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transfer, rather than the quantity transferred. The correct answer is that Energy is transferred when a force causes movement."},{"text":"Force is transferred.","correct":false,"explanation":"A force is an interaction that can cause movement, not the quantity transferred by that movement. The correct answer is that Energy is transferred."},{"text":"Distance is transferred.","correct":false,"explanation":"Distance measures how far the object moves, but it is not transferred. The movement caused by the force transfers energy, so Energy is transferred."}]},{"id":"573ee3aa-55aa-4fb4-8dc6-e9acabd05bef","cardType":"cloze","points":null,"front":"Work done is energy transferred when a [ ] moves an object through a distance in the [ ].","back":"Work done is energy transferred when a force moves an object through a distance in the direction of the force.","mcqOptions":[{"text":"Work done is energy transferred when a force moves an object through a distance in the direction of the force.","correct":true,"explanation":"Work done is defined as energy transferred when a force moves an object through a distance in the force’s direction. Only the force-parallel movement contributes to the work."},{"text":"Work done is energy transferred when a friction moves an object through a distance in the direction of the force.","correct":false,"explanation":"A frictional force can do work, but it is not the general definition. The correct statement uses a force moving an object in the direction of the force."},{"text":"Work done is energy transferred when a force moves an object through a distance perpendicular to the force.","correct":false,"explanation":"Movement perpendicular to a force does not involve a force component along the movement, so no work is done by that force. The correct direction is the direction of the force."},{"text":"Work done is energy transferred when a pressure moves an object through a distance in the direction of the force.","correct":false,"explanation":"Pressure is force per unit area, not the general agent in the work definition. Work involves a force moving an object in the direction of the force."}]},{"id":"8d56c39f-11be-415c-86c3-9b1d3b58c5d3","cardType":"basic","points":1,"front":"What equation calculates work done by a force?","back":"$$W = F \\times d$","mcqOptions":[{"text":"$$W = F \\times d$","correct":true,"explanation":"Work done equals force multiplied by the distance moved in the force’s direction, so the equation is $W = F \\times d$."},{"text":"$$W = F / d$","correct":false,"explanation":"Dividing force by distance gives pressure-like units, not energy. Work done is calculated by multiplying force by distance: $W = F \\times d$."},{"text":"$$W = d / F$","correct":false,"explanation":"Dividing distance by force does not produce the joule unit. The correct relationship multiplies force and distance: $W = F \\times d$."},{"text":"$$W = F + d$","correct":false,"explanation":"Adding force and distance combines different physical quantities and is not meaningful for calculating work. The correct equation is $W = F \\times d$."}]},{"id":"54fc3594-de1d-48f1-81b3-5ec083e6f081","cardType":"basic","points":1,"front":"In $W = F \\times d$, which distance should be used?","back":"The distance moved in the direction of the force.","mcqOptions":[{"text":"The distance moved in the direction of the force.","correct":true,"explanation":"Only the component of movement parallel to the force contributes to work, so the required distance is the distance moved in the direction of the force."},{"text":"The total distance travelled by the object.","correct":false,"explanation":"The total path length can include movement that is not parallel to the force. The correct distance is only the distance moved in the direction of the force."},{"text":"The distance moved perpendicular to the force.","correct":false,"explanation":"A perpendicular distance contributes no work from that force because there is no movement along the force’s direction. Use the distance moved in the direction of the force."},{"text":"The distance moved in the opposite direction to the force.","correct":false,"explanation":"Movement opposite to the force would give negative work, but the equation requires the relevant signed directional distance. The required distance is moved in the direction of the force."}]},{"id":"6a80df49-00dd-4c82-ad2f-a25034fbc735","cardType":"basic","points":1,"front":"If a force causes no movement in its own direction, how much work does it do?","back":"Zero work.","mcqOptions":[{"text":"Zero work.","correct":true,"explanation":"With no movement in the force’s direction, the distance contributing to work is zero. Since work equals force multiplied by directional distance, the force does Zero work."},{"text":"Positive work.","correct":false,"explanation":"A force can exist without doing work if it causes no movement in its direction. The correct result is Zero work, not positive work."},{"text":"Negative work.","correct":false,"explanation":"Negative work requires movement with a component opposite to the force, not simply no movement. With no directional displacement, the correct result is Zero work."},{"text":"Maximum work.","correct":false,"explanation":"There is no maximum work when the object does not move. The zero distance makes force multiplied by distance equal to Zero work."}]},{"id":"eacd2835-bb4e-42a5-9668-65c7c8174306","cardType":"cloze","points":null,"front":"One joule is the work done when a force of [ ] 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":"One joule is the work done when a force of 1 N moves an object 1 m in the direction of the force.","correct":true,"explanation":"One joule equals one newton-metre, so a force of 1 N moving an object 1 m in its direction transfers one joule of energy."},{"text":"One joule is the work done when a force of 1 N moves an object 1 cm in the direction of the force.","correct":false,"explanation":"Using centimetres instead of metres makes the displacement 100 times too small. One joule requires a force of 1 N moving an object 1 m."},{"text":"One joule is the work done when a force of 1 kg moves an object 1 m in the direction of the force.","correct":false,"explanation":"A kilogram measures mass, not force, so it cannot define the force component of a joule. The correct force is 1 N, with a displacement of 1 m."},{"text":"One joule is the work done when a force of 1 N moves an object 1 s in the direction of the force.","correct":false,"explanation":"Seconds measure time rather than distance, so this cannot produce the joule unit. One joule requires 1 N acting through 1 m."}]},{"id":"45876e6a-79dd-4c7b-abdb-7d30e9398246","cardType":"basic","points":2,"front":"Calculate the work done by an 80 N force moving a box 5.0 m.","back":"400 J","mcqOptions":[{"text":"400 J","correct":true,"explanation":"Work done is force multiplied by distance: $80 \\times 5.0 = 400$ joules. Therefore, the work done is 400 J."},{"text":"85 J","correct":false,"explanation":"Adding the force and distance gives 85, but work requires multiplication. Calculating $80 \\times 5.0$ gives the correct answer, 400 J."},{"text":"16 J","correct":false,"explanation":"Dividing the distance by the force gives 0.0625, while dividing force by distance gives 16; neither is the work equation. Multiplication gives 400 J."},{"text":"4000 J","correct":false,"explanation":"This treats 5.0 m as 50 m, an error in place value or unit conversion. Using the stated distance, $80 \\times 5.0$ gives 400 J."}]},{"id":"32b8d20e-13e4-499f-9b2f-16954d3c3c66","cardType":"cloze","points":null,"front":"Gravitational potential energy is energy stored because of an object's position in a [ ].","back":"Gravitational potential energy is energy stored because of an object's position in a gravitational field.","mcqOptions":[{"text":"gravitational field","correct":true,"explanation":"Gravitational potential energy is stored because an object’s position in a gravitational field gives it the potential to transfer energy as it moves."},{"text":"magnetic field","correct":false,"explanation":"A magnetic field stores magnetic potential energy through magnetic interactions, not gravitational potential energy. The correct term for height-related stored energy is gravitational field."},{"text":"electric field","correct":false,"explanation":"An electric field is associated with electric potential energy and charge, rather than gravitational potential energy. The correct term is gravitational field."},{"text":"gravitational force","correct":false,"explanation":"Gravitational force can act within a field, but the standard definition refers to an object’s position in a gravitational field, not in a force."}]},{"id":"45c83b45-130b-4c84-82f1-1c45d7da87c9","cardType":"basic","points":1,"front":"Near Earth, what value of $g$ is commonly used in IGCSE questions?","back":"$10\\ \\text{N/kg}$, unless another value is given.","mcqOptions":[{"text":"$10\\ \\text{N/kg}$, unless another value is given.","correct":true,"explanation":"IGCSE questions commonly use $g=10\\ \\text{N/kg}$ near Earth unless the question specifies a different value, such as $9.8\\ \\text{N/kg}$."},{"text":"$9.8\\ \\text{N/kg}$, unless another value is given.","correct":false,"explanation":"$$9.8\\ \\text{N/kg}$ is a more precise value for Earth’s gravitational field strength, but IGCSE questions commonly use $10\\ \\text{N/kg}$ unless another value is given."},{"text":"$1\\ \\text{N/kg}$, unless another value is given.","correct":false,"explanation":"The value is not $1\\ \\text{N/kg}$; that can result from confusing gravitational field strength with a unit conversion. IGCSE questions commonly use $10\\ \\text{N/kg}$ near Earth."},{"text":"$100\\ \\text{N/kg}$, unless another value is given.","correct":false,"explanation":"The factor of ten is too large, often caused by confusing newtons per kilogram with a percentage or by shifting a decimal place. The standard IGCSE value is $10\\ \\text{N/kg}$."}]},{"id":"358cae15-5644-4f0e-adf4-7cd320401ffb","cardType":"basic","points":1,"front":"What equation calculates gravitational potential energy?","back":"$$GPE = m \\times g \\times h$","mcqOptions":[{"text":"$$GPE = m \\times g \\times h$","correct":true,"explanation":"Gravitational potential energy equals mass multiplied by gravitational field strength and vertical height gained, so the equation is $GPE = m \\times g \\times h$."},{"text":"$$GPE = m \\times v \\times h$","correct":false,"explanation":"This substitutes speed for gravitational field strength. GPE depends on the object’s mass, $g$, and vertical height, so the correct equation is $GPE = m \\times g \\times h$."},{"text":"$$GPE = \\frac{1}{2}m \\times g \\times h$","correct":false,"explanation":"The factor of one-half belongs to the kinetic energy equation, not gravitational potential energy. The correct relationship is $GPE = m \\times g \\times h$."},{"text":"$$GPE = m \\times g \\div h$","correct":false,"explanation":"Height is multiplied because gaining more height increases GPE; dividing by height would predict less energy at greater height. The correct equation is $GPE = m \\times g \\times h$."}]},{"id":"736f1967-c13d-47b8-8b00-4628b8361f77","cardType":"basic","points":1,"front":"For GPE, what height should be used on a ramp?","back":"The vertical height gained, not ramp length.","mcqOptions":[{"text":"The vertical height gained, not ramp length.","correct":true,"explanation":"GPE depends on the vertical increase in position, because gravity acts vertically. The distance measured along the ramp is irrelevant to the final GPE change."},{"text":"The distance along the ramp, not vertical height.","correct":false,"explanation":"Using ramp length confuses the distance travelled with the vertical rise. GPE uses the vertical height gained, so the correct choice is not the ramp’s length."},{"text":"The horizontal distance travelled, not vertical height.","correct":false,"explanation":"Horizontal movement does not increase gravitational potential energy. The relevant measurement is the vertical height gained between the starting and finishing positions."},{"text":"The total distance travelled, not vertical height.","correct":false,"explanation":"GPE is determined by the change in vertical position, not the complete path length. Even on a winding route, use only the vertical height gained."}]},{"id":"83ff86e7-5d96-443f-9788-66ad5c3a40e2","cardType":"basic","points":2,"front":"Calculate the GPE gained by lifting 15 kg through 2.0 m. Use $g=10\\ \\text{N/kg}$.","back":"300 J","mcqOptions":[{"text":"300 J","correct":true,"explanation":"Using $GPE=mgh$, the gain is $15\\times10\\times2.0=300\\ \\text{J}$. Mass, gravitational field strength, and vertical height are all multiplied."},{"text":"30 J","correct":false,"explanation":"This is ten times too small, usually because the gravitational field strength has been omitted or treated as $1\\ \\text{N/kg}$. The correct calculation gives $300\\ \\text{J}$."},{"text":"150 J","correct":false,"explanation":"This result comes from omitting the factor of two in the height or incorrectly inserting a factor of one-half. Applying $GPE=mgh$ gives $300\\ \\text{J}$."},{"text":"3000 J","correct":false,"explanation":"This is ten times too large, suggesting that $g=10$ has been applied twice or a zero has been added. The correct value from $15\\times10\\times2.0$ is $300\\ \\text{J}$."}]},{"id":"e0b0f28c-4e52-4849-a31f-bb7c69011c86","cardType":"basic","points":1,"front":"What equation calculates kinetic energy?","back":"$$KE = \\frac{1}{2}mv^2$","mcqOptions":[{"text":"$$KE = \\frac{1}{2}mv^2$","correct":true,"explanation":"Kinetic energy depends on mass and the square of speed, with a factor of one-half: $KE = \\frac{1}{2}mv^2$."},{"text":"$$KE = mv^2$","correct":false,"explanation":"The speed is correctly squared, but the factor of one-half is missing. That makes the result twice the kinetic energy; the correct equation is $KE = \\frac{1}{2}mv^2$."},{"text":"$$KE = \\frac{1}{2}mv$","correct":false,"explanation":"Speed must be squared because kinetic energy changes with the square of speed. Leaving out the square gives the wrong dependence; use $KE = \\frac{1}{2}mv^2$."},{"text":"$$KE = mgh$","correct":false,"explanation":"This is the equation for gravitational potential energy, which involves height. Kinetic energy involves motion and is calculated using $KE = \\frac{1}{2}mv^2$."}]},{"id":"35fbbe16-7f0b-4484-947d-8b2336fc81e6","cardType":"cloze","points":null,"front":"In $KE = \\frac{1}{2}mv^2$, only [ ] is squared, not [ ].","back":"In $KE = \\frac{1}{2}mv^2$, only speed is squared, not mass.","mcqOptions":[{"text":"In $KE = \\frac{1}{2}mv^2$, only speed is squared, not mass.","correct":true,"explanation":"The exponent applies to $v$, representing speed, while mass is outside the square. Therefore doubling speed has a much greater effect on kinetic energy than doubling mass."},{"text":"In $KE = \\frac{1}{2}mv^2$, only speed is squared, not time.","correct":false,"explanation":"Time is not one of the variables in the kinetic energy equation, so this does not identify the relevant contrast. Only speed is squared, not mass."},{"text":"In $KE = \\frac{1}{2}mv^2$, only mass is squared, not speed.","correct":false,"explanation":"The superscript $2$ is attached to $v$, the symbol for speed, not $m$, the symbol for mass. Thus only speed is squared, not mass."},{"text":"In $KE = \\frac{1}{2}mv^2$, only acceleration is squared, not mass.","correct":false,"explanation":"Acceleration does not appear in this equation, and the squared quantity is represented by $v$. Only speed is squared, while mass is not."}]},{"id":"9f6e4205-9647-4b66-b237-6ef7c4060800","cardType":"basic","points":2,"front":"If speed doubles, what happens to kinetic energy?","back":"It becomes four times larger.","mcqOptions":[{"text":"It becomes four times larger.","correct":true,"explanation":"Kinetic energy is proportional to the square of speed. Doubling speed gives a factor of $2^2=4$, so the kinetic energy becomes four times larger."},{"text":"It becomes two times larger.","correct":false,"explanation":"This treats kinetic energy as directly proportional to speed and forgets the square. Since speed is doubled, kinetic energy changes by $2^2$, becoming four times larger."},{"text":"It becomes half as large.","correct":false,"explanation":"Increasing speed cannot halve the kinetic energy. The inverse relationship is a misconception; squaring the doubled speed makes the kinetic energy four times larger."},{"text":"It becomes eight times larger.","correct":false,"explanation":"An eightfold change would involve cubing the speed, not squaring it. Kinetic energy depends on $v^2$, so doubling speed makes it four times larger."}]},{"id":"1784f324-b71a-4147-82a9-3d1ada59d6bd","cardType":"basic","points":2,"front":"Calculate the KE of a 70 kg cyclist moving at 6.0 m/s.","back":"1260 J","mcqOptions":[{"text":"1260 J","correct":true,"explanation":"Using $KE=\\frac{1}{2}mv^2$, $KE=\\frac{1}{2}\\times70\\times6.0^2=35\\times36=1260\\ \\text{J}$."},{"text":"420 J","correct":false,"explanation":"This result comes from using the speed rather than squaring it, effectively calculating $\\frac{1}{2}\\times70\\times12$. The correct calculation squares $6.0$ and gives $1260\\ \\text{J}$."},{"text":"2520 J","correct":false,"explanation":"The factor of one-half has been omitted, so the result is twice the correct value. Including it gives $\\frac{1}{2}\\times70\\times6.0^2=1260\\ \\text{J}$."},{"text":"360 J","correct":false,"explanation":"This can result from multiplying mass by speed and treating the remaining factors incorrectly, rather than using the squared speed. The correct value is $1260\\ \\text{J}$."}]},{"id":"cc392002-02d0-471f-871b-0eadcc846acf","cardType":"cloze","points":null,"front":"Conservation of energy means energy cannot be [ ] or [ ].","back":"Conservation of energy means energy cannot be created or destroyed.","mcqOptions":[{"text":"Conservation of energy means energy cannot be created or destroyed.","correct":true,"explanation":"Energy may move between stores or be transferred between objects, but the total quantity is conserved because energy can be neither created nor destroyed."},{"text":"Conservation of energy means energy cannot be transferred or destroyed.","correct":false,"explanation":"Energy can be transferred between stores and objects, so transfer is not prohibited. Conservation means energy cannot be created or destroyed, although its form and location can change."},{"text":"Conservation of energy means energy cannot be created or transferred.","correct":false,"explanation":"Energy can be created neither from nothing nor by changing its form. Conservation states that energy cannot be created or destroyed, only transferred between stores."},{"text":"Conservation of energy means energy cannot be stored or transferred.","correct":false,"explanation":"Energy can be stored and transferred, so neither process violates conservation. The principle specifically states that energy cannot be created or destroyed."}]},{"id":"ee398cab-0128-4537-b8f3-85a45a964bc4","cardType":"basic","points":1,"front":"Without air resistance, how are GPE loss and KE gain related during a fall?","back":"Loss of GPE = gain in KE.","mcqOptions":[{"text":"Loss of GPE = gain in KE.","correct":true,"explanation":"With no resistive forces, no energy is dissipated to the surroundings. Conservation of energy therefore makes the decrease in gravitational potential energy equal the increase in kinetic energy."},{"text":"Loss of GPE > gain in KE.","correct":false,"explanation":"A greater GPE loss would imply that some energy was transferred elsewhere, but air resistance is absent. The loss of GPE therefore equals the gain in KE."},{"text":"Loss of GPE < gain in KE.","correct":false,"explanation":"A KE gain larger than the GPE loss would require an additional energy source. During an unresisted fall, the loss of GPE exactly equals the gain in KE."},{"text":"Loss of GPE is unrelated to gain in KE.","correct":false,"explanation":"The two changes are directly related by conservation of energy. Without air resistance, all lost gravitational potential energy becomes kinetic energy, so their magnitudes are equal."}]},{"id":"cbbd5bc0-cf39-437b-814c-0a9f1a5f581e","cardType":"basic","points":1,"front":"With friction or air resistance, what does some energy become?","back":"Thermal energy of the surroundings.","mcqOptions":[{"text":"Thermal energy of the surroundings.","correct":true,"explanation":"Friction and drag dissipate mechanical energy by heating the moving object and its surroundings, increasing their thermal energy stores."},{"text":"Kinetic energy of the surroundings.","correct":false,"explanation":"Friction and air resistance do not principally increase the surroundings’ kinetic energy store. They dissipate mechanical energy as thermal energy of the surroundings."},{"text":"Gravitational potential energy of the surroundings.","correct":false,"explanation":"Gravitational potential energy depends on position in a gravitational field and is not the dissipated store. Resistive forces transfer energy to the surroundings thermally."},{"text":"Chemical energy of the surroundings.","correct":false,"explanation":"Chemical energy is associated with fuels, foods and batteries, not the heating caused by resistive forces. Friction and air resistance produce thermal energy in the surroundings."}]},{"id":"8745d38c-cc21-417b-b5a8-560dd441e4b5","cardType":"basic","points":3,"front":"A 50 kg child slides down 4.0 m and has 1600 J KE at the bottom. Use $g=10\\ \\text{N/kg}$. What work is done against friction?","back":"400 J","mcqOptions":[{"text":"400 J","correct":true,"explanation":"The child loses $mgh=50\\times10\\times4.0=2000$ J of GPE. Since 1600 J becomes KE, the remaining $2000-1600=400$ J is work against friction."},{"text":"1600 J","correct":false,"explanation":"This uses the final kinetic energy as the frictional work. The GPE loss is 2000 J, so subtracting the 1600 J KE gives 400 J against friction."},{"text":"2000 J","correct":false,"explanation":"This treats the entire 2000 J GPE loss as work against friction and ignores the 1600 J kinetic energy gained. Only the remaining 400 J is dissipated."},{"text":"3600 J","correct":false,"explanation":"This adds GPE loss and KE gain instead of applying energy conservation. The frictional work is their difference, $2000-1600=400$ J."}]},{"id":"5f3c9457-f635-4d2b-8993-d29646545fdf","cardType":"cloze","points":null,"front":"Power is the rate of transfer of [ ] or the rate of doing [ ].","back":"Power is the rate of transfer of energy or the rate of doing work.","mcqOptions":[{"text":"Power is the rate of transfer of energy or the rate of doing work.","correct":true,"explanation":"Power measures how quickly energy is transferred. Because work done is a measure of transferred energy, power is also the rate at which work is done."},{"text":"Power is the rate of transfer of power or the rate of doing work.","correct":false,"explanation":"Power itself is a rate, rather than the quantity transferred. It describes the rate of energy transfer or, equivalently, the rate of doing work."},{"text":"Power is the rate of transfer of energy or the rate of doing force.","correct":false,"explanation":"Force is not something described as being done; work is done when a force causes displacement. Power is the rate of energy transfer or doing work."},{"text":"Power is the rate of transfer of work or the rate of doing energy.","correct":false,"explanation":"Work is a way of transferring energy, but energy is not described as being done. Power is the rate of energy transfer or the rate of doing work."}]},{"id":"c9caf318-f36d-4b01-91a5-931745098ba4","cardType":"basic","points":1,"front":"What equation calculates power from work done and time?","back":"$$P = \\frac{W}{t}$","mcqOptions":[{"text":"$$P = \\frac{W}{t}$","correct":true,"explanation":"Power is the rate of doing work, so the work done $W$ is divided by the elapsed time $t$. This gives joules per second, or watts."},{"text":"$$P = Wt$","correct":false,"explanation":"Multiplying work by time gives units of joule-seconds, not watts. Power is work done per unit time, so the equation is $P = \\frac{W}{t}$."},{"text":"$$P = \\frac{t}{W}$","correct":false,"explanation":"Reversing the fraction gives seconds per joule, the reciprocal of power. Work must be divided by time, giving $P = \\frac{W}{t}$."},{"text":"$$P = \\frac{W}{t^2}$","correct":false,"explanation":"Squaring the time has no basis in the definition and produces incorrect units. Power is simply work divided by time: $P = \\frac{W}{t}$."}]},{"id":"6ee0d618-16c3-4389-a269-b62b97431fc3","cardType":"cloze","points":null,"front":"One watt means [ ]; one kilowatt is [ ].","back":"One watt means 1 J/s; one kilowatt is 1000 W.","mcqOptions":[{"text":"One watt means 1 J/s; one kilowatt is 1000 W.","correct":true,"explanation":"The watt is defined as one joule of energy transferred per second. The prefix kilo means one thousand, so one kilowatt equals 1000 watts."},{"text":"One watt means 1 J; one kilowatt is 1000 W.","correct":false,"explanation":"A joule measures energy, whereas a watt measures the rate of energy transfer. One watt is 1 J/s, while one kilowatt equals 1000 W."},{"text":"One watt means 1 J/s; one kilowatt is 100 W.","correct":false,"explanation":"The watt definition is correct, but kilo represents a factor of 1000 rather than 100. Therefore, one kilowatt is 1000 W."},{"text":"One watt means 1 J s; one kilowatt is 10 W.","correct":false,"explanation":"A watt is one joule per second, not one joule multiplied by a second. Also, kilo means 1000, so one kilowatt is 1000 W."}]},{"id":"731854a1-b398-49af-972a-e079a4f6f712","cardType":"basic","points":2,"front":"Calculate the power of a motor doing 12 kJ of work in 30 s.","back":"400 W","mcqOptions":[{"text":"400 W","correct":true,"explanation":"Convert 12 kJ to 12,000 J, then use $P=W/t$. Dividing $12{,}000$ J by 30 s gives 400 J/s, which is 400 W."},{"text":"0.4 W","correct":false,"explanation":"This divides 12 by 30 without converting kilojoules to joules. The work is 12,000 J, so the power is $12{,}000/30=400$ W."},{"text":"40 W","correct":false,"explanation":"This result is ten times too small because of an incorrect division or conversion. Converting first gives $12{,}000/30=400$ W."},{"text":"4000 W","correct":false,"explanation":"This result is ten times too large, typically from treating 12 kJ as 120,000 J. Since 12 kJ is 12,000 J, the power is 400 W."}]}]}],"examCountdown":{"examDate":"2027-05-25","paperLabel":"Paper 1"},"heading":"Work and power","title":"Work and power Flashcards","topbar":{"backHref":"/igcse/physics/edexcel/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"efdbbf6c-66f4-57ac-bc8c-1bd9399fc4b0","walled":"$L38"}],"$L39","$L3a"]