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Chapter 1 — Force, Work, Power and Energy

Class 10 · Physics

Overview

This unit covers the fundamental concepts of force, work, energy and power. It begins with the idea of force as an interaction that can change the motion or shape of an object, introduces different types of forces including contact and non-contact forces, and explains how forces are represented as vectors. The unit develops the ideas of equilibrium, friction and circular motion, and then moves to work as the transfer of energy by a force acting through a displacement. Kinetic and potential energy are defined, with the principle of conservation of mechanical energy explained and applied to simple systems such as a falling body and a spring. Power is introduced as the rate at which work is done, with practical calculations for machines and human activity. The unit emphasises problem solving using formulas for work, energy and power, and shows how energy conversions underpin machines, transport and everyday devices. Understanding this unit matters because it links observable pushes and pulls to quantitative measures of how things move and how energy is stored and used. These concepts are central to mechanics, help in analysing simple structures and machines, and prepare students for advanced physics topics and real-world applications like engines, brakes and energy-efficient design.

Learning Objectives

  • Define force, work, energy and power in physical terms and state their SI units.
  • Represent forces as vectors and resolve a force into components in two dimensions.
  • Apply Newtonian ideas of equilibrium to understand balanced and unbalanced forces.
  • Calculate work done by a constant force and by variable forces along a straight line.
  • Distinguish between kinetic and potential energy and compute values in simple systems.
  • Use the work-energy theorem and the conservation of mechanical energy to solve problems.
  • Explain frictional forces and compute the work done against friction and the resulting energy changes.
  • Define power, calculate average and instantaneous power in practical situations, and relate them to efficiency.

Topics in this chapter

18 topics · tap a topic title to jump straight to it.

💪1

Introduction to Force

What is a force?
In physics, a force is an interaction that can change the state of motion of an object or change its shape. Forces are vector quantities; they have both magnitude and direction. Everyday examples include the push of a door, the pull of a rope, the weight of an object due to gravity, and the electrostatic attraction between charged objects.

Effects of forces
A force can produce three basic effects: it can start motion (set a stationary object moving), it can stop motion (bring a moving object to rest), or it can change the direction or speed of motion (accelerate or decelerate). A force can also change the shape of an object, for example compressing a spring or stretching a rubber band.

Contact and non-contact forces
Contact forces arise when objects touch each other: friction, normal reaction, tension, and applied pushes or pulls. Non-contact forces act at a distance: gravitational force, electrostatic force, and magnetic force. Understanding whether a force requires contact helps in modelling physical situations.

Representation and measurement
Because force is a vector, we draw it as an arrow whose length represents the magnitude and whose direction shows the line of action. The SI unit of force is newton (N). One newton is that force which gives a mass of one kilogram an acceleration of one metre per second squared. Quantitative measurement uses instruments such as spring balances or force sensors calibrated in newtons.

Net force and motion
The net force on an object is the vector sum of all forces acting on it. If the net force is zero, the object remains in its current state: stationary or moving with constant velocity. If the net force is non-zero, the object accelerates in the direction of the net force. This idea links force to motion and prepares for later topics on Newton's laws and energy.

📌 Examples
  • A book of mass 0.5 kg is pushed horizontally with a force of 2 N. Describe qualitatively how the book will move.
  • A magnet attracts a steel pin across a small gap. Identify the type of force and explain why it acts without contact.
🧮 Formulas
  1. Unit of force: newton, N
  2. 1 N = 1 kg m s^-2
📊 Visual ideas
A free-body diagram of a block on a horizontal surface showing applied force, friction, and normal reaction.
A vector diagram showing two forces on an object and their resultant by the parallelogram or triangle rule.
🔬2

Newton's Laws and Equilibrium

Newton's first idea: inertia
Objects resist changes to their motion. A body at rest tends to remain at rest and a body moving at constant velocity tends to continue doing so unless acted on by a net external force. This resistance is inertia and is related to mass: more mass means more inertia. In practice, inertia explains why heavy objects need larger forces to change their speed or direction.

Second law: quantitative link
Newton's second law gives a quantitative relation between net force, mass and acceleration. For a particle, the vector equation F_net = m a holds. Here F_net is the vector sum of all forces on the body, m is the inertial mass and a is the acceleration vector. This law allows us to compute acceleration when forces are known, or to predict the required force to produce a given acceleration. It also gives the unit of force, the newton.

Third law: action and reaction
For every force that one body exerts on a second, the second exerts an equal and opposite force on the first. These are called action–reaction pairs. They act on different bodies so they do not cancel in the free-body diagram of a single object. Examples include a book pushing down on a table while the table pushes up with equal magnitude; a rocket pushing exhaust gases backwards while gases push the rocket forward.

Equilibrium conditions and types
An object is in equilibrium when the net force acting on it is zero. Static equilibrium refers to objects at rest; dynamic equilibrium refers to objects moving with constant velocity. For equilibrium in a plane, the vector condition ΣF_x = 0 and ΣF_y = 0 must both hold. For rigid bodies the net torque must also be zero to prevent rotation; hence conditions for translational and rotational equilibrium are applied together in many problems.

Free-body diagrams and solving problems
To apply these laws, begin with a clear free-body diagram showing all forces including weight, normal, friction, tension and applied forces. Choose axes conveniently and resolve forces into components. Use ΣF = m a or ΣF = 0 for equilibrium. For connected systems (strings, pulleys) relate accelerations and tensions by geometric constraints. Check units and directions; sign errors are common if axes are chosen carelessly.

Examples of common exercises
Problems often ask for tensions in strings at angles, normal reaction on inclined planes, acceleration of blocks under given forces, and conditions for tipping vs sliding. When several forces act, constructing component equations and solving simultaneous equations yields the unknowns. Understanding these basics is essential before tackling energy methods which frequently use the same free-body diagrams to identify work done by forces.

📌 Examples
  • A block hangs from two strings at angles 30° and 60°. Find the tensions if the block is in equilibrium.
  • A ladder leans against a wall with friction. Sketch forces and state equilibrium conditions.
🧮 Formulas
  1. Equilibrium conditions: ΣFx = 0, ΣFy = 0
  2. Net force: F_net = ma
📊 Visual ideas
Free-body diagram of a body under three coplanar forces showing vector sum zero.
Tension-force triangle for two strings holding a weight.
🛞3

Friction: Types and Laws

What is friction?
Friction is a resistive force that acts opposite to the relative motion (or tendency to move) between two surfaces in contact. It arises because surfaces are not perfectly smooth; microscopic irregularities interlock and resist sliding. Friction converts ordered kinetic energy into thermal energy, so it often opposes motion and causes energy dissipation.

Static and kinetic friction
Static friction acts when two surfaces are at rest relative to each other; it adjusts in magnitude up to a maximum to prevent motion. Kinetic (sliding) friction acts when they move relative to each other and typically has a smaller constant value than the maximum static friction. The coefficients of static and kinetic friction, µs and µk, depend on the pair of materials and surface conditions.

Laws and formulae
The maximum static friction force is approximately Fs,max = µs N, where N is the normal reaction. The kinetic frictional force is Fk = µk N. Both are proportional to N and do not depend strongly on the contact area for rigid bodies; however, they depend on surface roughness and normal force. Friction is almost independent of sliding speed at ordinary speeds but can vary in special cases.

Effects and applications
Friction enables walking, braking and gripping, but also causes wear and energy loss. Engineers may reduce friction via lubrication, smooth surfaces, or ball bearings, or increase friction by using ridged soles or rough surfaces for safety. In energy problems, work done against friction reduces mechanical energy and must be accounted for when using conservation laws.

Problem solving hints
Always draw the normal reaction explicitly, compute N (often equal to weight component perpendicular to plane), and then multiply by µ to find friction. Check whether static friction can hold before replacing it with kinetic friction. For inclined planes, find the component of weight parallel to the plane and compare with maximum static friction to determine motion.

📌 Examples
  • A block of mass 4 kg rests on a horizontal surface with µs = 0.4 and µk = 0.30. Find the minimum horizontal force to start motion and the frictional force once sliding.
  • A box on a 30° incline starts to slide. Given µk, calculate acceleration down the plane.
🧮 Formulas
  1. Maximum static friction: Fs,max = µs N
  2. Kinetic friction: Fk = µk N
📊 Visual ideas
Diagram of a block on an inclined plane showing weight components, normal reaction and friction direction.
Graphical comparison showing Fs rising with applied force until Fs,max, then dropping to Fk when motion starts.
💪4

Circular Motion and Centripetal Force

Uniform circular motion: qualitative view
When an object moves in a circle at constant speed, its direction constantly changes. Although the speed remains constant, velocity does not because velocity includes direction. This change in velocity means the object is accelerating; the acceleration is always directed towards the centre of the circle. This inward acceleration is called centripetal acceleration and is responsible for turning the motion around the circle.

Quantitative form of centripetal acceleration
For uniform circular motion with speed v and radius r, centripetal acceleration has magnitude a_c = v^2 / r. The direction is radial, pointing to the centre. This formula can be derived by considering the change in velocity vector over a small angle and dividing by the small time interval to find acceleration; the v^2/r result follows from geometry and limits.

Centripetal force: origin and examples
The required inward force to produce centripetal acceleration is called centripetal force and equals F_c = m v^2 / r. It is not an additional kind of force but the name given to the net inward force. Different physical forces can provide the necessary inward force: tension in a string for a whirled stone, friction between tyres and road for a turning car, normal force from a banked road component, gravity for planets and satellites.

Non-uniform circular motion
If the speed changes while moving in a circle, there is a tangential acceleration component in addition to the radial component. The total acceleration vector has both radial (v^2/r) and tangential (dv/dt) parts; they must be combined vectorially. Problems involving speed change require treating both components when applying Newton's laws.

Limits, safety and consequences
Since F_c ∝ v^2, doubling speed requires four times the centripetal force. This explains why vehicles must slow down on sharper curves. If the available frictional force is insufficient, the vehicle will skid outward. In vertical circles, maintaining contact requires minimum speed; if speed is too low at the top of a loop the normal force can become zero and contact is lost. Engineers design roads and banking angles to provide safe centripetal components without relying solely on friction.

Problem solving steps
Identify the source of inward force, draw forces acting on the object, write equations for radial components summing to m v^2/r, and include tangential components if speed changes. Check units and consider limiting conditions such as maximum static friction or tension breaking strength. Use free-body diagrams carefully since action–reaction pairs act on different bodies and should not be mixed in the same equilibrium equation.

📌 Examples
  • A 0.2 kg stone is whirled in a horizontal circle of radius 0.5 m at speed 6 m s^-1. Find the tension in the string neglecting gravity.
  • A car of mass 1000 kg takes a circular bend of radius 50 m at 18 m s^-1. Calculate the required centripetal force provided by friction.
🧮 Formulas
  1. Centripetal acceleration: a_c = v^2 / r
  2. Centripetal force: F_c = m v^2 / r
📊 Visual ideas
Sketch of uniform circular motion with velocity vector tangent to circle and centripetal acceleration arrow pointing to centre.
Free-body diagram for a car on a banked curve showing weight, normal reaction and friction components supplying centripetal force.
⚙️5

Work: Definition and Calculation

Work as energy transfer
Work is a measure of energy transfer due to a force acting through a displacement. If a constant force F acts on a body and the body undergoes a straight-line displacement s, the work done by the force equals the dot product of force and displacement: W = F s cosθ, where θ is the angle between the force vector and the displacement vector. This scalar product picks out only the component of force along the displacement. The SI unit of work is the joule (J), where 1 J = 1 N·m.

Signs of work and physical meaning
Work can be positive, negative or zero depending on direction. When the force has a component along the displacement (θ < 90°), work is positive and the body gains energy from the force. When the force opposes displacement (θ > 90°), work is negative and energy is taken away from the object (for example, friction does negative work). When the force is perpendicular to displacement (θ = 90°), work is zero; for instance, the normal reaction on a horizontally sliding block does no work if there is no vertical displacement.

Work by variable forces
When force varies with position along the path, the total work is the integral of force along the displacement: W = ∫ F(x) dx over the path. Graphically, this equals the area under the force–displacement curve. A common classroom example is the spring: force varies linearly with displacement so the triangular area under the F–x graph gives work. For piecewise constant forces, compute work for each segment and sum the results.

Path dependence and conservative vs non-conservative forces
Some forces, such as gravity and ideal spring force, are conservative: work done depends only on initial and final positions, not on the path taken. Non-conservative forces like friction do path-dependent work; the longer the path, the greater the energy dissipated. When solving problems, recognise which forces are conservative to use potential energy methods, and calculate work explicitly for non-conservative forces.

Practical computation tips
Always resolve forces into components parallel to the chosen displacement and apply W = F_parallel × s. Pay attention to sign conventions: choose consistent positive direction for displacement and for work contributions. Convert all quantities to SI units before calculating. Where a force acts at an angle or along an incline, first resolve it; for circular or curved paths, approximate by small straight segments and integrate or use energy methods for efficiency.

Common classroom problems
Typical problems require calculating work done lifting objects against gravity, work done moving objects along inclines against friction, work done by varying forces such as springs, and net work when several forces act. Use diagrams to visualise displacements and forces. In many situations, combining work calculations with energy principles (work–energy theorem) simplifies solving for speeds or distances.

📌 Examples
  • A horizontal force of 10 N displaces a box by 3 m in the same direction. Compute the work done.
  • A crate is lifted vertically upwards by 2 m with a constant upward force equal to its weight. Find the work done against gravity.
🧮 Formulas
  1. Work by constant force: W = F s cosθ
  2. Unit of work: joule (J), 1 J = 1 N m
  3. Work by variable force: W = ∫ F(x) dx
📊 Visual ideas
Force–displacement graph for a spring showing linear variation; area under curve equals work done.
Diagram showing a force at angle θ to the displacement vector and resolution of force into parallel component.
6

Kinetic Energy and Work–Energy Theorem

Kinetic energy defined
Kinetic energy is the energy possessed by an object due to its motion. For a body of mass m moving at speed v, its kinetic energy (KE) is defined as KE = 1/2 m v^2. The SI unit is the joule. KE is a scalar and always non-negative. A stationary object has zero kinetic energy.

Work–energy theorem
The work–energy theorem links the net work done on an object to the change in its kinetic energy. It states that the net work done by all forces acting on a particle equals the change in its kinetic energy: W_net = Δ(1/2 m v^2). This theorem is useful to find velocities or distances when forces act and can be simpler than using equations of motion when forces are variable.

Derivation sketch and meaning
Using Newton's second law, F = ma, and writing acceleration as v dv/dx for motion along a line, integrating F dx gives ∫ m v dv = 1/2 m v^2 change. Thus work equals change in kinetic energy. Physically, positive net work increases kinetic energy (speed up), while negative net work reduces it (slow down).

Applications
The work–energy theorem is used to calculate stopping distance when brakes apply frictional force, the speed gained by a falling object when starting from rest, and the speed of a vehicle after applying a driving force over a distance. When non-conservative forces do work (like friction), that work appears as a negative contribution to net work and reduces kinetic energy.

Problem solving hints
Compute net work by summing works from each force along the displacement, then equate to change in kinetic energy. Ensure signs are correct: forces opposite to displacement give negative work. Remember to include work done by gravity or normal reaction if they have non-zero components along displacement.

📌 Examples
  • A 2 kg body accelerates from 3 m s^-1 to 7 m s^-1. Find the change in kinetic energy and net work done.
  • A car of mass 1000 kg slows from 20 m s^-1 to rest under braking force of 4000 N. Find stopping distance using work–energy theorem.
🧮 Formulas
  1. Kinetic energy: KE = 1/2 m v^2
  2. Work–energy theorem: W_net = Δ(1/2 m v^2)
📊 Visual ideas
Graph showing kinetic energy versus speed (parabolic dependence) for a fixed mass.
Force–displacement diagram for braking showing area equal to negative work and change in KE.
7

Gravitational Potential Energy

Potential energy near Earth's surface
Gravitational potential energy (GPE) represents the energy stored in an object because of its position in a gravitational field. Near Earth's surface, for modest heights where g may be treated constant, lifting a mass m by height h increases its potential energy by ΔU = m g h, measured relative to a chosen reference level. The reference (zero) can be ground, floor or any convenient level; only changes in potential energy are physically significant.

Sign conventions and work by gravity
If the upward vertical direction is taken as positive, then raising an object increases its gravitational potential energy and the work done by gravity is negative: W_gravity = -m g Δh. Conversely, when an object falls, gravity does positive work equal to the decrease in potential energy. Keep signs consistent: ΔU = U_final - U_initial and W_gravity = -ΔU. This connection simplifies many calculations of motion.

Conservative nature and path independence
Gravity is a conservative force: the work done by gravity between two points depends only on their vertical positions and not on the path taken. Because of this, gravitational potential energy can be defined unambiguously (once a zero is chosen) and energy methods can replace force integration in many problems. This property is especially useful in complex geometries where calculating the work directly would be cumbersome.

Using gravitational potential energy in problem solving
To analyse motion involving changes in height, combine kinetic and potential energies in the total mechanical energy: E = 1/2 m v^2 + m g h. For conservative systems, E remains constant. If non-conservative forces like friction act, include their work as an energy loss: E_initial - W_nc = E_final. This method allows straightforward computation of speeds at different heights without solving differential equations of motion.

Applications and everyday examples
GPE explains why hydroelectric dams store energy at height, why a ball gains speed when dropped, and how potential energy in elevated masses can do useful work later. In designs such as roller-coasters or elevators, potential energy changes determine required speeds and power ratings. In physics problems, choose the zero of potential energy to simplify calculations; often the lowest point in motion is taken as zero.

Numerical considerations and approximations
Use g = 9.8 m s^-2 for accurate results or g = 10 m s^-2 for quick estimates when permitted. For large heights comparable to Earth's radius, g changes with height and the simple m g h formula must be replaced by the general gravitational potential energy expression; this is beyond Class 10 requirements but useful to know.

📌 Examples
  • A 0.5 kg mass is lifted from the floor to a shelf 1.5 m high. Find the increase in its gravitational potential energy (take g = 9.8 m s^-2).
  • A roller coaster car of mass 500 kg is at a height 20 m. Compute its potential energy relative to ground.
🧮 Formulas
  1. Gravitational potential energy near Earth: PE = m g h
  2. Change in PE: ΔPE = m g Δh
📊 Visual ideas
Diagram of object raised to height h from reference level showing work done and increase in PE.
Energy bar chart comparing kinetic and potential energy at two positions during vertical motion.
8

Elastic Potential Energy and Springs

Hooke's law: spring force
Many elastic objects such as ideal springs exert a restoring force proportional to displacement from equilibrium for small deformations. Hooke's law gives this relation: F = -k x, where k is the spring constant (stiffness) and x is the extension or compression measured from the natural length. The negative sign indicates the restoring nature: the force acts opposite to the displacement.

Work done in stretching a spring
Because the spring force varies with displacement, the work done to stretch or compress a spring from x = 0 to x = X is found by integrating the force over displacement: W = ∫_0^X k x dx = 1/2 k X^2. This area under the F–x graph is a triangle with base X and height k X, giving the same result. This stored energy is called elastic potential energy and equals 1/2 k x^2 at displacement x.

Energy exchange in mass–spring systems
In a frictionless mass–spring system, energy oscillates between kinetic energy of the mass and elastic potential energy of the spring. At maximum extension or compression the kinetic energy is zero and the elastic potential is maximum. At equilibrium position the spring potential is zero and the kinetic energy is maximum. The total mechanical energy E = 1/2 m v^2 + 1/2 k x^2 remains constant if no non-conservative forces act, explaining simple harmonic motion behaviour at small amplitudes.

Limits of linear behaviour
Hooke's law holds only within the elastic limit of the material: beyond a certain displacement the relation becomes non-linear and permanent deformation can occur. Real springs also have damping and internal friction that convert some mechanical energy into heat over repeated cycles; ideal classroom springs ignore these effects unless specified.

Practical problem-solving and examples
Typical Class 10 problems include calculating energy stored for a given compression, finding maximum compression of a spring when a mass collides with it using energy conservation, and relating force required to a displacement using Hooke's law. When combining springs in series or parallel, effective spring constants change: in series 1/k_eq = 1/k1 + 1/k2, in parallel k_eq = k1 + k2. These relations allow designers to tune stiffness for desired response.

Applications in engineering and daily life
Springs are used in vehicle suspensions, watches, weighing scales, and shock absorbers. Understanding elastic energy helps in designing cushions and energy-absorbing structures. In many safety systems, springs help store and release energy smoothly to reduce peak forces and protect components.

📌 Examples
  • A spring with k = 200 N m^-1 is compressed by 0.1 m. Find the elastic potential energy stored.
  • A block of mass 0.5 kg compresses a spring and then moves off. Given initial speed, find maximum compression using energy conservation.
🧮 Formulas
  1. Hooke's law: F = -k x
  2. Elastic potential energy: PE_elastic = 1/2 k x^2
📊 Visual ideas
Force versus displacement graph for a spring: straight line from origin with slope k; area under curve equals 1/2 k x^2.
Mass-spring system diagram showing conversion between kinetic and elastic potential energy during oscillation.
9

Conservation of Mechanical Energy

Statement and scope
The principle of conservation of mechanical energy states that in a system where only conservative forces (such as gravity and ideal spring force) act, the total mechanical energy remains constant. Mechanical energy is the sum of kinetic and potential energies: E = K + U, where K = 1/2 m v^2 and U includes gravitational and elastic potential terms. This conservation holds provided no non-conservative forces like friction or air resistance do net work.

Using conservation in problem solving
To apply the principle, identify initial and final states and write E_initial = E_final, expanding into kinetic and potential terms as appropriate: 1/2 m v1^2 + m g h1 + 1/2 k x1^2 = 1/2 m v2^2 + m g h2 + 1/2 k x2^2. If non-conservative forces are present, include the work they do: E_initial - W_nc = E_final, where W_nc is work done by non-conservative forces (positive if they remove mechanical energy).

Advantages over force methods
Energy methods often simplify problems where forces vary or motion involves changes in height and spring compression. Instead of integrating forces or solving differential equations, equate energies to find unknown speeds or displacements. For example, the speed of a block sliding down a frictionless incline can be found using conversion of potential to kinetic energy without resolving forces along the plane.

Limitations and careful use
Conservation of mechanical energy applies only when non-conservative work is negligible or accounted for. When friction acts, simply equating KE and PE will give incorrect results; instead subtract energy lost to friction. Also, total mechanical energy conservation does not mean all energy forms are unchanged: mechanical energy may convert into internal energy (heat) even if total energy is conserved overall.

Common classroom applications
Problems include objects falling and reaching speed at bottom, pendulums swinging between heights, mass–spring oscillations, and blocks sliding on curved tracks. For systems combining springs and gravity, include both potential energies. Check units and choose convenient reference levels for potential energy to simplify algebra.

Example approach
To solve, list known quantities, choose zero of potential energy, write initial and final energy expressions, include any work by non-conservative forces, and solve algebraically for the unknown. This method is especially efficient in exam conditions for standard energy conversion problems.

📌 Examples
  • A 2 kg block slides down a frictionless 5 m high incline. Find its speed at the bottom using energy conservation.
  • A mass released from rest compresses a spring. Use energy conservation to find maximum compression if initial height and speed are known.
🧮 Formulas
  1. Total mechanical energy (conservative forces only): E = 1/2 m v^2 + m g h + 1/2 k x^2
  2. If friction does work W_f: E_initial - W_f = E_final
📊 Visual ideas
Energy bar diagram showing KE and PE at top and bottom of a fall.
Plot of total mechanical energy remaining constant while KE and PE exchange values over time.
⚙️10

Power: Rate of Doing Work

Definition and units
Power measures how quickly work is done or energy is transferred. Average power P_avg over a time interval Δt is defined as P_avg = W / Δt, where W is work done in that time. Instantaneous power is the limit as Δt → 0 and equals P = dW/dt. The SI unit of power is watt (W), where 1 W = 1 J s^-1. Larger practical units include kilowatt (kW) and horsepower (hp).

Relation to force and velocity
When a force F acts on an object moving with instantaneous velocity v, the instantaneous power delivered by the force is P = F · v, the scalar product of force and velocity. For motion in one dimension where force and velocity are parallel, this reduces to P = F v. This relation is useful when forces and speeds are given and power output must be found.

Average and instantaneous measures
Average power is useful for processes taking a finite time, like lifting a mass over several seconds. Instantaneous power applies to changing situations and can be found by differentiating work with time or using P = F v when force and velocity vary with time. In constant-speed situations, P = F v gives steady power required to overcome resistive forces like friction or air drag at that speed.

Efficiency and useful power
Machines convert input power to useful output power; efficiency η is the ratio η = P_out / P_in and is often expressed as a percentage. Losses due to friction, heat and sound reduce efficiency. For example, an electric motor rated at 1 kW may deliver less useful mechanical power if some energy becomes heat.

Applications and problem solving
Problems include computing power required to lift loads at given speeds, the power output of engines, and work done over time. Convert units carefully (1 hp ≈ 746 W). For human tasks, typical continuous power outputs are a few hundred watts during heavy exertion. Use P = W/Δt or P = F v as appropriate.

Examples for conversion
Relate rotational and translational power: for a rotating shaft P = τ ω where τ is torque and ω angular speed in rad s^-1. For a wheel of radius r, a tangential force F produces torque τ = F r and linear speed at rim v = ω r so P = F v, consistent across descriptions.

📌 Examples
  • A lift raises a 500 kg mass at constant speed 2 m s^-1. Find the power required to overcome gravity.
  • A motor delivers 1500 W and lifts 200 kg by 10 m. Find time taken ignoring losses.
🧮 Formulas
  1. Average power: P = W / Δt
  2. Instantaneous power: P = F · v
  3. Efficiency: η = P_out / P_in
📊 Visual ideas
Graph of power versus time during a machine cycle showing average and peak power values.
Sketch of force and velocity vectors showing P = F · v.
11

Energy Transformations and Conservation of Energy

Forms of energy and conversions
Energy appears in many forms: mechanical (kinetic and potential), thermal, chemical, electrical, nuclear, and radiant. Processes convert energy between these forms. For example, a battery converts chemical energy to electrical energy; a motor converts electrical energy to mechanical energy; and combustion converts chemical energy into thermal energy and mechanical work. Recognising the forms present in a situation helps in accounting for energy flows.

The general law of conservation of energy
The total energy of an isolated system remains constant: energy can neither be created nor destroyed, only transformed or transferred. This general principle includes mechanical energy conservation as a special case when only conservative forces act. In any process, if mechanical energy decreases, the missing energy appears in other forms such as heat, sound or chemical change.

Energy bookkeeping and practical examples
When solving problems, list each form of energy at initial and final times and include work done by external forces. For example, in braking, the kinetic energy lost by the vehicle is converted into thermal energy in the brakes and tyres and partly into sound. In electrical devices, input electrical energy is split into useful work and waste heat; the useful fraction determines efficiency. Drawing energy flow diagrams clarifies where energy goes and prevents double counting.

Efficiency, losses and real devices
No real device is 100% efficient; some input energy is always dissipated. Efficiency η = (useful energy output)/(input energy) quantifies performance. Engineers work to reduce losses by improving materials, lubrication, aerodynamics and insulation. In transport, reducing air resistance and rolling friction gives large savings because these losses grow with speed.

Everyday consequences and environmental relevance
Understanding energy conversions explains why power consumption matters, why insulation reduces heating costs, and how renewable sources change the dominant forms of available energy. For example, solar panels convert radiant energy to electrical energy, while wind turbines convert kinetic energy of air to mechanical and then electrical energy. Energy conservation laws allow estimation of how much useful output is possible from a given input and why large-scale energy efficiency gains are important for sustainability.

Practical problem solving tips
Identify all significant energy forms, include non-conservative work as losses, and use conservation of total energy when appropriate. For exam problems, clearly state assumptions (no friction, or include frictional work) and show calculations converting between joules, watts and other units as needed.

📌 Examples
  • A falling object loses potential energy; calculate the kinetic energy gained and thermal energy if there is friction.
  • A motor with 2 kW input power does useful work of 1.4 kW. Find its efficiency.
🧮 Formulas
  1. Conservation of total energy: E_total(initial) = E_total(final)
  2. Efficiency η = (useful energy output / energy input) × 100%
📊 Visual ideas
Energy flow diagram for a car showing chemical → thermal → mechanical → kinetic and heat losses.
Bar-chart showing distribution of input energy into useful output and losses for a device.
💪12

Work Done by Gravity and Conservative Forces

Work done by gravity
The work done by gravity when an object moves between two heights depends only on the vertical displacement, not on the specific path. If an object moves from height h1 to h2, the work done by gravity is W_gravity = m g (h1 - h2). This sign convention means that when the object falls (h2 < h1), W_gravity is positive and kinetic energy increases by that amount.

Conservative forces and potential energy
Conservative forces allow defining a potential energy function U such that the work done by the force when moving from A to B equals U(A) - U(B) = -ΔU. For gravity near Earth's surface, U = m g h. Because work by conservative forces is path independent, energy methods are powerful for analyzing systems with gravity and springs.

Closed-path property
For conservative forces, the work done around any closed path is zero. That is, moving a body in a closed loop returns the potential energy to its initial value with no net work done by the conservative force. This property simplifies calculations and explains why only potential differences matter physically.

Using gravity work in problems
When computing net work or energy changes, replace work by gravity with the negative change in potential energy: W_gravity = -ΔU = -m g Δh. In combined problems, include other works explicitly (friction etc.) and use the relation E_initial - W_nc = E_final to solve for unknown speeds or heights. This reduces computation by avoiding integration of the weight vector along complex paths.

Practical examples and checks
Carrying an object along a stairway then down a ramp returns the object to the original height; the net work by gravity in the closed loop is zero. When lifting objects slowly, the person does positive work equal to increase in potential energy; gravity does equal negative work. Keep units consistent and choose convenient zero of potential energy to simplify algebra in exam problems.

📌 Examples
  • A ball is carried along a curved path from height 0.5 m to 2.0 m. Compute work done by gravity.
  • A stone falls from 10 m to 3 m. Find the work done by gravity and change in potential energy.
🧮 Formulas
  1. Work done by gravity between heights h1 and h2: W = m g (h1 - h2)
  2. Gravitational potential energy: U = m g h
📊 Visual ideas
A path-independent illustration showing two different paths between same heights and equal work done by gravity.
Energy-level diagram marking initial and final potential energies and the work done.
🏃13

Power in Rotational and Translational Motion

Power for translational motion
For linear motion, instantaneous power delivered by a force is P = F · v. When the force is parallel to velocity, P = F v. This relation is widely used for moving objects, lifts and conveyor belts to compute required engine or motor ratings.

Rotational analogy
For rotational motion about a fixed axis, torque τ plays the role of force and angular velocity ω plays the role of linear velocity. The instantaneous power delivered to a rotating body equals P = τ ω. If a constant torque rotates an object at angular speed ω, the power is constant and equals product of torque and angular speed. This relation follows from power = torque × angular velocity because work done by torque in a small rotation dθ is dW = τ dθ and dividing by dt gives P = τ dθ/dt = τ ω.

Converting between linear and rotational forms
For a wheel of radius r, a tangential force F produces torque τ = F r and linear speed at rim v = ω r. Substituting gives P = F v, consistent with translational expression. This shows consistency between linear and rotational descriptions and is useful when analysing motors, gears and pulleys.

Applications and unit conversions
When given torque and rotational speed in revolutions per minute (RPM), convert RPM to angular speed using ω = 2π (RPM)/60. Then compute P = τ ω. Keep units: torque in N·m, ω in rad s^-1, power in W. For example, a motor providing constant torque at higher RPM delivers proportionally higher power until limits of torque or speed are reached.

Problem solving tips
Identify motion type and use P = F v for translation or P = τ ω for rotation. For combined systems, convert rotational quantities to linear equivalents at contact points to check consistency. When asked for power required to maintain constant speed against resistive torques or forces, multiply resisting force or torque by the constant speed or angular speed respectively.

📌 Examples
  • A motor provides torque 20 N m at 1500 RPM. Find the power output in kilowatts.
  • A force of 50 N acts tangentially on a wheel of radius 0.2 m causing it to rotate at speed 10 rad s^-1. Compute the power.
🧮 Formulas
  1. Rotational power: P = τ ω
  2. Relation between torque and tangential force: τ = F r
  3. Convert RPM to rad s^-1: ω = 2π (RPM) / 60
📊 Visual ideas
Diagram of a rotating wheel showing torque, angular velocity and tangential velocity at the rim.
Plot showing power versus angular speed for a machine at constant torque; linear relation.
⚙️14

Practical Machines: Simple Machines and Efficiency

What simple machines do
Simple machines—lever, pulley, inclined plane, wheel and axle, wedge and screw—do not create energy; they change the magnitude or direction of applied forces so tasks become easier. They allow trade-offs: a smaller input force acting through a larger distance can balance a larger load moved through a smaller distance. Understanding these trade-offs is central to solving practical problems where effort, distance and force are related.

Ideal machines and mechanical advantage
In an ideal (frictionless) machine work input equals work output: F_in × d_in = F_out × d_out. Therefore the ideal mechanical advantage (IMA) is F_out / F_in = d_in / d_out. For example, in a ramp of length L and height h, the ideal effort force to raise weight W is W × h / L, showing the familiar reduction of force required at the cost of larger distance travelled. Pulleys multiply rope segments supporting the load to reduce effort proportionally to the number of supporting strands in ideal cases.

Real machines and efficiency
Real machines suffer energy losses due to friction, deformation and other dissipative processes. Efficiency η is defined as useful work output divided by work input (or useful power out divided by power in). It is always less than 100% in practice. For a given output requirement, an inefficient machine demands higher input energy or power, and wastes more energy as heat or sound. Calculating efficiency helps compare devices and size motors or engines appropriately.

Design considerations and trade-offs
Engineers design machines balancing mechanical advantage, speed of operation, weight, size and efficiency. A gearbox may trade torque for speed, while a pulley system trades rope length for reduced effort. Sometimes a lower mechanical advantage is chosen to allow faster operation or to reduce structural stresses. Safety margins account for material strength and possible overloads.

Solving classroom problems
Identify input and output forces and distances, draw diagrams showing distances moved, apply ideal work equality to find ideal effort, then include efficiency to get realistic inputs: F_in × d_in = (F_out × d_out)/η. For pulleys and levers, count supporting strands or measure arm lengths. For inclined planes include frictional work separately as energy lost. Clearly state assumptions: ideal vs real, friction neglected or included, constant speed or accelerating.

Examples in everyday life
Inclined planes are used in ramps and roads; levers in crowbars and see-saws; pulleys in cranes; screws in jacks and bottle caps. Understanding how they alter force and distance allows practical calculations, for instance estimating effort needed to raise a load with a given ramp or finding required motor power to lift goods at a steady rate.

📌 Examples
  • An inclined plane of length 5 m and height 1 m is used to lift a 200 kg crate. Find ideal force required ignoring friction and compute work done.
  • A pulley system reduces effort by factor 4. If load is 400 N, what is ideal input force and rope displacement ratio?
🧮 Formulas
  1. Work input = Work output (ideal): F_in d_in = F_out d_out
  2. Efficiency: η = (useful work output / work input) × 100%
📊 Visual ideas
Diagram of a lever showing effort arm and load arm and relation for mechanical advantage.
Sketch of pulley system with number of supporting ropes indicated and mechanical advantage count.
15

Energy Losses: Heat, Sound and Deformation

Non-conservative forces and energy dissipation
In real systems non-conservative forces such as friction, air resistance and inelastic deformation convert organised mechanical energy into less organised forms like heat, internal energy and sound. These processes are called dissipation. While total energy is conserved, mechanical energy decreases and the lost portion appears as thermal energy or other non-mechanical forms, often increasing entropy.

Frictional heating and wear
When surfaces slide, microscopic asperities interlock and shear, converting kinetic energy into heat. The work done against friction equals the thermal energy generated (neglecting small losses to sound and light). Repeated friction leads to wear and material degradation. Understanding the amount of energy converted into heat is important for brake design, machining, and maintaining mechanical systems to avoid overheating.

Air resistance and aerodynamic drag
Objects moving through air suffer drag forces whose magnitude depends on speed, shape and fluid properties. At moderate to high speeds drag often scales approximately as v^2, so the power needed to overcome drag increases roughly with v^3 (since P = F v ≈ c v^3). This explains why vehicles become much less fuel-efficient at higher speeds: small increases in speed demand disproportionately large engine power to overcome air resistance.

Inelastic collisions and deformation
In collisions where bodies stick or deform plastically, some kinetic energy is irreversibly converted into heat and internal energy and may also cause permanent deformation. Energy absorbed in deformation is harnessed in safety devices like car crumple zones, which aim to convert kinetic energy into controlled deformation rather than letting forces transmit to occupants. Elastic collisions, by contrast, conserve kinetic energy (idealised case).

Accounting for losses in calculations
In energy balance problems include work done by non-conservative forces as subtractive terms: E_initial - W_nc = E_final. For braking, the stopping distance from initial speed v uses 1/2 m v^2 = F_brake × d where F_brake is average braking force and the left side becomes heat produced. For vehicles, include rolling resistance and aerodynamic drag to estimate fuel consumption and required engine power.

Mitigation and engineering responses
To reduce losses engineers use lubrication, streamlined shapes, low-friction coatings and bearings, and materials engineered for minimal deformation. Energy recovery systems such as regenerative braking convert part of the otherwise lost kinetic energy back into stored electrical energy, improving overall system efficiency. Recognising major loss pathways guides better design and energy-saving strategies.

📌 Examples
  • A block slides 10 m under a constant frictional force of 20 N. Compute thermal energy produced due to friction.
  • A cyclist rides at higher speed and experiences increased air resistance power loss proportional to v^3; explain qualitatively why energy cost rises rapidly.
🧮 Formulas
  1. Work done against friction: W_f = F_f × s
  2. Power to overcome drag (approx.): P_drag ∝ v^3 (for constant drag coefficient and frontal area)
📊 Visual ideas
Plot showing increasing power required versus speed due to air resistance.
Diagram of energy flow showing mechanical input, useful work and losses to heat and sound.
👑16

Applications: Braking, Springs and Collisions

Braking and energy dissipation
When brakes are applied, they exert frictional forces that do negative work on the moving vehicle, removing kinetic energy. The work done by the braking force over the stopping distance equals the initial kinetic energy lost, so 1/2 m v^2 = F_brake × d (using average braking force F_brake). This relation gives a direct way to estimate stopping distance for a given braking capability or the required braking force to stop within a given distance.

Design of brake systems
Brake systems are engineered to convert kinetic energy into heat efficiently and to dissipate that heat safely. Materials and ventilation prevent overheating. In addition, modern vehicles often use regenerative braking to recover part of the kinetic energy as electrical energy stored in batteries, reducing net energy loss and improving efficiency.

Springs as energy absorbers
Springs absorb kinetic energy by converting it into elastic potential energy, given by 1/2 k x^2. In impacts, a spring can reduce peak forces by stretching the time over which energy is absorbed. For a mass colliding with a spring, equating initial kinetic energy to elastic potential energy yields maximum compression: 1/2 m v^2 = 1/2 k x^2. Damped springs also convert some energy into heat to avoid excessive oscillations.

Collisions: elastic and inelastic
Collisions are analysed by conservation of momentum always, and by conservation of kinetic energy only in elastic collisions. In inelastic collisions some kinetic energy is converted to internal energy, heat, or deformation. For two-body collisions in one dimension, use momentum conservation and, if elastic, kinetic energy conservation to solve for final velocities. If completely inelastic, bodies stick together and move with common velocity determined by momentum conservation.

Practical problems and safety
Crashworthy structures are designed to deform in controlled ways, converting kinetic energy into deformation work to protect occupants. In sports, helmets and padding use materials that absorb energy to reduce impact forces. Understanding the energy transformations involved enables calculation of forces, distances and deformations relevant for safety design.

Problem solving approach
Identify whether collision is elastic or inelastic, decide which conservation laws apply, and write equations for momentum and energy accordingly. For braking and springs, use work–energy relations to link forces, distances and energies, and compute required parameters such as stopping distance or spring compression.

📌 Examples
  • A car of mass 1200 kg at 20 m s^-1 is braked to rest by constant force 6000 N. Find stopping distance.
  • A mass 0.2 kg moving at 5 m s^-1 collides with a spring k = 500 N m^-1. Find maximum compression assuming no losses.
🧮 Formulas
  1. Stopping distance from work–energy: d = (1/2 m v^2) / F_brake
  2. Maximum spring compression from KE: 1/2 m v^2 = 1/2 k x^2
📊 Visual ideas
Sketch of car kinetic energy vs distance during braking showing linear decrease if braking force constant.
Diagram of spring compression under impact showing conversion of kinetic to elastic potential energy.
🔬17

Estimations and Order-of-Magnitude Calculations

Purpose of estimation
Estimation and order-of-magnitude calculations give quick, approximate answers that help check detailed work, compare alternatives and understand the scale of physical quantities. They are particularly useful in exams for verifying whether a detailed calculation is reasonable and in real life when precise data are unavailable.

Techniques for quick estimates
Use rounded values for physical constants (for example g ≈ 10 m s^-2) and round masses and distances to one or two significant figures for rough work. Convert units into SI early and use dimensional analysis to ensure equations are consistent. For uncertainty, keep track of factors of 2 or 10; an order-of-magnitude estimate usually aims to be correct within a factor of about 3 or 10 depending on the context.

Common references and benchmarks
Memorise a few useful benchmarks: lifting 1 kg by 1 m requires about 10 J, an average human sustained power is a few hundred watts, a 100 W bulb uses 100 J every second, and a car at highway speed uses kilowatts of power. These help judge whether computed energies or powers are sensible. For example, if an appliance claims to do mechanical work equivalent to several kilowatts but draws only tens of watts, the claim is suspect.

Examples of estimation problems
Estimate energy to lift a class of 30 students by 1 m: take average mass 60 kg so total mass 1800 kg, energy ≈ m g h ≈ 1800 × 10 × 1 = 18,000 J. Estimate battery life: a 40 W fan running for 5 hours uses 40 × 5 × 3600 = 720,000 J which is 0.2 kWh. These simple computations give immediate practical insights.

Using approximations sensibly
Be explicit about assumptions: neglect air resistance, assume constant acceleration, use average forces, etc. Compare results using two different reasonable approximations to see sensitivity. If the estimate is used to decide engineering choices, refine the model in steps but keep the initial estimate as a sanity check.

Exam strategy
When time is limited, use estimation to check if answer is in the right ballpark and to catch arithmetic errors. Show steps and assumptions briefly to gain method marks. Estimation skills improve with practice and help build intuition for more complex physics problems.

📌 Examples
  • Estimate the energy required to lift 100 people by 2 m in a school assembly hall. Use mass per person ≈ 60 kg and g = 10 m s^-2.
  • Estimate the power needed for a small electric fan rated 40 W to run for 5 hours and the energy consumed.
🧮 Formulas
  1. Approximate GPE: E ≈ m g h with g ≈ 10 m s^-2 for quick estimates
  2. Power estimate: Energy ≈ Power × time
📊 Visual ideas
No diagram required beyond simple bar comparisons of energy magnitudes for everyday activities.
🔬18

Revision: Solving Combined Problems

Overview and strategy
Combined problems test the ability to use forces, work, energy and power together. The best approach is structured: read the problem carefully, list knowns with units, draw clear diagrams (free-body and energy-level), and decide which principles apply—Newton's laws for forces and acceleration, work–energy theorem for net work and change in kinetic energy, or conservation of mechanical energy when only conservative forces act.

Choosing the simplest method
Often energy methods are simpler when the question asks for speed, height, or maximum compression because they avoid integrating forces or solving time-dependent equations. Use Newton's laws when acceleration or tensions are required at particular instants. For problems involving dissipative forces like friction, combine energy and work methods: write E_initial - W_nc = E_final, where W_nc is work by non-conservative forces computed from frictional force times distance or by integrating a varying resistive force.

Step-by-step problem solving
1. Draw clear diagrams: free-body for forces, energy bars for initial and final energies.
2. Choose coordinate axes and reference levels for potential energy.
3. Identify conservative and non-conservative forces.
4. Apply appropriate equations: ΣF = m a, W_net = ΔK, or E_initial - W_nc = E_final. Solve algebraically for unknowns, keeping units consistent. Use approximations only when justified and state them.

Checking and interpreting results
After solving, check dimensions and limiting cases: results should reduce sensibly when a parameter goes to zero (e.g., zero friction should recover frictionless values). Estimate order of magnitude to ensure answer is realistic. Also verify sign conventions: speeds must be non-negative and energies should not be negative if the reference is set sensibly.

Worked example types
Examples include a block sliding down an incline with friction where one calculates speed at bottom (use energy minus friction work), a mass released compressing a spring (use conservation between KE and spring PE), connected masses and pulleys where tensions and accelerations are found by Newton's laws and then speeds by work–energy, and lifting problems where power and time are asked using P = W/t.

Exam technique and presentation
Write each step clearly: state the physical law used, show substitutions with units, and give final answers with appropriate significant figures and units. Label diagrams and justify approximations. This structured approach gains method marks even if arithmetic slips occur and builds confidence for varied exam questions.

📌 Examples
  • A 5 kg block slides down 4 m of an incline, with µk = 0.2, starting from rest. Find speed at bottom using energy methods including work done by friction.
  • A motor lifts 300 kg load at constant speed through 10 m in 20 s. Calculate required power and energy consumed.
🧮 Formulas
  1. Work–energy: W_net = Δ(1/2 m v^2)
  2. Include non-conservative work: E_initial - W_nonconservative = E_final
📊 Visual ideas
Combined free-body and energy diagram showing forces, displacements and energy conversions for a sliding block problem.

Key Concepts

Force
An interaction that can change an object's motion or shape, represented as a vector with SI unit newton.
Work
The product of component of force along displacement and the displacement, measured in joules.
Energy
A scalar quantity representing the capacity to do work, existing in kinetic, potential and other forms.
Power
The rate at which work is done or energy is transferred, measured in watts.
Kinetic Energy
Energy of motion given by 1/2 m v^2 for a mass m moving at speed v.
Potential Energy
Energy due to position; gravitational potential energy near Earth is m g h.
Conservative Force
A force for which work done is path independent and a potential energy function exists.
Friction
A non-conservative contact force opposing relative motion, characterized by coefficients µs and µk.
Centripetal Force
The net inward force causing circular motion, magnitude m v^2 / r.
Hooke's Law
For small deformations of a spring, restoring force F = -k x where k is the spring constant.
Work–Energy Theorem
The net work done on a body equals its change in kinetic energy.
Efficiency
The ratio of useful output energy or power to input, usually expressed as a percentage.
Mechanical Advantage
The factor by which a machine multiplies the input force in ideal conditions.
Elastic Potential Energy
Energy stored in a spring given by 1/2 k x^2 for displacement x.
Normal Reaction
The contact force perpendicular to a surface that supports an object against gravity.

Practice Questions

  1. A force of 15 N acts on a 3 kg box and moves it through 4 m in the direction of the force. Calculate the work done. / 15 N का बल 3 kg के एक डिब्बे पर लगता है और इसे बल की दिशा में 4 m तक ले जाता है। किया गया कार्य कितने जूल है?
    Show answer

    Work = F s = 15 × 4 = 60 J / कार्य = F s = 15 × 4 = 60 J

  2. A block of mass 2 kg falls from height 5 m. Find its speed just before hitting the ground neglecting air resistance. / 2 kg का एक ब्लॉक 5 m की ऊँचाई से गिरता है। हवा के प्रतिरोध की उपेक्षा करते हुए जमीन से ठीक पहले उसकी गति बताइए।
    Show answer

    Using energy conservation: m g h = 1/2 m v^2, v = sqrt(2 g h) ≈ sqrt(2 × 9.8 × 5) ≈ 9.9 m s^-1 / ऊर्जा संरक्षण से: v = sqrt(2 g h) ≈ sqrt(2 × 9.8 × 5) ≈ 9.9 m s^-1

  3. A car of mass 800 kg takes a circular bend of radius 40 m at speed 20 m s^-1. Calculate the centripetal force needed. / 800 kg की कार 40 m त्रिज्या के गोल मोड़ को 20 m s^-1 की गति से लेती है। आवश्यक केन्द्राभिमुख बल ज्ञात कीजिए।
    Show answer

    F_c = m v^2 / r = 800 × 20^2 / 40 = 800 × 400 / 40 = 8000 N / F_c = m v^2 / r = 800 × 400 / 40 = 8000 N

  4. A spring with k = 250 N m^-1 is compressed by 0.08 m. Find the potential energy stored. / k = 250 N m^-1 वाले एक स्प्रिंग को 0.08 m तक संपीडित किया जाता है। इसमें संग्रहीत संभाव्य ऊर्जा कितनी होगी?
    Show answer

    Elastic PE = 1/2 k x^2 = 0.5 × 250 × (0.08)^2 = 125 × 0.0064 = 0.8 J / PE = 1/2 k x^2 = 0.8 J

  5. A worker lifts a 50 kg crate vertically by 1.5 m in 4 s. Calculate the power developed (take g = 9.8 m s^-2). / एक कर्मी 50 kg का डिब्बा 1.5 m ऊर्ध्वाधर रूप से 4 s में उठाता है। विकसित शक्ति ज्ञात कीजिए (g = 9.8 m s^-2 लें)।
    Show answer

    Work = m g h = 50 × 9.8 × 1.5 = 735 J. Power = W / t = 735 / 4 = 183.75 W ≈ 184 W / कार्य = 735 J, शक्ति = 735 / 4 ≈ 184 W

  6. A 5 kg block slides down a 6 m long incline of height 3 m with coefficient of kinetic friction 0.2. Find its speed at bottom. / 5 kg का एक ब्लॉक 6 m लंबी तिरछी सतह से 3 m ऊँचाई तक सरकता है, जहाँ µk = 0.2 है। तिरछे के नीचे इसकी गति ज्ञात कीजिए।
    Show answer

    Loss of PE = m g h = 5 × 9.8 × 3 = 147 J. Work done against friction = F_f × s, normal reaction N = m g cosθ. Here sinθ = 3/6 = 0.5, cosθ = sqrt(1-0.25)=0.866. N = 5 × 9.8 × 0.866 ≈ 42.4 N. Friction = µk N ≈ 0.2 × 42.4 ≈ 8.48 N. Work against friction = 8.48 × 6 ≈ 50.9 J. Available KE = 147 - 50.9 = 96.1 J. So 1/2 m v^2 = 96.1, v^2 = 2 × 96.1 / 5 = 38.44, v ≈ 6.2 m s^-1 / ΔPE = 147 J, घर्षण के विरुद्ध कार्य ≈ 50.9 J, शेष KE = 96.1 J, अतः v ≈ 6.2 m s^-1

  7. A motor delivers 2 kW of useful power with an efficiency of 80%. What is the electrical power input? / एक मोटर 80% दक्षता पर 2 kW उपयोगी शक्ति देता है। विद्युत् इनपुट शक्ति कितनी होगी?
    Show answer

    Efficiency η = P_out / P_in. So P_in = P_out / η = 2000 W / 0.8 = 2500 W = 2.5 kW / P_in = 2.5 kW

  8. Explain why walking on a rough surface is possible though friction opposes motion. / रगड़ वाली सतह पर चलना संभव क्यों है जबकि घर्षण गति का विरोध करता है?
    Show answer

    When we walk, our foot pushes backward on the ground; static friction from the ground acts forward on our foot, propelling the body forward. Static friction prevents slipping and provides the forward push even though friction resists relative motion; thus friction enables walking. / जब हम चलते हैं, हमारा पैर जमीन पर पीछे की ओर दबाता है; जमीन से प्राप्त स्थैतिक घर्षण हमारे पैर पर आगे की दिशा में बल देता है जो शरीर को आगे बढ़ाता है। स्थैतिक घर्षण फिसलने से रोकता है और इसी कारण चलना संभव होता है।

  9. A cyclist of effective mass 70 kg rides steadily at 10 m s^-1 and overcomes resistive forces totaling 50 N. Find power output. / प्रभावी द्रव्यमान 70 kg वाला एक साइकिल चालक 10 m s^-1 की स्थिर गति पर चल रहा है और कुल प्रतिरोधी शक्ति 50 N है। शक्ति ज्ञात कीजिए।
    Show answer

    Power = F v = 50 × 10 = 500 W. / शक्ति = F v = 500 W

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