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24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/as-level/physics/edexcel/as-work-energy-and-power/flashcards","label":"Work, energy and power","subjectCode":"as_level_physics","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"27adb224-4eb9-4093-8649-7bc102dcab3a","title":"Work, energy and power Flashcards","topicId":"b284815f-1808-5297-b7eb-c6e8315dbf2b","topicSlug":"as-work-energy-and-power","cards":[{"id":"8d73165e-07a1-486f-8693-6e59375e9c8f","cardType":"basic","points":1,"front":"What does work done represent in terms of energy?","back":"The energy transferred by a force.","mcqOptions":[{"text":"The energy transferred by a force.","correct":true,"explanation":"Work done measures the energy transferred when a force causes displacement. One joule of work corresponds to one joule of energy transferred."},{"text":"The energy stored inside a force.","correct":false,"explanation":"A force is an interaction, so it does not itself contain stored energy. Work done represents the energy transferred by a force as it causes displacement."},{"text":"The power supplied by a force.","correct":false,"explanation":"Power is the rate at which work is done, measured in watts, rather than an amount of energy. Work done is the energy transferred by a force."},{"text":"The momentum transferred by a force.","correct":false,"explanation":"Momentum transfer is associated with impulse, which equals force multiplied by time. Work done instead represents the energy transferred by a force through a displacement."}]},{"id":"21206236-27a1-4a94-8a0c-eb8b8cacb31a","cardType":"basic","points":1,"front":"State the formula for work done when a force acts at an angle to the direction of motion.","back":"$$W = Fs\\cos\\theta$","mcqOptions":[{"text":"$$W = Fs\\cos\\theta$","correct":true,"explanation":"Only the component of force parallel to the displacement does work. That component is $F\\cos\\theta$, so multiplying it by displacement gives $W = Fs\\cos\\theta$."},{"text":"$$W = Fs\\sin\\theta$","correct":false,"explanation":"Using sine selects the force component perpendicular to the displacement when $\\theta$ is measured between force and motion. The parallel component gives $W = Fs\\cos\\theta$."},{"text":"$$W = Fs$","correct":false,"explanation":"This assumes that the force acts entirely in the direction of displacement, so it is valid only when $\\theta=0^\\circ$. In general, $W = Fs\\cos\\theta$."},{"text":"$$W = \\frac{Fs}{\\cos\\theta}$","correct":false,"explanation":"Dividing by $\\cos\\theta$ incorrectly makes work increase as the force becomes more perpendicular. The parallel force component is found by multiplication, giving $W = Fs\\cos\\theta$."}]},{"id":"752978bc-19e3-4016-8859-ea787c83b59b","cardType":"cloze","points":null,"front":"One joule is the work done when a force of [ ] moves an object 1 m in the direction of the force.","back":"One joule is the work done when a force of 1 N moves an object 1 m in the direction of the force.","mcqOptions":[{"text":"1 N","correct":true,"explanation":"Using $W=Fs$ for motion parallel to the force gives $W=(1\\,\\text{N})(1\\,\\text{m})=1\\,\\text{J}$. Therefore the required force is 1 N."},{"text":"2 N","correct":false,"explanation":"A force of 2 N acting through 1 m would do 2 J of work, not 1 J. One joule requires a force of 1 N."},{"text":"0.5 N","correct":false,"explanation":"A force of 0.5 N acting through 1 m would transfer only 0.5 J. One joule is produced by a force of 1 N over this distance."},{"text":"10 N","correct":false,"explanation":"A force of 10 N acting through 1 m would do 10 J of work. The definition of one joule uses a force of 1 N."}]},{"id":"dfed311e-f341-4223-88ee-ba859a28726e","cardType":"cloze","points":null,"front":"One joule is the work done when a force of 1 N moves an object [ ] in the direction of the force.","back":"One joule is the work done when a force of 1 N moves an object 1 m in the direction of the force.","mcqOptions":[{"text":"1 m","correct":true,"explanation":"For a force parallel to motion, $W=Fs$. A 1 N force acting through 1 m therefore does $1\\,\\text{N}\\times1\\,\\text{m}=1\\,\\text{J}$ of work."},{"text":"2 m","correct":false,"explanation":"A 1 N force acting through 2 m would do 2 J of work, because $W=Fs$. One joule requires a displacement of 1 m."},{"text":"0.5 m","correct":false,"explanation":"A 1 N force acting through 0.5 m would do only 0.5 J of work. The displacement required for one joule is 1 m."},{"text":"1 cm","correct":false,"explanation":"Confusing centimetres with metres makes the displacement 100 times too small. A 1 N force acting through 1 cm does 0.01 J, whereas one joule requires 1 m."}]},{"id":"b95d097d-5354-4daa-baae-7d7cd69f219e","cardType":"basic","points":2,"front":"Why does a force acting perpendicularly to the displacement do no work?","back":"

The component of force in the direction of motion is zero.
$\\cos 90^\\circ = 0$.

","mcqOptions":[{"text":"

The component of force in the direction of motion is zero.
$\\cos 90^\\circ = 0$.

","correct":true,"explanation":"Work depends on the force component parallel to displacement. At $90^\\circ$ this component is zero because $F\\cos90^\\circ=0$, so no energy is transferred mechanically."},{"text":"

The component of force in the direction of motion is maximum.
$\\cos 90^\\circ = 1$.

","correct":false,"explanation":"The parallel component is maximum at $0^\\circ$, not $90^\\circ$. Since $\\cos90^\\circ=0$, the component along the displacement is zero and no work is done."},{"text":"

The component of force in the direction of motion is equal to the force.
$\\sin 90^\\circ = 0$.

","correct":false,"explanation":"This confuses the parallel and perpendicular components and also misstates the sine value. At $90^\\circ$, $\\cos90^\\circ=0$, so the force component along the motion is zero."},{"text":"

The component of force in the direction of motion is negative.
$\\cos 90^\\circ = -1$.

","correct":false,"explanation":"A negative parallel component occurs for angles greater than $90^\\circ$, with $\\cos180^\\circ=-1$ at the opposite direction. At $90^\\circ$, the component is zero because $\\cos90^\\circ=0$."}]},{"id":"66e54e91-0df8-4981-8d58-b843fc185c64","cardType":"basic","points":1,"front":"What does a negative value for work done indicate?","back":"The force opposes the direction of motion.","mcqOptions":[{"text":"The force opposes the direction of motion.","correct":true,"explanation":"Negative work occurs when the force has a component opposite to the displacement, so energy is transferred away from the moving object, as with friction or braking."},{"text":"The force assists the direction of motion.","correct":false,"explanation":"A force component in the direction of motion gives positive work because $\\cos\\theta$ is positive. Negative work means the force has a component opposing the motion."},{"text":"The force matches the direction of motion.","correct":false,"explanation":"When force and displacement have the same direction, $\\theta=0^\\circ$ and work is positive. A negative value indicates that the force opposes the direction of motion."},{"text":"The force is perpendicular to the direction of motion.","correct":false,"explanation":"A perpendicular force gives zero work because $\\cos90^\\circ=0$, not negative work. Negative work requires a force component acting opposite to the displacement."}]},{"id":"d0acf1f3-b768-45b4-9cb7-0c765066240d","cardType":"basic","points":1,"front":"How is work done calculated from a force-displacement graph where the force varies?","back":"Find the area under the graph.","mcqOptions":[{"text":"Find the area under the graph.","correct":true,"explanation":"Each small displacement contributes work $dW=F\\,ds$, so the total work is $\\int F\\,ds$. This integral is represented by the signed area under a force-displacement graph."},{"text":"Find the gradient of the graph.","correct":false,"explanation":"The gradient gives the rate at which force changes with displacement and may represent a spring constant. Work done is instead the area under the force-displacement graph."},{"text":"Find the final force on the graph.","correct":false,"explanation":"The final force alone ignores how the force varied during the earlier displacement. Work done is found by summing $F\\,ds$, represented by the area under the graph."},{"text":"Find the maximum height of the graph.","correct":false,"explanation":"The maximum height gives only the greatest force and does not include the distance over which forces act. Work done is the area under the force-displacement graph."}]},{"id":"fcbd550f-78c9-4dd9-8458-b4a76905b754","cardType":"cloze","points":null,"front":"When a spring is stretched, the elastic potential energy stored is the area under the [ ] graph.","back":"When a spring is stretched, the elastic potential energy stored is the area under the force-extension graph.","mcqOptions":[{"text":"force-extension","correct":true,"explanation":"The area under a force-extension graph equals $\\int F\\,dx$, which is the work done stretching the spring and therefore the elastic potential energy stored."},{"text":"force-time","correct":false,"explanation":"The area under a force-time graph gives impulse, or change in momentum, rather than energy. Elastic potential energy is the area under the force-extension graph."},{"text":"extension-time","correct":false,"explanation":"An extension-time graph describes how the spring’s extension changes over time, but its area has no direct meaning as stored energy. The required graph is force-extension."},{"text":"stress-strain","correct":false,"explanation":"The area under a stress-strain graph gives energy stored per unit volume, not the total elastic potential energy of a spring. Total energy uses the force-extension graph."}]},{"id":"9f1a3001-48d8-47bc-8561-9604cc880c8d","cardType":"basic","points":1,"front":"State the formula for kinetic energy ($E_k$).","back":"$$E_k = \\frac{1}{2}mv^2$","mcqOptions":[{"text":"$$E_k = \\frac{1}{2}mv^2$","correct":true,"explanation":"Kinetic energy equals half the product of an object’s mass and the square of its speed. Squaring speed means kinetic energy increases rapidly as speed rises."},{"text":"$$E_k = mv^2$","correct":false,"explanation":"Omitting the factor of one-half doubles the calculated energy. The correct classical expression for translational kinetic energy is $E_k = \\frac{1}{2}mv^2$."},{"text":"$$E_k = \\frac{1}{2}m^2v$","correct":false,"explanation":"This incorrectly squares the mass rather than the speed. Kinetic energy is proportional to mass and to speed squared, so $E_k = \\frac{1}{2}mv^2$."},{"text":"$$E_k = \\frac{1}{2}mv$","correct":false,"explanation":"This fails to square the speed and is dimensionally equivalent to momentum rather than energy. The correct formula is $E_k = \\frac{1}{2}mv^2$."}]},{"id":"4a93b5f8-fa71-4217-a4ee-4cc6f932bdbe","cardType":"basic","points":1,"front":"State the formula for the change in gravitational potential energy.","back":"$$\\Delta E_p = mg\\Delta h$","mcqOptions":[{"text":"$$\\Delta E_p = mg\\Delta h$","correct":true,"explanation":"Near Earth’s surface, the change in gravitational potential energy equals mass multiplied by gravitational field strength and the change in vertical height."},{"text":"$$\\Delta E_p = m\\Delta h$","correct":false,"explanation":"Leaving out gravitational field strength gives the wrong units and ignores the strength of the gravitational field. The correct relation is $\\Delta E_p = mg\\Delta h$."},{"text":"$$\\Delta E_p = \\frac{mg}{\\Delta h}$","correct":false,"explanation":"Dividing by the height change reverses the required relationship. Gravitational potential energy change is proportional to height change, giving $\\Delta E_p = mg\\Delta h$."},{"text":"$$\\Delta E_p = mg(\\Delta h)^2$","correct":false,"explanation":"Height change is not squared in a uniform gravitational field. The energy change increases linearly with vertical displacement, so $\\Delta E_p = mg\\Delta h$."}]},{"id":"24e61530-5013-40f0-9a75-0656a1ebbbd0","cardType":"basic","points":1,"front":"What is the principle of conservation of energy?","back":"Energy cannot be created or destroyed, only transferred between stores.","mcqOptions":[{"text":"Energy cannot be created or destroyed, only transferred between stores.","correct":true,"explanation":"The total energy of a closed system remains constant. Energy may move between stores or spread into the surroundings, but the total quantity is conserved."},{"text":"Energy cannot be transferred or destroyed, only created within stores.","correct":false,"explanation":"Energy transfers occur routinely, while energy creation would violate conservation. Energy cannot be created or destroyed, only transferred between stores."},{"text":"Energy can be created but not destroyed, only transferred between stores.","correct":false,"explanation":"Treating an energy input as newly created energy is a common misconception. Energy cannot be created or destroyed; it can only be transferred between stores."},{"text":"Energy can be destroyed when it is transferred into less useful stores.","correct":false,"explanation":"Dissipated energy may become less useful, but it has not been destroyed. Conservation means energy can only be transferred between stores, not created or destroyed."}]},{"id":"7596362b-8c7e-4007-8c66-04f6153ca144","cardType":"basic","points":1,"front":"What happens to energy that is dissipated in a system?","back":"It is transferred to less useful stores, typically heating the surroundings.","mcqOptions":[{"text":"It is transferred to less useful stores, typically heating the surroundings.","correct":true,"explanation":"Dissipation spreads energy into stores from which it is difficult to recover, commonly the thermal stores of the surroundings, while total energy remains conserved."},{"text":"It is transferred to more useful stores, typically cooling the surroundings.","correct":false,"explanation":"Dissipation reduces the energy available for useful work rather than increasing it. The energy enters less useful stores, usually by heating the surroundings."},{"text":"It is removed from all energy stores, typically disappearing into the surroundings.","correct":false,"explanation":"Energy does not disappear, because total energy is conserved. Dissipated energy is transferred to less useful stores, commonly the thermal stores of the surroundings."},{"text":"It is retained in useful energy stores, typically increasing the system’s efficiency.","correct":false,"explanation":"Retaining energy in useful stores would reduce dissipation and raise efficiency. Dissipated energy instead transfers to less useful stores, typically heating the surroundings."}]},{"id":"eea8e425-94f3-447c-93f9-3a81e860c953","cardType":"basic","points":1,"front":"State the work-energy principle relating resultant force to kinetic energy.","back":"Work done by the resultant force = $\\Delta E_k$.","mcqOptions":[{"text":"Work done by the resultant force = $\\Delta E_k$.","correct":true,"explanation":"The net work done by all forces equals the object’s change in kinetic energy. Positive net work increases kinetic energy, while negative net work decreases it."},{"text":"Work done by the resultant force = $E_k$.","correct":false,"explanation":"Net work determines the change in kinetic energy, not necessarily its final value. The work-energy principle is work done by the resultant force $= \\Delta E_k$."},{"text":"Work done by the driving force = $\\Delta E_k$.","correct":false,"explanation":"Using only the driving force ignores resistive forces that also do work. It is the work done by the resultant force that equals $\\Delta E_k$."},{"text":"Work done by the resultant force = $\\Delta E_p$.","correct":false,"explanation":"The work-energy principle relates net work specifically to the change in kinetic energy. Gravitational potential energy changes concern work associated with gravity, not every resultant force."}]},{"id":"43dc93d5-17d4-4369-b3a3-b245de8900cd","cardType":"cloze","points":null,"front":"In a real mechanics system with friction: $\\text{initial energy} = \\text{final useful energy} + \\text{[...]}$.","back":"In a real mechanics system with friction: $\\text{initial energy} = \\text{final useful energy} + \\boxed{\\text{energy dissipated}}$.","mcqOptions":[{"text":"$$\\text{energy dissipated}$","correct":true,"explanation":"Friction transfers some initial energy into less useful stores, usually thermal stores. Therefore, initial energy equals final useful energy plus the energy dissipated."},{"text":"$$\\text{energy created}$","correct":false,"explanation":"Friction cannot create energy, because energy is conserved. The missing quantity is energy dissipated, representing energy transferred from useful stores into less useful stores."},{"text":"$$\\text{energy destroyed}$","correct":false,"explanation":"Dissipation does not destroy energy; it transfers energy to less useful stores. The conservation equation therefore includes energy dissipated rather than energy destroyed."},{"text":"$$\\text{energy gained}$","correct":false,"explanation":"Adding a general energy gain would not account for frictional losses from useful stores. The missing term is energy dissipated, which completes the conservation balance."}]},{"id":"7832187e-e7c7-4fb6-a80e-72d77b7a7d1f","cardType":"basic","points":1,"front":"Define power in terms of energy.","back":"The rate of energy transfer (or work done).","mcqOptions":[{"text":"The rate of energy transfer (or work done).","correct":true,"explanation":"Power measures how quickly energy is transferred or work is done. It equals energy transferred divided by time and is measured in watts."},{"text":"The total amount of energy transferred (or work done).","correct":false,"explanation":"This describes energy or work, not power, because it omits the time taken. Power is the rate of energy transfer or work done."},{"text":"The force causing energy transfer (or work done).","correct":false,"explanation":"Force may cause work, but it does not by itself define power. Power is the rate at which energy is transferred or work is done."},{"text":"The distance over which energy is transferred (or work done).","correct":false,"explanation":"Distance contributes to calculating work but does not define power. Power is the rate of energy transfer or work done, so time must be considered."}]},{"id":"ab55fd7c-cb20-404f-8e48-0164e2ce0c91","cardType":"basic","points":2,"front":"State the two main formulas for power ($P$).","back":"

$P = \\frac{W}{t}$
$P = Fv$

","mcqOptions":[{"text":"

$P = \\frac{W}{t}$
$P = Fv$

","correct":true,"explanation":"Average power is work done divided by time. For a force acting parallel to an object’s velocity, instantaneous mechanical power is the product $Fv$."},{"text":"

$P = Wt$
$P = \\frac{F}{v}$

","correct":false,"explanation":"Both relationships reverse the role of time or velocity and have incorrect dimensions for power. The correct formulas are $P = \\frac{W}{t}$ and $P = Fv$."},{"text":"

$P = \\frac{t}{W}$
$P = \\frac{v}{F}$

","correct":false,"explanation":"These are reciprocal-style errors and do not produce units of watts. Power is work divided by time, and for parallel force and velocity it is $Fv$."},{"text":"

$P = \\frac{W}{t^2}$
$P = Fv^2$

","correct":false,"explanation":"Unnecessarily squaring time and velocity gives incorrect dimensions and relationships. The standard power formulas are $P = \\frac{W}{t}$ and $P = Fv$."}]},{"id":"d6386401-8bda-4f24-95cf-dd28d2107ba0","cardType":"basic","points":1,"front":"What is the unit of power, and what is its equivalent in base units?","back":"The watt (W), equivalent to $1 \\text{ J s}^{-1}$.","mcqOptions":[{"text":"The watt (W), equivalent to $1 \\text{ J s}^{-1}$.","correct":true,"explanation":"Power is the rate of energy transfer, so one watt equals one joule transferred per second, written as $1 \\text{ J s}^{-1}$."},{"text":"The watt (W), equivalent to $1 \\text{ J s}$.","correct":false,"explanation":"Multiplying joules by seconds does not describe a rate of energy transfer. Power is measured in watts, with $1 \\text{ W}=1 \\text{ J s}^{-1}$."},{"text":"The joule (J), equivalent to $1 \\text{ N m}$.","correct":false,"explanation":"A joule is the unit of energy or work, not power. Power is measured in watts, and one watt is equivalent to $1 \\text{ J s}^{-1}$."},{"text":"The newton (N), equivalent to $1 \\text{ kg m s}^{-2}$.","correct":false,"explanation":"A newton is the unit of force, so newtons per second do not give the standard unit of power. Power is measured in watts, equivalent to $1 \\text{ J s}^{-1}$."}]},{"id":"0384ebdf-af90-4871-a3c0-2bee7dd44760","cardType":"basic","points":2,"front":"Under what conditions is $P = Fv$ applied?","back":"

Constant force.
Force is in the direction of motion (or using the parallel component).

","mcqOptions":[{"text":"

Constant force.
Force is in the direction of motion (or using the parallel component).

","correct":true,"explanation":"Since power is the rate of work done, a constant force parallel to the velocity gives $P=Fv$. If the force is angled, its parallel component must be used."},{"text":"

Constant force.
Force is perpendicular to the motion (or using the perpendicular component).

","correct":false,"explanation":"A perpendicular force does no work because there is no force component along the displacement. The relation requires a constant force in the direction of motion, or its parallel component."},{"text":"

Constant acceleration.
Force is in the direction of motion (or using the parallel component).

","correct":false,"explanation":"Constant acceleration is not the required condition and usually means velocity is changing. The stated form is applied for a constant force directed along the motion, using its parallel component if necessary."},{"text":"

Constant velocity.
Force is in the opposite direction to motion (or using the antiparallel component).

","correct":false,"explanation":"A force opposite to motion gives negative power under the usual sign convention rather than the stated positive product. The relation requires constant force along the direction of motion, or its parallel component."}]},{"id":"48e3608f-40fc-4999-8f3c-ca6923d40429","cardType":"cloze","points":null,"front":"Efficiency is calculated as the ratio of [ ] energy to total input energy.","back":"Efficiency is calculated as the ratio of useful output energy to total input energy.","mcqOptions":[{"text":"useful output","correct":true,"explanation":"Efficiency compares the energy transferred into the intended, useful form with all the energy supplied to the system, so useful output energy forms the numerator."},{"text":"total output","correct":false,"explanation":"Total output can include both useful and wasted energy, so it cannot be the numerator of efficiency. The numerator must be useful output energy."},{"text":"wasted output","correct":false,"explanation":"Wasted output is specifically the energy not transferred into the desired form. Efficiency uses useful output energy divided by total input energy."},{"text":"useful input","correct":false,"explanation":"Input energy is what is supplied, not the desired result obtained from the system. Efficiency compares useful output energy with total input energy."}]},{"id":"7b89c849-26c6-4862-ac6f-23aa1ba37f94","cardType":"cloze","points":null,"front":"Efficiency is calculated as the ratio of useful output energy to [ ] energy.","back":"Efficiency is calculated as the ratio of useful output energy to total input energy.","mcqOptions":[{"text":"total input","correct":true,"explanation":"The denominator must include all energy supplied to the system, including portions later transferred usefully and wastefully. Therefore efficiency divides useful output by total input energy."},{"text":"useful input","correct":false,"explanation":"Using only the useful part of the input would omit energy eventually wasted and could make efficiency appear too high. The denominator is total input energy."},{"text":"total output","correct":false,"explanation":"Output energy belongs on the result side of the process, whereas efficiency compares the useful result with everything supplied. The denominator must be total input energy."},{"text":"wasted output","correct":false,"explanation":"Wasted output is the undesired part of the energy transfer, not the full quantity supplied. Efficiency divides useful output energy by total input energy."}]},{"id":"0e358638-2f69-458c-ba48-3924b6169f47","cardType":"basic","points":2,"front":"How do you distinguish between average power and instantaneous power?","back":"

Average: Total work done divided by total time ($W/t$).
Instantaneous: Force multiplied by velocity at a specific moment ($Fv$).

","mcqOptions":[{"text":"

Average: Total work done divided by total time ($W/t$).
Instantaneous: Force multiplied by velocity at a specific moment ($Fv$).

","correct":true,"explanation":"Average power spreads the total work over the whole time interval. Instantaneous power is the rate at one moment and, for parallel force and velocity, equals $Fv$."},{"text":"

Average: Force multiplied by velocity at a specific moment ($Fv$).
Instantaneous: Total work done divided by total time ($W/t$).

","correct":false,"explanation":"This reverses the two meanings: $W/t$ over the full interval gives average power, while $Fv$ using the velocity at a particular moment gives instantaneous power."},{"text":"

Average: Total work done divided by total displacement ($W/s$).
Instantaneous: Force multiplied by time at a specific moment ($Ft$).

","correct":false,"explanation":"Dividing work by displacement gives force, not average power, while multiplying force by total time gives impulse, not instantaneous power. The correct expressions are $W/t$ and $Fv$."},{"text":"

Average: Total work done divided by average velocity ($W/v$).
Instantaneous: Force multiplied by acceleration at a specific moment ($Fa$).

","correct":false,"explanation":"Work divided by velocity does not give power, and $Fa$ has the wrong dimensions for power. Average power is $W/t$, while instantaneous power is $Fv$ for parallel quantities."}]},{"id":"f7f5020a-8c06-4592-80ca-820e6ba0410f","cardType":"basic","points":2,"front":"If a motor lifts a mass at constant speed, why is the force $F$ in $P=Fv$ equal to $mg$?","back":"The upward lifting force must balance the weight of the object for zero acceleration.","mcqOptions":[{"text":"The upward lifting force must balance the weight of the object for zero acceleration.","correct":true,"explanation":"Constant speed means zero acceleration, so the resultant force is zero. The upward lifting force must therefore equal the downward weight $mg$."},{"text":"The upward lifting force must exceed the weight of the object for zero acceleration.","correct":false,"explanation":"A lifting force greater than the weight would produce an upward resultant force and acceleration. At constant speed, the lifting force instead balances the weight $mg$."},{"text":"The upward lifting force must be less than the weight of the object for zero acceleration.","correct":false,"explanation":"A lifting force smaller than the weight would leave a downward resultant force, causing acceleration. Zero acceleration requires the upward force to balance the weight $mg$."},{"text":"The upward lifting force must replace the weight of the object for zero acceleration.","correct":false,"explanation":"Constant speed does not mean that no forces act; it means their resultant is zero. The upward lifting force must balance the downward weight $mg$."}]}]}],"examCountdown":{"examDate":"2027-05-12","paperLabel":"Paper 1"},"heading":"Work, energy and power","title":"Work, energy and power Flashcards","topbar":{"backHref":"/as-level/physics/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"b284815f-1808-5297-b7eb-c6e8315dbf2b","walled":"$L39"}],"$L3a","$L3b"]