-conservative force, such as sliding friction, does not satisfy the conservative-force condition. Energy transferred through friction cannot generally be recovered simply by reversing the path.

How is energy stored in an ideal spring?

An ideal spring exerts a restoring force proportional to its displacement from equilibrium. Equilibrium here is the unstretched position where its force vanishes. A restoring force acts towards that position. The spring is treated as light enough for its mass to be neglected.

Fₛ = −kx

This is Hooke's law. Here Fₛ is the spring-force component, x is extension or compression measured from equilibrium, and k is the spring constant, the force magnitude per unit displacement. The negative sign shows that force opposes displacement from equilibrium.

The SI unit of the spring constant is N m⁻¹, and its dimensions are [MT⁻²]. A larger k means a stiffer spring. Use the relation provided the spring obeys Hooke's law; it describes the ideal spring assumed here.

Derivation: Potential energy of a spring

Choose U = 0 at x = 0. Stretch the spring slowly enough that the attached body's change in kinetic energy is negligible.

1. The spring force is −kx. The balancing external pulling force is +kx.
2. The small external work in extending it by dx is dW = kx dx.
3. Integrate from zero extension to x: W = ∫ kx dx = ½kx².
4. This work is stored as elastic potential energy, so U = ½kx².

U = ½kx²

Compression also stores positive energy relative to equilibrium because x is squared. Work done by the spring itself from equilibrium to x is −½kx². Between arbitrary endpoints its work is ½kxᵢ² − ½kx𝒻², confirming that it depends only on endpoints.

A spring joins a wall to a block on a horizontal surface. Separate drawings show zero, positive and negative displacement. A descending straight force-displacement line passes through the origin; its shaded triangle on the positive-displacement side lies below the axis.

See Fig. 5.7 in your NCERT textbook

How is mechanical energy conserved in falling and oscillating systems?

Mechanical energy, E, is the sum of kinetic and potential energies. Its SI unit is joule and its dimensions are [ML²T⁻²]. The law of conservation of mechanical energy states that this total remains constant provided the forces doing work are conservative.

E = K + U

The work-energy theorem gives ΔK = W for the net work. When this work is conservative, W = −ΔU, so ΔK + ΔU = 0. Equivalently, initial kinetic plus potential energy equals final kinetic plus potential energy.

How does a freely falling body's energy change?

For a body released from rest at height H, take the ground as zero potential energy and neglect air resistance. With constant g, total energy is mgH. At an intermediate height h, its energy is ½mv² + mgh.

v² = 2g(H − h)

Here H is the initial height and v is speed at height h. At release all the energy is potential. Just before reaching the ground it is kinetic. The two forms change, while their sum remains the same.

How do spring energies vary with displacement?

A block attached to an ideal spring on a smooth horizontal surface oscillates, moving repeatedly on either side of equilibrium. Let A be its amplitude, the greatest displacement magnitude. If released from rest at x = A, its energy is ½kA².

K = ½k(A² − x²)

The sign ± means either positive or negative. At an extreme the block momentarily stops before reversing direction. At equilibrium speed is greatest. Thus E equals maximum kinetic energy when U = 0, and maximum potential energy when K = 0.

Energy is plotted vertically against displacement. An upward-opening potential-energy parabola and a downward-opening kinetic-energy parabola lie beneath a horizontal total-energy line over the allowed displacement interval. The printed potential-energy label is V, equivalent to U here; the extremes are labelled −xₘ and xₘ, corresponding to −A and A.

See Fig. 5.8 in your NCERT textbook

What happens to energy when mechanical energy changes?

Conservation of mechanical energy is a conditional statement. When friction does work, some mechanical energy is transferred into other forms, including internal energy, the energy associated with microscopic motions and interactions. A decrease in mechanical energy does not imply that total energy has disappeared.

Let Wₙ꜀ mean the total work of non-conservative forces, and Eᵢ and E𝒻 the initial and final mechanical energies. Combining the work-energy theorem with the potential-energy relation gives:

E𝒻 − Eᵢ = Wₙ꜀

For sliding against friction on a stationary surface, this work is negative and mechanical energy decreases. In an energy balance for a system exchanging no energy with its surroundings, all forms must be included. Energy can change form while the total remains constant.

What does mass-energy equivalence mean?

Mass-energy equivalence relates mass to an equivalent amount of energy. The relation is E = mc², where E now denotes the energy equivalent of mass m, and c is the speed of light in vacuum, approximately 3 × 10⁸ m s⁻¹.

The relation expresses mass as a form of energy. It is distinct from the classical kinetic-energy expression. Energy conservation in a reaction must include energy associated with mass as well as other energy forms.

Worked example 5. Calculate the energy equivalent of 1 g of matter. Use c = 3 × 10⁸ m s⁻¹.

Formula: E = mc².

Substitute: m = 10⁻³ kg, so E = 10⁻³ × (3 × 10⁸)².

Answer: E = 9 × 10¹³ J, or 90 000 000 000 000 J. This is the energy equivalent of the given mass.

How is power related to work, force and velocity?

Power is the rate at which work is done or energy is transferred. Average power, Pₐᵥ, is work W divided by the time interval t. Instantaneous power, P, is the rate at a particular instant.

Pₐᵥ = W/t

P = dW/dt

Since dW = F⃗ · ds⃗ and ds⃗/dt is velocity v⃗, instantaneous power is P = F⃗ · v⃗. Its magnitude and sign depend on the force component along the velocity. If force and velocity point in the same direction, P = Fv.

The SI unit of power is the watt, symbol W: 1 W = 1 J s⁻¹. Its dimensions are [ML²T⁻³]. The W following a numerical value denotes watt; W used as a quantity in W/t denotes work.

Which units measure energy and which measure power?

A kilowatt hour is the energy transferred by one kilowatt over one hour. The prefix kilo means a thousand. Multiplying a rate of energy transfer by time gives energy, not another unit of power.

Worked example 6. An elevator and passengers have total mass 1800 kg. They rise at constant speed 2 m s⁻¹ against a frictional force of 4000 N. Take g = 10 m s⁻². Find the minimum motor power delivered to the elevator.

Formula: F = mg + f; P = Fv. Here F is the upward motor force and f the magnitude of downward friction.

Substitute: F = 1800 × 10 + 4000 = 22000 N; P = 22000 × 2.

Answer: P = 44000 W, approximately 59 hp. Constant speed requires the upward force to balance weight and friction.

How are collisions in one dimension analysed?

A collision is an interaction during which bodies exert forces on one another and their velocities change. A one-dimensional collision has initial and final velocities along the same straight line. Choose one direction as positive and retain signed velocities.

Linear momentum, p⃗ = mv⃗, is the product of mass and velocity. Its SI unit is kg m s⁻¹ and its dimensions are [MLT⁻¹]. Total momentum is conserved provided the system has no net external impulse, meaning no net momentum transfer from external forces over the collision.

The colliding bodies exert equal opposite internal forces at each instant. Their momentum changes therefore cancel in the total. Individual momenta need not remain constant. Impulse has the same unit and dimensions as momentum.

How do elastic and inelastic collisions differ?

During contact, even an elastic collision can temporarily store energy in deformation, a change of shape. Kinetic energy need not stay constant at each intermediate instant. In ordinary inelastic impacts, some initial kinetic energy becomes other forms, such as heat and sound.

Derivation: Elastic collision with a stationary target

Let m₁ and m₂ be the two masses. The first approaches with initial velocity u₁; the target has initial velocity u₂ = 0. Let v₁ and v₂ be their final velocities. Assume an elastic collision and negligible net external impulse.

1. Momentum conservation gives m₁u₁ = m₁v₁ + m₂v₂, or m₁(u₁ − v₁) = m₂v₂.
2. Kinetic-energy conservation gives m₁(u₁² − v₁²) = m₂v₂² after multiplying by two.
3. Factor the difference of squares and divide by the momentum relation for a non-zero momentum transfer: u₁ + v₁ = v₂.
4. Substitute v₂ = u₁ + v₁ into the momentum equation and solve for the final velocities.

v₁ = [(m₁ − m₂)/(m₁ + m₂)]u₁

v₂ = [2m₁/(m₁ + m₂)]u₁

The symbols ≫ and ≈ mean “much greater than” and “approximately equal to”. The unequal-mass limits are approximations. A negative final velocity indicates reversal relative to the chosen positive direction.

What changes if the bodies stick together?

For a completely inelastic collision with the same stationary target, let v be the common final velocity. Momentum conservation gives v = m₁u₁/(m₁ + m₂). Do not also impose kinetic-energy conservation.

The kinetic energy lost is ½[m₁m₂/(m₁ + m₂)]u₁². It follows by subtracting the final energy ½(m₁ + m₂)v² from the initial energy ½m₁u₁². Total energy is accounted for by including the non-kinetic forms produced.

How are oblique collisions in two dimensions analysed?

An oblique collision sends bodies away in different directions in a plane. Momentum must be conserved separately along two perpendicular axes when external impulse is negligible. The kinetic-energy equation remains a scalar equation.

Take the incoming direction of mass m₁ as the positive x-axis, with mass m₂ initially stationary. Choose the perpendicular y-axis in the plane of motion. Let θ₁ and θ₂ be positive outgoing angles above and below the x-axis, respectively; here v₁ and v₂ denote final speeds.

m₁u₁ = m₁v₁ cos θ₁ + m₂v₂ cos θ₂

The initial y-momentum is zero. The final components have opposite signs, so 0 = m₁v₁ sin θ₁ − m₂v₂ sin θ₂, where sin denotes sine. For an elastic collision there is also ½m₁u₁² = ½m₁v₁² + ½m₂v₂².

The incoming mass m₁ moves horizontally towards m₂. After collision, arrows point above and below the horizontal x-axis. They are labelled with the outgoing velocities, and angles θ₁ and θ₂ are marked relative to that axis.

See Fig. 5.10 in your NCERT textbook

Why is additional information needed?

With masses and incoming speed known, four quantities remain unknown: two outgoing speeds and two angles. The two momentum equations and one elastic-energy equation provide three relations. An additional datum, such as an outgoing angle, is needed to solve a general problem.

For equal masses in a glancing elastic collision, with one initially at rest and both moving afterwards, the outgoing velocities are perpendicular. Squaring the vector momentum relation and comparing it with kinetic-energy conservation makes their scalar product zero.

Thus, for the billiard-ball case with the target directed at 37° to the incoming direction, the other ball leaves at 53° on the opposite side. This assumes an elastic collision, with friction and rotational motion unimportant.

Glossary

• Work — Energy transfer measured by the force component along displacement multiplied by that displacement, for a constant force.
• Scalar product — Product of two vector magnitudes and the cosine of their included angle, giving a scalar.
• Kinetic energy — Energy associated with motion, equal to half the mass multiplied by speed squared in classical mechanics.
• Potential energy — Energy associated with the position or configuration of a system under a conservative interaction.
• Conservative force — Force whose work depends only on the endpoints, with zero work around a closed path.
• Non-conservative force — Force for which work can depend on the path rather than solely on initial and final positions.
• Mechanical energy — Sum of kinetic and potential energies, conserved when the forces doing work are conservative.
• Spring constant — Magnitude of restoring force per unit displacement for an ideal spring obeying Hooke's law.
• Amplitude — Greatest magnitude of displacement from equilibrium reached by an oscillating body during its motion.
• Power — Rate at which work is done or energy is transferred, measured in watts.
• Elastic collision — Collision for which the total kinetic energies before and after the interaction are equal.
• Inelastic collision — Collision in which total kinetic energy differs between the initial and final states.
• Completely inelastic collision — Collision in which the interacting bodies move together with a common velocity afterwards.
• Mass-energy equivalence — Relation identifying an energy equivalent of mass, given by mass multiplied by the squared speed of light.

Common errors and misconceptions

• Misconception: Exerting a large force necessarily means doing work. Correct: Work also requires displacement with a component along the force; an unmoving rigid wall receives no mechanical work from the push.
• Misconception: Friction always does negative work. Correct: Determine the force direction relative to displacement. The negative-work result here applies to the specified sliding or resistive forces opposing motion.
• Misconception: Work by any single force equals the whole kinetic-energy change. Correct: The work-energy theorem uses net work, including all contributing forces.
• Misconception: Losing 90% of kinetic energy means losing 90% of speed. Correct: At fixed mass, kinetic energy depends on speed squared; use a square root to recover speed.
• Misconception: A compressed spring has negative potential energy relative to its natural length. Correct: Its energy is ½kx², which is positive for either non-zero extension or compression.
• Misconception: Mechanical energy is conserved whenever total energy is conserved. Correct: Mechanical energy can become internal energy through friction while the complete energy balance remains valid.
• Misconception: A kilowatt hour measures power. Correct: It measures energy; watt and horsepower measure power.
• Misconception: Every inelastic collision makes the bodies stick together. Correct: Sticking together characterises a completely inelastic collision; other inelastic collisions can leave the bodies moving separately.

Exam-style questions with model answers

Q1. Define work done by a constant force and explain why gravity does no work on a block displaced horizontally on a table. [2 marks]

1. Work is W = Fs cos θ, where F is force magnitude, s is displacement magnitude and θ is the angle between them.
2. Gravity acts vertically while the displacement is horizontal. Their angle is 90°, so the force component along the displacement, and hence gravitational work, is zero.

Q2. A bullet of mass 50.0 g enters plywood 2.00 cm thick at 200 m s⁻¹. It emerges with 10% of its initial kinetic energy. Calculate its initial and final kinetic energies and emergent speed. [3 marks]

1. Convert the mass to m = 0.0500 kg. With initial speed vᵢ = 200 m s⁻¹, the initial kinetic energy is Kᵢ = ½mvᵢ² = ½ × 0.0500 × 200² = 1000 J.
2. The final kinetic energy K𝒻 is the stated fraction of the initial energy: K𝒻 = 0.10 × 1000 = 100 J.
3. Using K𝒻 = ½mv𝒻², the emergent speed is v𝒻 = √(2K𝒻/m) = √(200/0.0500) = 63.2 m s⁻¹.

Q3. A woman pushes a trunk with 100 N for 10 m. During the next 10 m her force decreases linearly to 50 N. Friction is a constant 50 N opposing motion throughout. Calculate the work done by the woman and by friction using a force-displacement graph. [4 marks]

1. Work equals signed area under the graph of the force component along displacement. The first applied-force region is a rectangle with area 100 × 10 = 1000 J.
2. The second region is a trapezium. Its area is ½(100 + 50) × 10 = 750 J.
3. Add the two positive contributions: the woman's total work is 1000 + 750 = 1750 J.
4. Friction acts opposite displacement, so its graph lies below the axis. Its work over 20 m is −50 × 20 = −1000 J.

Q4. A raindrop of mass 1.00 g falls from rest through 1.00 km and reaches 50.0 m s⁻¹. Only gravity and air resistance do work. Take constant g = 10 m s⁻². State the work-energy theorem and find gravitational work and resistive work. Explain the sign of the latter. [5 marks]

1. The work-energy theorem states that net work equals the change in kinetic energy. Here the net work is the sum of gravitational work and work by air resistance.
2. Convert mass to m = 10⁻³ kg and height fallen to h = 10³ m. Gravitational work is Wg = mgh = 10.0 J.
3. The drop starts from rest, so its initial kinetic energy is zero. Its increase is ΔK = ½ × 10⁻³ × 50.0² = 1.25 J.
4. If Wr denotes resistive work, ΔK = Wg + Wr. Therefore Wr = 1.25 − 10.0 = −8.75 J.
5. The negative sign shows that air resistance removes kinetic energy relative to gravity's contribution. It opposes the drop's downward displacement.

Q5. An ideal massless spring obeys Fₛ = −kx, where k is its constant and x its displacement from equilibrium. Derive its potential energy, choosing zero at x = 0. A block attached to it then oscillates without friction with amplitude A. State the energies at equilibrium and at the extremes. [5 marks]

1. Stretch the spring slowly so the block's change in kinetic energy is negligible. The balancing external force is +kx, opposite to the restoring spring force.
2. For an infinitesimal additional extension dx, external work is dW = kx dx. Integrating from 0 to x gives W = ½kx².
3. This work is stored as potential energy U = ½kx². Squaring x gives positive stored energy for both extension and compression relative to equilibrium.
4. At either extreme x = ±A, the block momentarily stops. Kinetic energy is zero and potential energy, equal to total energy E, is ½kA².
5. At equilibrium x = 0, potential energy is zero and kinetic energy is maximum, ½kA². With no friction, their sum remains constant throughout the oscillation.

Q6. A body of mass m₁ with initial velocity u₁ strikes a stationary body of mass m₂ in a one-dimensional elastic collision. Net external impulse is negligible. Derive the final velocities v₁ and v₂ and state the result for equal masses. [6 marks]

1. Take the incoming direction as positive. Conservation of total momentum gives m₁u₁ = m₁v₁ + m₂v₂, since the second body's initial velocity is zero.
2. Elasticity supplies a second equation: ½m₁u₁² = ½m₁v₁² + ½m₂v₂². This compares total kinetic energy before and after the collision.
3. Rewrite these as m₁(u₁ − v₁) = m₂v₂ and m₁(u₁² − v₁²) = m₂v₂². Factoring and dividing for non-zero momentum transfer gives u₁ + v₁ = v₂.
4. Substitute this relation into momentum conservation: m₁u₁ = m₁v₁ + m₂(u₁ + v₁). Rearranging gives v₁ = [(m₁ − m₂)/(m₁ + m₂)]u₁.
5. Insert this expression into v₂ = u₁ + v₁. The target's final velocity is v₂ = [2m₁/(m₁ + m₂)]u₁.
6. For m₁ = m₂, these become v₁ = 0 and v₂ = u₁. The incoming body stops and the target takes its initial velocity.

Q7. An elevator plus passengers has mass 1800 kg and rises at constant speed 2 m s⁻¹. A frictional force of 4000 N acts downwards. Take g = 10 m s⁻² and 1 hp = 746 W. Calculate the minimum motor power delivered to the elevator in watts and horsepower. [4 marks]

1. Constant speed means zero acceleration, so the resultant force vanishes. The motor's upward force must balance both weight and downward friction.
2. The required force is F = mg + f = 1800 × 10 + 4000 = 22000 N, where f is the frictional-force magnitude.
3. Force and velocity point upwards together. Instantaneous power is therefore P = Fv = 22000 × 2 = 44000 W.
4. Using the supplied conversion, power in horsepower is 44000/746, approximately 59 hp. This is mechanical power delivered to the elevator.

Q8. Two equal-mass billiard balls undergo a glancing elastic collision, with one initially stationary and both moving afterwards. Friction and rotation are negligible. The target leaves at 37° to the incoming direction. Show that the outgoing directions are perpendicular and find the other ball's angle. [3 marks]

1. Dividing vector momentum conservation by the common mass gives u⃗ = v⃗₁ + v⃗₂, where u⃗ is the incoming velocity and v⃗₁, v⃗₂ are outgoing velocities. Their magnitudes are u, v₁ and v₂. Squaring gives u² = v₁² + v₂² + 2v⃗₁ · v⃗₂.
2. Kinetic-energy conservation gives u² = v₁² + v₂². Thus v⃗₁ · v⃗₂ = 0, making the two non-zero outgoing velocities perpendicular.
3. The outgoing angles therefore sum to 90°. The other ball's angle is 90° − 37° = 53°, on the opposite side of the incoming direction.

Key takeaways

• Work uses the force component along displacement; its sign depends on whether that component assists or opposes the displacement.
• For a variable force, integrate the force component over displacement or calculate signed area under its force-displacement graph.
• The work-energy theorem equates net work with kinetic-energy change and applies to both constant and variable forces.
• Conservative work depends only on endpoints and equals the decrease in potential energy; its value around a closed path is zero.
• Ideal spring energy is ½kx²; during frictionless oscillations, kinetic and potential energies exchange while their sum remains constant.
• Mechanical energy can decrease through friction while a complete energy balance includes the other energy forms produced.
• Power measures how quickly work is done; watt is a power unit, whereas kilowatt hour is an energy unit.
• With negligible external impulse, collisions conserve total momentum; elastic collisions additionally conserve total kinetic energy between initial and final states.

Test yourself

Why does a force perpendicular to displacement do no work?

Its component along displacement is zero, because cos 90° = 0; therefore the scalar product vanishes.

What does an area below the displacement axis on a force-displacement graph represent when displacement increases?

It represents negative work by that force component over the corresponding interval of displacement.

Which work appears in the work-energy theorem?

The net work of all forces on the body equals its change in kinetic energy.

Why must a potential-energy zero be used consistently?

Energy differences must refer to the same chosen zero throughout a calculation, although that zero can initially be chosen conveniently.

Where is a frictionlessly oscillating spring-block system's speed greatest?

At equilibrium, potential energy is zero and kinetic energy is maximum, so the speed is greatest.

In E = mc², what does c represent?

It is the speed of light in vacuum, approximately 3 × 10⁸ m s⁻¹.

Does total kinetic energy remain constant at every instant of an elastic collision?

No. Energy can temporarily be stored in deformation during contact; total kinetic energy is equal before and after the collision.

Why are two momentum equations needed for an oblique collision?

Momentum is a vector, so its components along both perpendicular axes in the collision plane must be conserved.