Power Power Power is about speed — not speed of motion, but speed of doing work. If two students lift identical bags to the same height, both do the same work, yet the one who finishes sooner has delivered more power. In mechanics, work measures energy transfer from a force acting through a displacement. Power measures the rate of that transfer. This “rate” idea matters in everyday life: a 1000 W electric kettle boils water faster than a 500 W one; a 3 kW motor can pull an elevator more briskly than a 1 kW motor (for the same load). In motion, a force that lines up with the velocity pumps energy into a body; a force opposite to the velocity drains energy out. That is why accelerating a car at high speed demands a lot more power than at low speed — because power depends on both force and instantaneous velocity. Thinking in terms of power makes many problems simpler: you can balance input power from an engine against power needed to climb a hill and fight friction; or relate braking power to how quickly a vehicle’s kinetic energy is being reduced. Units and ratings then come naturally: the watt (W) is joule per second, kilowatt (kW) scales this for machines, and horsepower (hp) is a legacy unit engineers still use (1 hp ≈ 746 W). Big-picture: Work tells how much energy changed; Power tells how fast it changed. Same work in half the time means double the power. remember Power Rate of doing work or transferring energy. If work changes by W in time t , average power P avg = W t ; instantaneous power P= d W d t . Total work done divided by total time interval: P avg = W t . Useful for coarse comparisons over an interval. Average Power Power at an instant: P= d W d t = F v . It tells how fast energy is flowing right now. Instantaneous Power Horsepower (hp) A legacy unit of power. 1 , hp =746 , W approximately. Efficiency How much of input power becomes useful output power. = P out P in 100 % . SI unit of power: 1 , W =1 , J/s . Watt The designed operating power of a machine or appliance (e.g., 750 W mixer, 2 kW motor). It often indicates how quickly the device can do its job under nominal conditions. Rating (of a device) We care about both average and instantaneous views. Average power tells how demanding a task is over a whole interval — like your average study pace in a day. Instantaneous power is like your pace at a particular minute. Many mechanics questions ask for instantaneous power because forces and speeds change with time. The bridge between force and power comes from the definition of work: if a small displacement d r occurs while a force F acts, the small work is d W= F d r . Dividing by d t gives P= F d r d t = F v . The dot product captures alignment: only the component of force along velocity contributes to power. Average power is the total work done per total time interval. Average power Instantaneous power and force–velocity relation Instantaneous power equals the dot product of force and velocity. tip Applicability: P= F v uses the instantaneous F and v of the same particle at the same instant. If force and motion are not along one line, take the component of F along v . If they are perpendicular (e.g., ideal centripetal force), the instantaneous power is zero. Work element: ( d W = F d r ). Velocity: ( v = d r d t ). Force and displacement are defined for the same particle and instant. P = F v P = F v Average power over an interval is total work divided by time: P avg = W t . Units: J/s = W . In a 1D collision, e= v 2 - v 1 u 1 - u 2 (component along line of impact). Cross-link: power is about rate; collisions change energy suddenly, so average power over the short impact is large, even though we mainly track momentum and restitution. e = v 2 - v 1 u 1 - u 2 e = v 2 - v 1 u 1 - u 2 1D collision or components along the line of impact. Relative velocities measured along the line of impact. No external impulse along line of impact during collision (isolated pair). When force and velocity make an angle , only the along-motion component F contributes. Then P=Fv . Positive power adds energy (force component along velocity); negative power removes energy (opposing component), as in braking or kinetic friction. Only the component of force along the velocity does work per unit time. Power with oblique force Use the angle to determine if the force component is adding energy (positive power) or removing energy (negative power). neet-alert Sign trap: If >90 , is negative, so power is negative (energy is being taken out of the body). Many errors come from forgetting this sign in P=Fv . Examples of positive power: a motor pulling a lift up, a runner accelerating forward. Negative power: kinetic friction acting on a sliding block, air drag on a fast bike, brakes on a car. Zero power: a perfect centripetal force in uniform circular motion (force is radial, velocity is tangential, so F v ). Power Rate of doing work d W d t W (J/s) Instantaneous via F v Average Power P avg W t Over a time interval Energy E, W Capacity to do work 1 , kWh =3.6 10 6 , J Horsepower hp Legacy unit of power 1 , hp =746 , W Efficiency P out P in Often quoted in % Quantity Symbol Definition SI Unit Notes Where power shows up in mechanics Climbing or lifting against gravity: P mgh/ t (average), or P=mgv (instant). Towing or driving at steady speed with resistive forces: P=F res v . Braking and frictional losses: negative power equals rate of mechanical energy loss. Power-limited acceleration: if engine power is capped at P 0 , then a= P 0 mv in straight-line motion without other losses. Energy consumption is often given in kilowatt-hour (kWh). This is a unit of energy, not power. If a device of constant power P runs for time t , the energy used is E=Pt . For example, a 1000 W heater running for 2 h uses 2 , kWh =7.2 10 6 , J . Distinguish carefully: power is the rate now; energy accumulates over time. This formula is valid when the power output (P) is constant over the entire time interval (t). For variable power, the integral form W = Energy is the time integral of power; for constant power, it is the simple product. Energy–power relation A useful result for straight-line motion under a power-limited engine is a= P mv . It follows from P=Fv with F=ma . It explains why vehicles pick up speed rapidly at low v but gain slowly at high v when power is fixed. With fixed available power P and negligible losses, acceleration falls inversely with v . Acceleration under power cap This formula applies when the power supplied to an object is held constant, causing the acceleration to be inversely proportional to the obj Boundary conditions for a= P mv : assumes straight-line motion, power delivered entirely to translation (no rotation), and negligible resistive forces. As v 0 , the formula suggests very large a , but in reality traction limits and engine torque curves cap the force. tip Average power against gravity A 60 kg student runs up a 5.0 m high staircase in 10 s. Find the average power developed against gravity. Take g=9.8 , m/s 2 . m = 60 , kg h = 5.0 , m g = 9.8 , m/s 2 t = 10 , s Work done against gravity is W=mgh . Average power P avg = W t . easy Even though the student’s instantaneous power output likely varied during the run, the average over 10 s is 294 W. If the same climb is done in 5 s, the average power would double to about 590 W, even though the work (mgh) is unchanged. A 1000 kg car accelerates uniformly from rest to 20 m/s in 10 s on a level road with negligible resistance. Find the instantaneous power delivered by the tractive force at t=10 , s . Instantaneous power at 10 s medium With uniform acceleration: a= v-u t . Instantaneous power P=Fv= m a v . m = 1000 , kg u = 0, ; v = 20 , m/s t = 10 , s Road level, resistances negligible control derived P = F v At rest, P = 0 Example operating point F×20 20 Power P (at constant F) custom At constant F, power demand grows linearly with speed. P increases linearly with v for a fixed applied force along motion. Speed v m/s 2D PLOT Power vs speed at constant force P = F v Applied force The linear P – v graph at constant force explains why towing or pushing at higher speed requires more power. If resistive force itself grows with speed (e.g., air drag v 2 ), then P v 3 — power demand becomes very steep at high speeds. For power-limited motion: a= P mv and d v d t = P mv . Rearranging: d t= m P v , d v and integrate. hard m = 1200 , kg P = 60 , kW = 6.0 10 4 , W v 1 = 10 , m/s , ; v 2 = 25 , m/s Level road, no resistive forces Time to increase speed from 10 m/s to 25 m/s A vehicle of mass 1200 kg is powered by an engine that can supply a constant 60 kW to translation on a level road (ignore losses). Starting from 10 m/s, how long will it take to reach 25 m/s? Notice how the time depends on the difference of squared speeds, not on the average acceleration. A power-limited vehicle gains speed quickly at low v but slows its rate of increase at higher v because each unit of speed requires more power. Power depends on both force and velocity: P=Fv . For a given power, the force you can apply falls as speed rises, and vice versa. Power and force are the same. More power means more force. Instantaneous power can be nonzero even if average work over an interval is zero. Also, in uniform circular motion, force does no work and instantaneous power is zero, yet the force is nonzero. If work is zero, power must be zero. neet-alert Average vs instantaneous trap: P avg = W t applies over a time interval. Do not plug average speed into P=Fv unless the force is constant and aligned throughout and you explicitly want a time-averaged estimate. Power–Work–Time triangle: cover the quantity you want. P = W/t, W = P×t, t = W/P. Write P on top, W and t at the bottom corners. Machines have efficiencies less than 100%. If a motor draws P in from the source and delivers P out to the load, then = P out P in . The rest is lost as heat, sound, or vibration. Always distinguish input vs output power in questions. Applies to any system undergoing energy conversion, such as machines, electrical circuits, or engines, where useful output power is derived Fraction of input power that becomes useful output. Efficiency in power terms A 2.0 kW motor of efficiency 80% lifts a 200 kg load vertically at constant speed. Find the speed of lifting. Take g=9.8 , m/s 2 . Lifting speed v medium Output power P out = P in =mgv . Solve for v . m/s P in = 2.0 , kW = 2000 , W = 80 % = 0.80 m = 200 , kg g = 9.8 , m/s 2 Constant speed (no acceleration) Checklist for power problems Decide: average or instantaneous power? If instantaneous, identify the forces and the velocity at that instant. Project the force along the velocity: use P=Fv . Include efficiency if device ratings are given (input vs output). For power-limited motion, use a= P mv and integrate if needed. For energy comparisons, use W= P , d t or K= F d r and connect to power. Use the visualizer to feel P= F v . Keep F aligned with motion and increase v : power rises. Flip F against motion: power turns negative. Set F v : measured power hovers near zero (up to numerical noise). W/2 Half the time → double power W/5 Slower task → lower power W (fixed) control P avg = W/Δt derived Time taken Δt Average Power P avg for fixed W Average power vs time for a fixed amount of work. custom Hyperbola: as Δt increases, P avg decreases inversely. Fixed work spread over more time means lower average power. Many NEET questions change only the time to finish the same task; then P avg scales inversely with the time factor. In uniform circular motion, instantaneous power of the centripetal force is zero because F v . Yet maintaining the motion in the real world costs power due to non-idealities (air drag, bearing friction). remember Collision note (cross-link): During a short impact, forces are large and time is tiny. The average power over that tiny interval can be enormous, even though in solving collisions we mainly conserve momentum and use restitution e . Power is still the rate-of-energy lens on the same event. Edge cases to watch At v=0 , P= F v =0 even if force is huge; power rises only when motion begins. If =180 (purely opposing force), P=-Fv : energy drains at rate Fv . If P is fixed and v 0 , the formula a= P mv suggests very large a ; in reality traction and torque limits cap the actual force. Electrical side note: for a DC circuit element with voltage V and current I , electrical power is P=VI . If the element is a motor of efficiency , the mechanical output power is VI . This ties electric ratings directly to mechanical performance. Electrical input power equals product of voltage and current in DC. Electrical power (DC) This formula accurately describes the instantaneous power dissipated in a DC circuit where voltage and current are measured simultaneously. Device ratings help compare performance: kettles (1–2 kW), room heaters (1–2 kW), mixer grinders (500–750 W), window ACs (1–2 kW input but higher cooling power in kW of heat removal). Mechanical systems like elevators list motor kW and load capacity — from which lifting speed can be estimated using P=mgv/ . neet-alert Do not confuse kW and kWh. kW is a unit of power (rate); kWh is a unit of energy (amount). Electricity bills charge per kWh (units of energy), not per kW. Worked-energy view vs power view: The work–energy theorem says K= F d r . Differentiating in time gives d K d t = F v =P . Power is thus the time rate of change of kinetic energy due to the net force. When potential energies are present, mechanical power balances can include storage terms as well. Climbing at constant speed v against gravity needs P=mgv (ignoring losses). On an incline of angle , the component against you is mg , so P=(mg )v at constant speed. Add rolling resistance or air drag by summing forces along motion before applying P=Fv . This formula calculates the mechanical power required to lift an object of mass m up an incline of angle at a constant velocity v. Climb power on incline Minimum power to maintain steady climb (no acceleration, no other losses). Braking power quantifies how fast kinetic energy is being removed. If a car of mass m decelerates with constant a at speed v , the braking force is F b=ma opposite to motion, so instantaneous braking power is P b=-F b v=-m a v (negative). Brake design must dissipate this as heat. Negative sign indicates energy removal from the moving vehicle. Braking power This formula calculates the instantaneous power dissipated by a resistive force (like friction or air resistance) acting opposite to the dir Household example: A 60 W bulb used for 5 h consumes 60 , W 5 , h =300 , Wh =0.30 , kWh . A 1500 W geyser for 20 minutes uses 1.5 , kW ( 1 3 , h )=0.5 , kWh . Same energy as running five 100 W bulbs for an hour. High-speed vehicles face air drag power losses scaling roughly as P drag v 3 (since F drag v 2 ). Doubling speed needs about eight times drag power. This is why top speed is primarily power-limited rather than torque-limited. In rotating systems, an analogous formula is P= , , where is torque and is angular speed (radians per second). Although rotation is a later topic, this mirrors P=Fv and is widely used for motors and turbines. Rotational analog of P=Fv (for a rigid body rotating about a fixed axis). Rotational power (preview) This formula applies to any object undergoing rotational motion where the torque and angular velocity are defined relative to a fixed axis. When a force varies with time, power varies too. If F(t)=kt acts along the motion of a mass starting from rest (no other forces), then a(t)= F m = kt m and v(t)= k t 2 2m . The instantaneous power is P(t)=Fv= k 2 t 3 2m , which grows rapidly with time. Conversely, if power is prescribed as a function of time, P(t) , you can integrate to find work: W= 0 T P(t) , d t . In many thermal/engine contexts, this is how total energy output is estimated from a time-varying power curve. Dimensional check: Power has dimensions [M L 2 T -3 ] . From P=Fv , dimensions are [F][v]=[M L T -2 ][L T -1 ]=[M L 2 T -3 ] , consistent. From P= W t , we get [M L 2 T -2 ]/[T]=[M L 2 T -3 ] , also consistent. Graph interpretation skill: If you plot kinetic energy K vs time t , the slope d K d t at a point equals the net power into translation at that instant. A steeper slope means higher power input; a negative slope means net power loss (e.g., braking). Practical estimation: To raise an elevator (500 kg cabin + 300 kg load) at 1.0 m/s with 75% efficiency, minimum motor input power is P in = mgv = (800)(9.8)(1.0) 0.75 10.4 , kW . Such back-of-the-envelope checks are common in NEET numericals. Rating headroom: Devices are rated for continuous operation at their nominal power without overheating. Short bursts above rating may be possible but can cause thermal stress. In problems, unless stated, assume devices operate at their given rating steadily. Comparison strategy: If two machines do the same work W but in times t 1 and t 2 , then P 1 P 2 = t 2 t 1 . Halving the time doubles the average power; reducing time by a factor of 3 multiplies average power by 3, and so on. Unit conversions to memorize: 1 , kW =10 3 , W ; 1 , MW =10 6 , W ; 1 , hp =746 , W ; 1 , kWh =3.6 10 6 , J . Often a question is won or lost on these quick conversions. Mechanical vs electrical: Mechanical output power can be steady while electrical input fluctuates (or vice versa) due to control systems or losses changing with load. In idealized NEET problems, when not specified, treat given ratings as constant. Key terms recap Rate of energy transfer: P= d W d t = F v . Power rate of doing work P avg Average Power P avg = W t over a time interval t . Instantaneous Power Power at an instant; depends on alignment of force and velocity. Horsepower hp Legacy unit of power: 1 , hp =746 , W . Efficiency = P out P in 100 % .