Overview
This unit explains the physical concepts of work, energy and power and their relationships. You will learn how force acting through a displacement produces work, how work changes kinetic energy, and how potential energy stores energy in systems like stretched springs or raised masses. The unit covers conservative and non-conservative forces, the work-energy theorem, and the principle of conservation of mechanical energy. You will also study power as the rate of doing work and efficiency of energy transfer in machines. These ideas form the basis for analysing motion, machines, and energy transformations across mechanics and are essential for solving problems in kinematics, dynamics, and real-world engineering. Understanding these concepts helps in practical contexts like calculating the energy needed to lift objects, the power rating of motors, losses due to friction, and behaviour of oscillating systems such as springs and pendulums.
Learning Objectives
- Define work, energy and power and state their SI units.
- Calculate work done by constant and variable forces using geometry and calculus.
- Apply the work-energy theorem to relate net work to change in kinetic energy.
- Distinguish conservative and non-conservative forces and compute potential energy for common systems.
- Use conservation of mechanical energy to solve problems involving gravitational and elastic potential energy.
- Derive and use expressions for elastic potential energy in springs and gravitational potential energy near Earth's surface.
- Compute instantaneous and average power and evaluate efficiency for simple machines.
- Analyse energy transformations in systems with friction and determine energy dissipated as heat.
Topics in this chapter
15 topics · tap a topic title to jump straight to it.
Physical quantity: Work — definition and units
What is work?
Work is the quantitative measure of energy transfer that takes place when a force moves its point of application through a displacement. It tells us how much energy is passed from one object or system to another by mechanical action. Work is a scalar quantity: it has magnitude and sign but no direction. The sign shows whether energy is supplied to the system (positive work) or removed from it (negative work).
Mathematical definition for constant force
For a constant force F acting on a body which undergoes a straight-line displacement s, the work done by the force is W = F s cosθ, where θ is the angle between the force vector and displacement vector. This formula captures three situations: when θ = 0° (force and displacement parallel), W = F s and work is maximum and positive; when θ = 90° (force perpendicular to motion), W = 0; when θ = 180° (force opposite displacement), W = −F s and work is negative.
Units, dimensions and practical examples
The SI unit of work is the joule (J), defined as one newton metre (1 J = 1 N·m). The dimensional formula is ML2T−2. In practice, we often use kilojoule (kJ) and megajoule (MJ) for larger amounts. Example situations: pushing a box horizontally with a force in the direction of motion produces positive work; the weight of an object does negative work when you lift it; the normal force does zero work when it remains perpendicular to displacement.
Scalar nature, sign conventions and energy bookkeeping
Because work is scalar, you add algebraically when multiple forces act. Careful sign convention is essential: if you compute work done by each force, the algebraic sum equals net work. In many problems you may compute work done by external agent, by gravity, by friction etc., and keep track of energy transfers: work done on a system increases its energy, while work done by a system reduces it. The convention that work done by conservative forces can be related to potential energy is especially useful for energy conservation methods.
Limitations and context
Note that the simple formula W = F s cosθ applies only for constant forces and straight-line displacement. For variable forces or curved paths, work must be calculated by integrating small contributions along the path. This leads to line integrals in vector form and emphasizes the geometric interpretation of work as area under a force-displacement curve for one-dimensional forces.
- A horizontal force of 10 N pushes a box 3 m. Work = 10 × 3 = 30 J.
- A force of 20 N at 60° to the horizontal moves an object 2 m horizontally. Work = 20 × 2 × cos60° = 20 J.
- A satellite moving in a circular orbit experiences centripetal force perpendicular to motion; the centripetal force does zero work.
- Lifting a 2 kg book vertically by 1.5 m against gravity: work by you = m g h = 2 × 9.8 × 1.5 = 29.4 J (approx).
- W = F s cosθ
- Unit: 1 J = 1 N m
- Dimensional formula: ML2T−2
Work by a variable force — calculus approach
Why variable force matters
In many physical situations the force acting on an object changes with position or time. Examples include a spring force that varies with displacement, gravitational force that varies with distance for large separations, or a drag force that depends on speed. For such variable forces the simple product W = F s is not valid. Instead we break the path into infinitesimal pieces and sum the small contributions of work using integration.
Elemental work and integral definition
Consider a one-dimensional motion along the x-axis with force F(x) that varies with x. Over a tiny displacement dx, the small work dW = F(x) dx (if force and displacement are collinear). The total work moving from x1 to x2 is the definite integral W = ∫x1x2 F(x) dx. This formula is the direct limit of Riemann sums and gives the exact area under the curve F(x) from x1 to x2.
Vector form and curved paths
For motion along a curved path C in space with vector force F(r) and displacement vector dr, the differential work is dW = F ⋅ dr and the total work is the line integral W = ∫C F ⋅ dr. This form handles forces that vary in direction as well as in magnitude, and it reduces to the one-dimensional integral when motion is along a single axis.
Graphical interpretation and sign
On a graph of F(x) versus x, the work equals the algebraic area between the curve and the x-axis between x1 and x2. Areas below the axis represent negative contributions to work. Pay attention to limits: reversing the path swaps signs. For force laws that change sign, positive and negative contributions can partly cancel, as happens when a spring pushes then pulls during a motion cycle.
Techniques and examples
Compute integrals analytically when F(x) is given by a simple algebraic expression. For complex functions or measured data, use numerical integration (trapezoidal rule or Simpson’s rule). For piecewise forces integrate over each piece. Typical examples include: work done by spring force W = ∫(−k x) dx giving (−1/2) k x^2 change, or work done by inverse-square forces leading to gravitational potential energy expression. In problems, check units and use symmetry where possible to simplify integrals.
- Work by a spring force from x1 to x2: W = ∫x1x2 (−k x) dx = −(1/2)k(x2^2 − x1^2).
- If F(x) = a x^2, work from 0 to b is ∫0b a x^2 dx = (a b^3)/3.
- Graphical: area under F(x)=3x between x=0 and x=2 equals ∫0^2 3x dx = 6 J.
- Numerical: approximate ∫0^1 F(x) dx by dividing interval into small segments and summing F(x_i) Δx.
- W = ∫x1x2 F(x) dx
- Line integral: W = ∫C F ⋅ dr
Work done by gravity and near-Earth approximation
Gravity near Earth's surface
Close to Earth’s surface, the gravitational acceleration g is nearly constant. The gravitational force on a mass m can be written as Fg = −m g j (taking upward as positive y). For a vertical displacement from y1 to y2, the work done by gravity is Wg = ∫y1y2 (−m g) dy = −m g (y2 − y1). Thus if an object is raised by h = y2 − y1, gravity does negative work −m g h; if it falls by h, gravity does positive work +m g h.
Potential energy near surface
Because gravity is conservative, we can define a gravitational potential energy U(y) = m g y (choice of zero arbitrary). The change in potential between two heights equals negative of work done by gravity: ΔU = U(y2) − U(y1) = m g (y2 − y1) = −Wg. This relation helps convert between work and stored potential energy and simplifies many calculations.
General gravitational force and exact expression
For large separations where variation of g matters, use Newton’s law of gravitation: F(r) = −G M m / r^2 radial. Work done in moving mass m from r1 to r2 is W = ∫r1r2 (−G M m / r^2) dr = −G M m (1/r2 − 1/r1). From this, gravitational potential energy is defined as U(r) = −G M m / r (with U → 0 as r → ∞). Near the Earth's surface where r = R + h and h ≪ R, U(r) ≈ −G M m / R + m g h; the constant term is often ignored, yielding U ≈ m g h as used in many problems.
Applications, sign attention and problem examples
Use the near-surface formula for common lifting problems, pendulums and simple energy conversions, but use the general inverse-square expression for satellite motion and escape velocity problems. Always check signs: the work done by gravity is negative when lifting and positive when falling. When working with energy conservation, include the appropriate potential energy term so that mechanical energy K + U remains conserved in absence of non-conservative forces.
- Lifting 5 kg crate by 2 m: work by gravity = −m g h = −5 × 9.8 × 2 = −98 J.
- Dropping the crate: gravity does +98 J work as it falls 2 m, increasing kinetic energy.
- Work to move mass from 2R to 3R from Earth's centre using inverse-square formula: W = −G M m (1/3R − 1/2R) = −G M m (−1/6R) = +G M m/(6R) depending on orientation.
- Relating U = −G M m / r to near-surface U ≈ m g h by expanding for r = R + h and using g = G M / R^2.
- Near Earth: Wg = −m g Δy
- Gravitational potential energy: U(r) = −G m1 m2 / r
- Work for inverse square: W = −G m1 m2 (1/r2 − 1/r1)
Kinetic energy and theorem of work and kinetic energy
Definition of kinetic energy
Kinetic energy is the energy associated with the motion of a particle or system. For a particle of mass m moving with speed v, kinetic energy is defined by the scalar quantity K = (1/2) m v^2. This quadratic dependence on speed means doubling the speed increases kinetic energy by a factor of four. K is always non-negative and is measured in joules.
Derivation of the work–energy theorem
The work–energy theorem states that the net work done on a particle equals the change in its kinetic energy. Start from Newton’s second law F = m a. For motion along a path, small work dW = F ⋅ dr. Using a = dv/dt and dr = v dt, we get dW = m (dv/dt) ⋅ v dt = m v ⋅ dv. Integrating from initial velocity v1 to final v2 yields Wnet = ∫v1v2 m v dv = (1/2) m (v2^2 − v1^2) = ΔK.
Physical meaning and usefulness
This theorem provides a direct link between forces and energy change. It is especially useful when you are interested only in speeds and energy changes without solving for time-dependent acceleration. The theorem applies regardless of whether forces are conservative or not, as long as you include all forces when calculating net work. For example, if friction acts, its negative work reduces kinetic energy by exactly the amount of energy dissipated.
Extension to systems and rigid bodies
For a system of particles, the net work done by external forces equals the change in total kinetic energy of the system (internal forces often cancel). For extended bodies, kinetic energy includes translational and rotational parts: Ktotal = (1/2) M Vcm^2 + (1/2) I ω^2 where Vcm is centre-of-mass speed and I ω^2 accounts for rotation about centre of mass. The work–energy relation can be applied to these forms by including torques and angular displacements in rotational cases.
Problem-solving tips
Use Wnet = ΔK when forces and displacement are given or when you can compute work by integration. In collision or braking problems, equate work done by retarding forces to change in kinetic energy to find stopping distances or required force. Remember to include sign for work by each force and be cautious with reference frames; kinetic energy depends on speed relative to chosen inertial frame.
- A 1 kg block accelerated from 2 m/s to 6 m/s by net work W = (1/2)×1×(36−4) = 16 J.
- A car of mass 1000 kg slowed from 20 m/s to 10 m/s by brakes: work done by brakes = (1/2)×1000×(100−400) = −150000 J (energy removed).
- A variable net force does work of 50 J on a 2 kg object initially at rest; final speed v = sqrt(2W/m) = sqrt(50) = 7.07 m/s.
- Using work-energy theorem avoids computing acceleration if force-distance data are available.
- K = (1/2) m v^2
- Work–energy theorem: Wnet = ΔK = (1/2) m (v2^2 − v1^2)
Potential energy and conservative forces
Conservative versus non-conservative forces
Forces are called conservative if the work they do when moving a particle between two points is independent of the path taken. Equivalently, the work done by a conservative force around any closed path is zero. Gravity and ideal spring forces are common examples. Non-conservative forces, such as kinetic friction or air drag, dissipate mechanical energy and their work depends on the path length.
Defining potential energy
When a force F is conservative, we can associate a scalar function U called potential energy such that the work done by the force from point a to b equals −(U(b) − U(a)). In one dimension this relation is expressed as F(x) = −dU/dx. In three dimensions the force field is the negative gradient of the potential: F = −∇U. Potential energy is defined up to an arbitrary constant because only differences in U have physical meaning.
Choosing reference and practical examples
Selecting a convenient zero of potential simplifies calculations. For gravitational potential near Earth's surface, we choose U = m g y with zero at y = 0; for a spring, U = (1/2) k x^2 with zero at equilibrium x = 0. For gravitational potential at large distances U(r) = −G M m / r, zero is chosen at infinity. Use the reference that makes arithmetic easy for the given problem.
Relation to energy conservation
Because conservative forces can be written in terms of potential energy, mechanical energy E = K + U remains constant when only conservative forces act. Changes in kinetic energy are balanced by opposite changes in potential energy. When non-conservative forces are present, mechanical energy changes by the amount of work done by those forces: Δ(K + U) = Wnc.
Detecting conservative forces and field properties
Mathematically, a force is conservative if its curl is zero in a simply connected region (∇ × F = 0) and it can be expressed as the negative gradient of a scalar function. In practical problems, check whether work around closed paths vanishes or whether a potential function can be found. If yes, energy methods are powerful for solving motion and equilibrium questions.
- For gravitational force near Earth, F = −m g and U = m g y. Moving upward increases U.
- Spring force F = −k x is conservative; potential energy U = (1/2) k x^2 with minimum at x = 0.
- Electric forces in electrostatics are conservative and have associated potential energy functions.
- Closed-path work: move mass in a loop under gravity only; net work = 0, confirming conservativeness (if path returns to same height).
- F = −dU/dx (one dimension)
- F = −∇U (three dimensions)
- Work by conservative force = −ΔU
Elastic potential energy and Hooke’s law
Hooke's law and restoring force
An ideal spring exerts a restoring force proportional to its displacement from the equilibrium position. Mathematically, this is written as F = −k x, where k is the spring constant, x is displacement measured from equilibrium, and the negative sign shows the force points opposite to displacement. Hooke's law holds only within the elastic limit of the material; beyond that the relationship need not be linear.
Deriving elastic potential energy
Because the spring force is conservative, it can be derived from a potential energy function U(x). Using F = −dU/dx, integrate the force: dU = −F dx = k x dx. Integrating from 0 to x gives U(x) − U(0) = (1/2) k x^2. Choosing U(0) = 0 at equilibrium yields U(x) = (1/2) k x^2. This formula gives the stored elastic energy when a spring is stretched or compressed by distance x.
Work done in compressing or stretching
Work done by an external agent to compress or stretch a spring quasistatically from 0 to x equals the stored energy (1/2) k x^2. The work done by the spring force when moving from x1 to x2 is Wspring = ∫x1x2 (−k x) dx = −(1/2) k (x2^2 − x1^2) = −ΔU. This sign convention shows that when the spring returns to equilibrium, it does positive work on other bodies by converting stored U to kinetic energy.
Series and parallel springs and energy distribution
When springs are combined, equivalent spring constants determine how energy is shared. For springs in series, 1/keq = 1/k1 + 1/k2 etc., and for parallel keq = k1 + k2. The elastic energy for a given overall displacement uses keq in U = (1/2) keq x_total^2. Practical systems like vehicle suspensions and archery bows store elastic energy and release it as kinetic energy.
Limits, oscillations and energy conversion
Energy stored in a spring is the source for simple harmonic motion when attached to a mass. In a mass-spring oscillator, elastic potential converts to kinetic energy and back, and total mechanical energy remains constant for no damping. Remember real materials dissipate some energy as heat; hence ideal expressions apply within elastic limits and for conservative ideal springs.
- Spring constant k = 200 N/m, compressed by 0.05 m: U = (1/2) × 200 × (0.05)^2 = 0.25 J.
- Work by spring from x1 = 0.1 m to x2 = 0 m: W = (1/2) k (0.1^2 − 0^2) = (1/2) k × 0.01.
- Two springs in series with k1 and k2: equivalent k = (k1 k2)/(k1 + k2); elastic energy = (1/2) keq x^2.
- Stretching a bow stores elastic potential energy used to launch an arrow.
- Hooke's law: F = −k x
- Elastic potential energy: U = (1/2) k x^2
- Work by spring: W = −ΔU = (1/2) k (x1^2 − x2^2)
Work done by non-conservative forces — friction and dissipative forces
Nature of non-conservative forces
Non-conservative forces such as kinetic friction, rolling resistance, air drag and viscous forces convert mechanical energy into internal energy (typically thermal) and are path-dependent. The work done by these forces depends on the actual route taken between two points: longer paths usually involve more energy loss. For these reasons one cannot define a scalar potential energy for non-conservative forces that would restore path independence.
Kinetic friction and simple expression
For a block sliding on a surface with coefficient of kinetic friction μk, the frictional force magnitude is fk = μk N where N is the normal reaction. This frictional force opposes motion and thus does negative work: Wf = −fk s = −μk N s for displacement s in direction of motion. If the surface is horizontal, N = m g and Wf = −μk m g s. This expression gives energy converted to heat by friction along the path.
Velocity-dependent resistive forces
Air resistance and viscous drag often depend on speed; common models are Fdrag ∝ v (laminar regime) or Fdrag ∝ v^2 (high-speed turbulent regime). For such forces the work done over a distance requires integrating F(v) along the trajectory, often converting dx to dt using v = dx/dt and integrating power P = F v over time. The total energy dissipated equals the time integral of drag power or the space integral of drag force along path.
Energy bookkeeping with non-conservative work
When non-conservative forces do work Wnc, mechanical energy changes: Δ(K + U) = Wnc. For dissipative forces Wnc is negative and mechanical energy reduces; the lost energy appears as heat, sound or deformation. In problem solving, compute conservative work through potentials and then include Wnc explicitly to find final energy states. This method avoids solving forces dynamically when only energy outcomes are required.
Examples and practical consequences
Stopping distances are found by equating initial kinetic energy to work done by friction. Damping in oscillators removes energy cycle by cycle, reducing amplitude. Engineers must account for dissipative work to design brakes, insulation and aerodynamic shapes that minimise energy losses for efficiency. Always remember non-conservative work is path dependent and accumulates with longer or rougher paths.
- A 10 kg box slides 5 m with μk = 0.2 and N = mg; friction work = −μk m g s = −0.2 × 10 × 9.8 × 5 = −98 J.
- A car experiencing air drag force proportional to v^2 over distance s loses energy equal to ∫ Fdrag dx.
- Stopping distance for a vehicle with initial kinetic energy K0 and constant braking force Fbrake: Fbrake s = −K0, so s = K0 / |Fbrake|.
- A damping force F = −b v reduces amplitude of oscillator; energy lost per cycle equals work done by damping over that cycle.
- Work by friction: Wf = −μk N s
- Energy change due to non-conservative forces: ΔEmech = Wnc
Conservation of mechanical energy
Statement and origin
The principle of conservation of mechanical energy states that for a closed system acted upon only by conservative forces, the total mechanical energy E = K + U remains constant in time. This follows directly from the work–energy theorem and the definition of potential energy: work done by conservative forces equals negative change in potential, so net work by conservatives changes kinetic energy by −ΔU, leading to Δ(K + U) = 0.
How to use conservation
To apply conservation, identify all forms of mechanical energy present: kinetic energy (translational and rotational) and potential energies due to gravity, springs or other conservative fields. Write Einitial = Efinal: K1 + U1 = K2 + U2. This equation allows solving for unknown speeds, heights or displacements without directly computing forces or accelerations. It is especially powerful when motion converts energy between kinetic and potential forms, as in pendula, roller coasters and mass-spring systems.
Including non-conservative work
If non-conservative forces (friction, air resistance) do work Wnc, mechanical energy changes by that amount: K1 + U1 + Wnc = K2 + U2. Here Wnc is typically negative for dissipative forces. This modification allows one to account for energy losses when only the net work done by non-conservative forces is easier to compute than detailed force distributions.
Examples and common problem types
Typical problems include computing speed at a lower point after descent from height h, maximum compression of springs when a mass collides and compresses a spring, and determining amplitude of oscillations. In orbital mechanics, conservation of mechanical energy determines speeds at different radii. In each case choose convenient zero of potential to simplify algebra. Be careful to include rotational kinetic energy where objects roll without slipping.
Practical advice and pitfalls
Always check whether external non-conservative work is present; if so, do not apply conservation without modification. Remember that potential energy is defined up to an additive constant — choose zero levels smartly. Verify results by dimensional analysis and limiting cases. Use conservation to simplify multi-stage problems by equating energies between stages rather than solving motion equations stepwise.
- A 0.5 kg mass falls from height 2 m; ignoring air resistance, initial U = m g h = 0.5×9.8×2 = 9.8 J, initial K = 0, so speed at bottom v = sqrt(2 g h) ≈ 6.26 m/s.
- Mass-spring: from amplitude A to equilibrium, potential energy (1/2) k A^2 converts into kinetic energy (1/2) m v^2 at equilibrium; v_max = A sqrt(k/m).
- A pendulum released from small angle: maximum potential at amplitude converts to kinetic at lowest point giving v = sqrt(2 g h).
- Including friction: if friction does −20 J work, then mechanical energy decreases by 20 J and final K + U = initial K + U − 20 J.
- Conservation: K1 + U1 = K2 + U2 (if only conservative forces act)
- With non-conservative forces: K1 + U1 + Wnc = K2 + U2
Power: instantaneous and average
Definition and physical meaning
Power measures how quickly work is done or energy is transferred. Average power Pavg over a time interval Δt is Pavg = ΔW/Δt, where ΔW is the work done in that time. Instantaneous power P is the time derivative of work: P = dW/dt. Power tells us the rate at which energy flows into, out of, or within a system and is essential for sizing motors, engines and power supplies.
Relation to force and velocity
If a force F acts on a body moving with velocity v, the instantaneous power delivered by the force is P = F ⋅ v, the dot product of force and velocity vectors. For motion along the direction of force this reduces to P = F v. If force is perpendicular to velocity there is no power transfer (P = 0), which explains why centripetal forces do not change kinetic energy despite doing no work.
Units and common scales
The SI unit of power is the watt (W), with 1 W = 1 J s−1. Practical units include kilowatt (kW) and horsepower (1 hp ≈ 746 W). When a device runs with power P for time t, the energy consumed is E = P t. Electric energy bills use kilowatt-hours (kWh), where 1 kWh = 3.6 MJ.
Applications: lifting, drag and machines
Power required to lift a mass m at constant speed v vertically is P = m g v. For resistive forces like air drag, instantaneous power dissipated is Pdrag = Fdrag v; if Fdrag ∝ v^2 then Pdrag ∝ v^3, which explains the rapid rise in power needed at high speeds. For rotating systems, power delivered by a torque τ at angular speed ω is P = τ ω, directly analogous to P = F v in translation.
Average versus peak power and efficiency
Real devices have varying power demands; average power over a duty cycle determines energy consumption, while peak power determines sizing for transient loads. Efficiency relates useful power output to input: η = Pout / Pin. When calculating required input for a mechanical task, divide required mechanical power by efficiency to find electrical input or fuel rate. Always check units and time intervals when converting between energy and power.
- Lifting 50 kg at constant speed 0.4 m/s: required power P = m g v ≈ 50 × 9.8 × 0.4 = 196 N m/s ≈ 196 W.
- A force of 10 N pushes an object with instantaneous speed 3 m/s in same direction: instantaneous power = 10 × 3 = 30 W.
- Car engine rated 100 kW can in ideal case deliver 100 kJ per second; actual usable power depends on transmission losses.
- If friction does 200 J of work per second, power dissipated as heat is 200 W.
- Average power: Pavg = ΔW / Δt
- Instantaneous power: P = dW/dt = F ⋅ v
- Unit: 1 W = 1 J s−1
Mechanical energy in oscillations — simple harmonic oscillator
Mass-spring system basics
A mass m attached to an ideal spring of spring constant k oscillates about its equilibrium position when displaced. For small displacements and negligible damping, the restoring force F = −k x leads to simple harmonic motion (SHM) with angular frequency ω = sqrt(k/m). The motion is sinusoidal and energy oscillates between kinetic and potential forms without net loss in the ideal case.
Total mechanical energy
The total mechanical energy of an undamped mass-spring oscillator is constant and equals E = (1/2) k A^2 where A is the amplitude of oscillation. At maximum displacement x = ±A, velocity is zero and energy is entirely elastic potential U = (1/2) k A^2. At equilibrium x = 0, potential is zero and kinetic energy is maximum Kmax = (1/2) k A^2. At intermediate positions the energies partition so that K(x) + U(x) = E always.
Time dependence of energies
With x(t) = A cos(ω t + φ) and v(t) = −A ω sin(ω t + φ), the instantaneous potential energy U(t) = (1/2) k x^2 = (1/2) k A^2 cos^2(ω t + φ) and kinetic energy K(t) = (1/2) m v^2 = (1/2) m A^2 ω^2 sin^2(ω t + φ). Since k = m ω^2 these add to constant E. Graphs of K and U versus time are out of phase by π/2, showing smooth energy exchange.
Damping and forced oscillations
When damping (non-conservative forces) is present, mechanical energy decreases over time and amplitude decays, with energy converted to heat. In forced oscillations with driving force, steady-state amplitude depends on driving frequency and damping; energy input from driver balances dissipative losses in steady state. Energy methods help compute power absorbed and dissipated per cycle and explain resonance phenomena where energy transfer is most efficient.
Applications and problem solving
SHM energy ideas apply to pendulums (small angles), LC circuits (electrical analogues), molecular vibrations and mechanical resonators. Use energy methods to find maximum speed, amplitude relations, and energy dissipation without solving differential equations explicitly when possible.
- Mass m = 0.2 kg, spring k = 50 N/m, amplitude A = 0.1 m: total energy E = (1/2) k A^2 = 0.5 × 50 × 0.01 = 0.25 J.
- At x = A/2, potential U = (1/2) k (A/2)^2 = E/4, kinetic K = E − U = 3E/4.
- Angular frequency ω = sqrt(k/m) for numbers above = sqrt(50/0.2) = sqrt(250) ≈ 15.81 s−1.
- A lightly damped oscillator loses small fraction of E each cycle to heat: model energy decay E(t) ≈ E0 e^(−γ t) for small damping constant γ.
- Angular frequency: ω = sqrt(k/m)
- Total energy: E = (1/2) k A^2 = (1/2) m A^2 ω^2
- Instantaneous energies: U = (1/2) k x^2, K = (1/2) m v^2
Energy in a simple pendulum
Geometry and energies
A simple pendulum consists of a mass m suspended from a fixed point by a massless string of length L. When the bob is displaced to an angle θ from vertical, its height above the lowest position is h = L(1 − cosθ). The potential energy relative to the lowest point is U = m g h = m g L(1 − cosθ). Kinetic energy at any instant is K = (1/2) m v^2 where v = L dθ/dt is the tangential speed.
Small-angle approximation and SHM
For small oscillation angles (θ small, in radians), cosθ ≈ 1 − θ^2/2 so h ≈ (1/2) L θ^2 and U ≈ (1/2) m g L θ^2. The restoring torque is proportional to θ and the motion approximates simple harmonic with angular frequency ω = sqrt(g/L). The total mechanical energy for small amplitude θmax is E = (1/2) m g L θmax^2, which equals maximum potential at amplitude and maximum kinetic at lowest point.
Exact energy relations for larger amplitudes
For larger amplitudes the small-angle linearisation fails and the motion is nonlinear. Nevertheless energy conservation still holds: E = (1/2) m L^2 (dθ/dt)^2 + m g L(1 − cosθ) is constant if damping is absent. Solving for dθ/dt gives an expression that can be integrated to find period in terms of elliptic integrals; energy methods give speeds at positions without solving the full equation of motion.
Practical calculations and examples
To find maximum speed at the lowest point given amplitude θmax, use energy conservation: m g L(1 − cosθmax) = (1/2) m v_max^2 so v_max = sqrt(2 g L (1 − cosθmax)). For small angles this reduces to v_max ≈ L ω θmax. Energy analysis simplifies many pendulum problems like finding heights, speeds, and effects of small damping.
Applications and limitations
Pendulum energy ideas are used in clocks, seismographs and experiments to measure g. Remember that air resistance and friction at the pivot convert mechanical energy to heat and must be included as non-conservative work if precision is required. The small-angle SHM model is an excellent approximation for many school-level problems.
- Pendulum length L = 1 m, release angle 5°: approximate maximum speed v_max ≈ sqrt(2 g L (1 − cos5°)) ≈ sqrt(2 × 9.8 × 1 × (1 − 0.9962)) ≈ 0.31 m/s.
- Total energy for small θmax ≈ (1/2) m g L θmax^2; for m = 0.2 kg, L = 1 m, θmax = 0.1 rad, E ≈ 0.5 × 0.2 × 9.8 × 1 × 0.01 = 0.0098 J.
- Using conservation: height change for small θ from 0.1 rad gives h ≈ 0.5 × L × θ^2 ≈ 0.005 m; U = m g h.
- For larger amplitude, calculate energies using exact expression U = m g L (1 − cosθ).
- Potential energy: U = m g L (1 − cosθ)
- Small-angle ω = sqrt(g/L)
- Maximum speed: v_max = sqrt(2 g L (1 − cosθmax))
Power in rotating systems and relation to torque
Rotational work and angular displacement
In rotational motion about a fixed axis, the analogue of linear displacement is angular displacement θ. When a torque τ acts and the body rotates through angle dθ, the small work done is dW = τ dθ. Integrating over an angular interval gives W = ∫ τ dθ. This mirrors the translational expression W = ∫ F dx but uses rotational quantities.
Power delivered by torque
Instantaneous power due to torque while the body rotates at angular speed ω is P = τ ω. This is directly analogous to P = F v because for tangential forces τ = r F_tangential and v = ω r, so τ ω = F_tangential v. In rotating machines like motors and turbines, torque and angular speed determine mechanical power output or input.
Rotational kinetic energy and work–energy
Rotational kinetic energy of a rigid body rotating about a fixed axis is Krot = (1/2) I ω^2 where I is the moment of inertia. The rotational work–energy theorem states that net work done by torques equals change in rotational kinetic energy: Wnet = ΔKrot. For combined translation and rotation the total kinetic energy includes both translational (1/2) M Vcm^2 and rotational parts, and work done may change both.
Power transmission and efficiency
In mechanical transmissions, power ideally remains the same through gears (Pin = Pout) while torque and angular speed trade off according to gear ratio. Real systems have losses due to friction and heat so output power is lower by efficiency factors. For a motor with torque τ and speed ω, shaft power is Pshaft = τ ω and electrical input power must cover shaft power divided by motor efficiency.
Applications and examples
Use τ ω to compute power in engines, machine shafts, wind turbines, and exercise equipment. When accelerating a flywheel from ω1 to ω2, the work required equals change in rotational energy (1/2) I (ω2^2 − ω1^2). For rolling objects, include rotational energy when computing acceleration from energy methods. Energy methods simplify many rotational dynamics problems compared with torque–angular acceleration integration.
- A motor provides torque τ = 5 N m at angular speed ω = 100 rad/s: power P = τ ω = 500 W.
- A flywheel with I = 0.2 kg m^2 speeds up from ω1 = 10 to ω2 = 20 rad/s; change in rotational energy ΔK = (1/2) I (ω2^2 − ω1^2) = 0.1 × (400 − 100) = 30 J.
- Rolling wheel with v = ω R has total kinetic energy K = (1/2) m v^2 + (1/2) I ω^2. Power to accelerate includes both contributions.
- In ideal gearbox, if torque increases by factor n, angular speed reduces by 1/n and power remains same: P_in = P_out.
- Rotational power: P = τ ω
- Rotational kinetic energy: Krot = (1/2) I ω^2
- Work by torque: W = ∫ τ dθ
Efficiency and energy transformations in machines
Definition and interpretation
Efficiency measures how effectively a machine converts input energy or power into useful output. Expressed as a ratio η = (useful output / input) it is often written as a percentage: η% = (useful output / input) × 100. Efficiency cannot exceed 100%; real machines have efficiencies much lower because of unavoidable losses such as friction, heat, sound and leakage.
Energy and power forms
Efficiency can be applied to energy over a time interval (energy efficiency) or to instantaneous power (power efficiency). For steady operations, energy and power efficiencies coincide: η = Eout / Ein = Pout / Pin. For cyclic or transient processes, compute energy per cycle and divide to find cycle efficiency, or consider average powers over the cycle.
Sources of losses and practical considerations
Loss mechanisms include mechanical friction in bearings and gears, viscous losses in fluids, electrical resistance in windings, and aerodynamic drag. These losses convert useful mechanical or electrical energy into heat, which may be removed by cooling systems. System designers aim to reduce losses through lubrication, streamlining, better materials and tighter tolerances. Overall system efficiency is the product of stage efficiencies; multiple stages with moderate efficiencies can result in low overall efficiency if not designed carefully.
Calculations with work and power
To determine whether a machine is adequate, compute the required useful power Puse (for example Puse = m g v for lifting mass at speed v). Given machine efficiency η, input power required is Pin = Puse / η. For energy over time t, required energy Ein = Euse / η. Use these relations to size motors, estimate fuel consumption and compare alternatives. Remember to include safety margins and continuous versus peak power ratings.
Examples and interpretation
Electric motors often have high efficiencies (70–95%), while internal combustion engines typically have lower thermal efficiencies. For compound systems like pumps and gearboxes multiply stage efficiencies to get overall. Realistic assessment of cost and performance depends on both efficiency and power rating: a highly efficient but underpowered device is unsuitable, while a powerful but inefficient one wastes energy and increases operating costs.
- Motor lifts 100 kg at constant speed 0.2 m/s consuming electrical power 300 W. Useful mechanical power = m g v = 100 × 9.8 × 0.2 = 196 W. Efficiency η = 196/300 × 100% ≈ 65.3%.
- If a machine output is 400 J while consuming 500 J, efficiency = 400/500 × 100% = 80%.
- Two-stage gearbox: stage efficiencies 0.95 and 0.9. Overall efficiency = 0.95 × 0.9 = 0.855 or 85.5%.
- Design problem: need to choose motor power such that Pout/η ≥ required mechanical power.
- Efficiency: η = (useful output / input) × 100%
- Power relation: Pin × η = Pout
Work done in lifting and moving objects — examples and problem strategies
Common lifting problems and gravity
Many standard exercises ask for the work required to lift a mass by a vertical height h or to move it along an incline. For lifting at constant speed the external agent must supply force equal to weight mg, so work done equals W = m g h. If acceleration changes the kinetic energy, include the change in kinetic energy using the work–energy theorem: total work by agent = ΔK + m g h.
Inclined plane calculations
When an object is moved up an incline of angle α and length s corresponding to vertical rise h = s sinα, the gravitational work depends only on vertical rise and equals m g h. Frictional work along the incline is path-dependent: Wf = −μk N s where N = m g cosα. Total external work must overcome both gravity and friction: Wext = m g s sinα + μk m g cosα s if motion is quasistatic at constant speed.
Strategies for mixed problems
- Draw a clear diagram and free-body forces; identify displacements and angles.
- Decide whether to use W = ∫ F ⋅ dr, conservation of energy, or work–energy theorem depending on what is known and what is required.
- Account for non-conservative work explicitly as negative contributions when using energy conservation.
- Keep track of sign conventions and units, and check limiting cases for reasonableness.
Composite systems with springs and friction
Problems often combine springs, gravity and friction: for example pulling a block up an incline compresses a spring; use energy balance K1 + Ugravity1 + Uspring1 + Wnc = K2 + Ugravity2 + Uspring2. Compute each term carefully: gravitational potential uses vertical heights, spring potential uses (1/2) k x^2 and non-conservative work includes friction over the actual path.
Practical tips and verification
Check dimensions and whether results make sense (e.g., work should scale with mass and height). Compare with simple special cases: zero friction should recover m g h only, and zero displacement should give zero work. Use energy methods to avoid integrating equations of motion when only final speeds or displacements are required.
- Box up an incline: m = 10 kg, α = 30°, height h = 2 m, find work against gravity = m g h = 10 × 9.8 × 2 = 196 J. If moved along plane of length s = h / sin30° = 4 m with μk = 0.1, friction work = −μk m g cosα s = −0.1 × 10 × 9.8 × 0.866 × 4 ≈ −34.0 J. Total external work = 196 + 34 ≈ 230 J.
- Lift and accelerate: lift 2 kg mass by 1 m and increase speed from 0 to 3 m/s. Work to raise = m g h = 19.6 J. Work for kinetic increase = (1/2) m v^2 = 9 J. Total work by you = 28.6 J.
- Pulling sled on rough ground: compute work by tension minus work by friction to find change in kinetic energy over a given distance.
- Take path independence: gravitational work only depends on vertical difference; moving along zigzag path same as direct lift for gravity alone.
- Work against gravity for lift: W = m g h
- Friction along incline: Wf = −μk m g cosα s
- Gravitational work along incline: Wg = −m g s sinα
Work done by variable gravitational force — escape velocity and orbital energies
Energy in gravitational fields
For a mass m in the gravitational field of a large mass M, the gravitational potential energy at distance r is U(r) = −G M m / r when zero is chosen at infinity. The negative sign indicates bound systems: the potential energy is lower (more negative) near the attracting mass. Kinetic energy plus this potential gives the total mechanical energy of the system E = K + U.
Work for radial motions and relation to potential
The work done by gravity when moving from r1 to r2 is W = ∫r1r2 (−G M m / r^2) dr = −G M m (1/r2 − 1/r1). This expression shows that the work depends only on end radii, confirming gravity is conservative. If an external slowly-applied force moves the mass outward from r1 to infinity, the external agent must do work equal to +G M m / r1 to overcome attraction.
Orbit energies and circular motion
For a circular orbit of radius r, centripetal balance gives m v^2 / r = G M m / r^2, so v^2 = G M / r. Kinetic energy is K = (1/2) m v^2 = (1/2) G M m / r and potential U = −G M m / r. Therefore K = −(1/2) U and total energy E = K + U = −(1/2) G M m / r. Negative total energy indicates a bound circular orbit and the magnitude of E is the binding energy.
Escape velocity and physical meaning
Escape velocity is the minimum speed needed at distance r for the mass to reach infinity with zero kinetic energy remaining, ignoring other forces. Using energy conservation (1/2) m v_esc^2 + U(r) = 0 gives v_esc = sqrt(2 G M / r). For Earth surface this gives the well-known value near 11.2 km/s. Note escape speed is independent of direction and depends only on r.
Applications and estimates
These expressions are essential for satellite mechanics, launch energy estimates and understanding gravitational binding. For example, the work required to move a satellite from one circular orbit to another can be evaluated from changes in total energy, and mission planning uses these energy budgets to estimate fuel requirements and transfer manoeuvres.
- Escape velocity from Earth surface: v_esc ≈ 11.2 km/s using v_esc = sqrt(2 g R) with R ≈ 6.37×10^6 m and g ≈ 9.8 m/s^2.
- Total energy of satellite in circular orbit at altitude h: E = −(1/2) G M m / (R + h).
- Work by gravity moving from r = 2R to r = 3R: W = −G M m (1/3R − 1/2R) = +G M m/(6R) indicating gravity did positive work while moving outward between those radii depending on direction chosen.
- Energy needed to move mass m from Earth's surface to infinity ignoring atmosphere: ΔE = +G M m / R.
- Gravitational potential energy: U(r) = −G M m / r
- Escape velocity: v_esc = sqrt(2 G M / r)
- Total energy in circular orbit: E = −(1/2) G M m / r
Key Concepts
- Work
- Scalar quantity equal to force times displacement component in direction of force, W = F·s.
- Kinetic energy
- Energy of motion of a body, defined for a particle as K = (1/2) m v^2.
- Potential energy
- Energy stored in a system due to configuration in a conservative force field, defined up to an additive constant.
- Conservative force
- A force for which work between two points is path independent and whose work over any closed path is zero.
- Non-conservative force
- A force like friction whose work depends on path and typically converts mechanical energy into thermal energy.
- Work–energy theorem
- Statement that net work done on a particle equals change in its kinetic energy, Wnet = ΔK.
- Power
- Rate at which work is done or energy is transferred; instantaneous power P = dW/dt = F·v.
- Joule
- SI unit of energy and work, equal to one newton metre (1 J = 1 N m).
- Watt
- SI unit of power, equal to one joule per second (1 W = 1 J s−1).
- Hooke's law
- Linear relation for an ideal spring: restoring force F = −k x where k is spring constant.
- Elastic potential energy
- Energy stored in a spring: U = (1/2) k x^2 measured from equilibrium.
- Gravitational potential energy
- Near Earth's surface U = m g h; for general separation U(r) = −G m1 m2 / r.
- Escape velocity
- Minimum speed needed to move from distance r to infinity without further propulsion: v_esc = sqrt(2 G M / r).
- Mechanical energy
- Sum of kinetic and potential energies of a system: E = K + U.
- Efficiency
- Ratio of useful energy or power output to input, often expressed as a percentage.
- Rotational power
- Power delivered by a torque τ at angular velocity ω: P = τ ω.
- Binding energy
- Magnitude of negative total energy required to remove a bound particle to infinity.
Practice Questions
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A 3 kg block is pushed 4 m along horizontal floor by a constant horizontal force of 12 N. What is the work done by the force? / एक 3 किलोग्राम ब्लॉक को 12 N के क्षैतिज निरंतर बल से क्षैतिज फर्श पर 4 मीटर धकेला जाता है। बल द्वारा किया गया कार्य कितना है?
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Work W = F s = 12 N × 4 m = 48 J. / कार्य W = F s = 12 N × 4 m = 48 J।
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A force F(x) = 5x N acts on a particle along x-axis, where x is in metres. Calculate work done by this force from x = 0 to x = 3 m. / F(x) = 5x N नामक बल x-अक्ष के साथ क्रिया करता है, जहाँ x मीटर में है। x = 0 से x = 3 m तक इस बल द्वारा किया गया कार्य गणना कीजिए।
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W = ∫0^3 5x dx = (5/2) x^2|0^3 = (5/2) × 9 = 22.5 J. / W = ∫0^3 5x dx = (5/2) x^2|0^3 = (5/2) × 9 = 22.5 J।
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A mass m = 2 kg is released from rest at height 5 m. Neglect air resistance. What is its speed just before hitting the ground? / एक द्रव्यमान m = 2 kg को ऊँचाई 5 m पर विश्राम से छोड़ा जाता है। वायु प्रतिरोध न मानें। जमीन से ठीक पहले इसकी वेग कितनी होगी?
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Using energy conservation: m g h = (1/2) m v^2 ⇒ v = sqrt(2 g h) = sqrt(2 × 9.8 × 5) ≈ sqrt(98) ≈ 9.90 m/s. / ऊर्जा संरक्षण: m g h = (1/2) m v^2 ⇒ v = sqrt(2 g h) = sqrt(2 × 9.8 × 5) ≈ sqrt(98) ≈ 9.90 m/s।
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A spring with k = 200 N/m is compressed by 0.1 m. How much work is required to compress it quasistatically from natural length? / k = 200 N/m वाला एक स्प्रिंग 0.1 m से संकुचित किया जाता है। प्राकृतिक लंबाई से इसे क्वासिस्टेटिक रूप से संकुचित करने में कितना कार्य चाहिए?
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Elastic potential U = (1/2) k x^2 = 0.5 × 200 × (0.1)^2 = 1.0 J. Work required = 1.0 J. / लोचीय संभावित ऊर्जा U = (1/2) k x^2 = 0.5 × 200 × (0.1)^2 = 1.0 J। आवश्यक कार्य = 1.0 J।
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A car of mass 1200 kg climbs a hill at constant speed 10 m/s. If the slope causes it to gain height at rate 0.5 m/s, find the power required to overcome gravity (ignore friction). / 1200 kg द्रव्यमान वाली एक कार 10 m/s की स्थिर गति से एक पहाड़ी पर चढ़ती है। ढलान के कारण ऊँचाई प्रति सेकेंड 0.5 m बढ रही है, गुरुत्वाकर्षण को पार करने के लिए आवश्यक शक्ति कितनी है (घर्षण न मानें)?
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Power P = m g (dh/dt) = 1200 × 9.8 × 0.5 = 5880 W ≈ 5.88 kW. / शक्ति P = m g (dh/dt) = 1200 × 9.8 × 0.5 = 5880 W ≈ 5.88 kW।
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A satellite of mass m in circular orbit radius r has kinetic energy K and potential energy U. Show relation between K and U and give total energy. / त्रिज्या r पर वृत्ताकार कक्ष में घूर्णन करने वाले द्रव्यमान m के उपग्रह की गतिज ऊर्जा K और संभावित ऊर्जा U हैं। K और U के बीच संबंध दर्शाइए और कुल ऊर्जा लिखिए।
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For circular orbit, v^2 = G M / r. Thus K = (1/2) m v^2 = (1/2) G M m / r. Potential U = −G M m / r. Hence K = −(1/2) U and total energy E = K + U = −(1/2) G M m / r. / वृत्ताकार कक्षा के लिए v^2 = G M / r। अतः K = (1/2) m v^2 = (1/2) G M m / r। संभावित ऊर्जा U = −G M m / r। इसलिए K = −(1/2) U और कुल ऊर्जा E = K + U = −(1/2) G M m / r।
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A force of 30 N acts at angle 60° above horizontal and moves an object 5 m horizontally. Calculate the work done by the force. / 30 N का एक बल क्षैतिज के 60° ऊपर क्रिया करता है और वस्तु को 5 m क्षैतिज रूप से स्थानांतरित करता है। बल द्वारा किया गया कार्य गणना कीजिए।
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Work W = F s cosθ = 30 × 5 × cos60° = 150 × 0.5 = 75 J. / W = F s cosθ = 30 × 5 × cos60° = 150 × 0.5 = 75 J।
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A 0.5 kg mass attached to spring k = 80 N/m oscillates with amplitude 0.1 m. Find maximum speed. / k = 80 N/m के स्प्रिंग से जुड़ा 0.5 kg द्रव्यमान 0.1 m व्यापकता के साथ दोलन करता है। अधिकतम वेग ज्ञात कीजिए।
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Total energy E = (1/2) k A^2 = 0.5 × 80 × 0.01 = 0.4 J. At equilibrium all energy is kinetic: (1/2) m vmax^2 = 0.4 ⇒ vmax = sqrt(2 × 0.4 / 0.5) = sqrt(1.6) ≈ 1.2649 m/s. / कुल ऊर्जा E = (1/2) k A^2 = 0.5 × 80 × 0.01 = 0.4 J। समतल पर पूरी ऊर्जा गतिज होगी: (1/2) m vmax^2 = 0.4 ⇒ vmax = sqrt(2 × 0.4 / 0.5) = sqrt(1.6) ≈ 1.2649 m/s।
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A truck applies brakes and does 1.2 × 10^5 J of work to stop from speed 20 m/s. Find truck mass. / एक ट्रक ब्रेक लगाकर 20 m/s की गति से रुकने के लिए 1.2 × 10^5 J काम करता है। ट्रक का द्रव्यमान ज्ञात कीजिए।
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Initial kinetic energy (1/2) m v^2 = 1.2 × 10^5 J ⇒ m = 2 × 1.2 × 10^5 / v^2 = 2.4 × 10^5 / 400 = 600 kg. / प्रारंभिक गतिज ऊर्जा (1/2) m v^2 = 1.2 × 10^5 J ⇒ m = 2 × 1.2 × 10^5 / v^2 = 2.4 × 10^5 / 400 = 600 kg।
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An electric motor rated 2 kW raises water at rate 0.02 m^3/s to a height of 15 m. If density of water is 1000 kg/m^3 and motor efficiency is 80%, determine whether motor is adequate. / 2 kW रेटेड एक इलेक्ट्रिक मोटर पानी को 0.02 m^3/s की दर से 15 m ऊँचाई तक उठाती है। पानी का घनत्व 1000 kg/m^3 और मोटर दक्षता 80% है, परखा जाए कि क्या मोटर पर्याप्त है।
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Mass flow rate ṁ = ρ Q = 1000 × 0.02 = 20 kg/s. Required mechanical power Preq = ṁ g h = 20 × 9.8 × 15 = 2940 W. Considering efficiency 80%, input power needed = Preq / η = 2940 / 0.8 = 3675 W ≈ 3.675 kW. Motor 2 kW is insufficient. / द्रव्यमान प्रवाह ṁ = ρ Q = 1000 × 0.02 = 20 kg/s। आवश्यक यांत्रिक शक्ति Preq = ṁ g h = 20 × 9.8 × 15 = 2940 W। दक्षता 80% को ध्यान में रखते हुए आवश्यक इनपुट शक्ति = Preq / η = 2940 / 0.8 = 3675 W ≈ 3.675 kW। 2 kW मोटर अपर्याप्त है।
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A block of mass 4 kg slides down a rough incline of height 3 m. Its speed at bottom is 6 m/s. Find work done by friction. / 4 kg का एक ब्लॉक 3 m ऊँचाई वाले खुरदरे तिरछे से नीचे फिसलता है। नीचे इसकी गति 6 m/s है। घर्षण द्वारा किया गया कार्य ज्ञात कीजिए।
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Initial energy Ui + Ki = m g h + 0 = 4 × 9.8 × 3 = 117.6 J. Final K = (1/2) m v^2 = 0.5 × 4 × 36 = 72 J. Work by friction Wf = ΔEmech = Kf + Uf − (Ki + Ui) = 72 + 0 − 117.6 = −45.6 J. So friction did −45.6 J (i.e., dissipated 45.6 J). / प्रारंभिक ऊर्जा Ui + Ki = m g h + 0 = 4 × 9.8 × 3 = 117.6 J। अंतिम K = (1/2) m v^2 = 0.5 × 4 × 36 = 72 J। घर्षण द्वारा काम Wf = ΔEmech = Kf + Uf − (Ki + Ui) = 72 + 0 − 117.6 = −45.6 J। अतः घर्षण ने −45.6 J किया (यानि 45.6 J नष्ट किया)।
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