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The forces can stretch, compress, bend or twist the object."},{"text":"A change in an object's mass caused by energy.","correct":false,"explanation":"This confuses deformation with a change in physical quantity caused by energy transfer. The correct answer is a change in an object's shape caused by forces."},{"text":"A change in an object's motion caused by friction.","correct":false,"explanation":"This describes an effect on movement rather than deformation. The correct answer is a change in an object's shape caused by forces."},{"text":"A change in an object's temperature caused by pressure.","correct":false,"explanation":"Pressure may sometimes produce heating, but temperature change is not the definition of deformation. The correct answer is a change in an object's shape caused by forces."}]},{"id":"ec955fc9-d2cb-4ce0-88d2-02d1d1c4049a","cardType":"basic","points":1,"front":"How do the forces act when an object is stretched?","back":"Forces pull outwards on different parts of the object.","mcqOptions":[{"text":"Forces pull outwards on different parts of the object.","correct":true,"explanation":"Stretching happens when forces act away from each other on different parts of an object, increasing its length or changing its shape."},{"text":"Forces pull inwards on different parts of the object.","correct":false,"explanation":"Pulling inwards describes compression, not stretching. The correct answer is that forces pull outwards on different parts of the object."},{"text":"Forces push outwards on different parts of the object.","correct":false,"explanation":"Pushing outwards is not the pulling action that stretches an object. The correct answer is that forces pull outwards on different parts of the object."},{"text":"Forces push inwards on different parts of the object.","correct":false,"explanation":"These forces would compress the object by pushing its parts together. Stretching requires forces that pull outwards on different parts of the object."}]},{"id":"e027bb80-0a20-4269-8786-3d4a0005e17d","cardType":"basic","points":1,"front":"How do the forces act when an object is compressed?","back":"Forces push inwards on different parts of the object.","mcqOptions":[{"text":"Forces push inwards on different parts of the object.","correct":true,"explanation":"Compression occurs when forces push towards each other on different parts of an object, reducing its length or changing its shape."},{"text":"Forces push outwards on different parts of the object.","correct":false,"explanation":"Pushing outwards does not force the parts together and is not the described compressive action. The correct answer is that forces push inwards on different parts."},{"text":"Forces pull outwards on different parts of the object.","correct":false,"explanation":"Pulling outwards stretches an object rather than compressing it. The correct answer is that forces push inwards on different parts of the object."},{"text":"Forces pull inwards on different parts of the object.","correct":false,"explanation":"Although the direction is inward, pulling is different from the pushing action needed for compression. The correct answer is that forces push inwards on different parts."}]},{"id":"e68ccde6-e122-44f1-9008-1f55bcff3369","cardType":"basic","points":2,"front":"Why must forces act at suitable positions to stretch an object?","back":"Forces acting at different positions can change the object's shape; forces pulling the same way move it instead.","mcqOptions":[{"text":"Forces acting at different positions can change the object's shape; forces pulling the same way move it instead.","correct":true,"explanation":"Forces at different positions can produce deformation by changing the object's shape. Forces pulling in the same direction produce movement rather than a shape change."},{"text":"Forces acting at different positions can change the object's speed; forces pulling the same way stretch it instead.","correct":false,"explanation":"Different positions can change shape, not necessarily speed, while forces in the same direction tend to move the object. The correct answer identifies deformation as the shape change."},{"text":"Forces acting at the same position can change the object's shape; forces pulling different ways move it instead.","correct":false,"explanation":"Forces need suitable different positions to deform an object, whereas forces pulling different ways can change its motion. The correct answer distinguishes position and direction correctly."},{"text":"Forces acting at different positions can change the object's mass; forces pushing the same way move it instead.","correct":false,"explanation":"Forces do not normally change an object's mass. The correct answer is that forces at different positions can change shape, while forces pulling the same way move it."}]},{"id":"bbd9c345-6da6-424c-ae0d-a42e745b1dee","cardType":"basic","points":2,"front":"Why must more than one force act to change the shape of a stationary object?","back":"

The forces must be balanced, giving a resultant force of zero.
They must act on different parts to cause deformation.

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

The forces must be balanced, giving a resultant force of zero.
They must act on different parts to cause deformation.

","correct":true,"explanation":"A stationary object remains stationary when the forces are balanced, but different points of application allow those forces to alter its shape and cause deformation."},{"text":"

The forces must be unbalanced, giving a resultant force of zero.
They must act on different parts to cause deformation.

","correct":false,"explanation":"An unbalanced force cannot have a resultant force of zero; it would accelerate the object. The correct answer requires balanced forces acting on different parts to deform it."},{"text":"

The forces must be balanced, giving a resultant force of zero.
They must act on the same part to cause deformation.

","correct":false,"explanation":"Balanced forces keep the object stationary, but acting on the same part does not create the required change in shape. The correct answer requires different parts to deform it."},{"text":"

The forces must be unbalanced, giving a resultant force forwards.
They must act on the same part to cause deformation.

","correct":false,"explanation":"An unbalanced forward resultant force causes acceleration, while forces on the same part do not provide the opposing actions needed for deformation. The correct answer uses balanced forces on different parts.resultant force, so it accelerates.","mcqOptions":[{"text":"The unbalanced force gives the object a resultant force, so it accelerates.","correct":true,"explanation":"An unbalanced force produces a non-zero resultant force. Newton's second law means the object accelerates, changing its motion rather than only changing shape."},{"text":"The unbalanced force gives the object a balanced force, so it accelerates.","correct":false,"explanation":"Balanced forces have a resultant force of zero and do not cause acceleration. The correct answer is that an unbalanced force gives the object a resultant force, so it accelerates."},{"text":"The unbalanced force gives the object a resultant force, so it stops.","correct":false,"explanation":"A resultant force causes acceleration, which may increase, decrease or change direction of velocity, rather than necessarily stopping the object. The correct answer says the object accelerates."},{"text":"The unbalanced force gives the object a deforming force, so it remains stationary.","correct":false,"explanation":"An unbalanced force cannot leave the object stationary because it creates a non-zero resultant force and acceleration. The correct answer links the resultant force to acceleration."}]},{"id":"77c0b70a-0788-4106-84bb-dadcb2b5b96e","cardType":"basic","points":1,"front":"What happens to an object after the deforming force is removed if its deformation is elastic?","back":"It returns to its original shape.","mcqOptions":[{"text":"It returns to its original shape.","correct":true,"explanation":"Elastic deformation is reversible. When the deforming force is removed, the object returns to its original shape, provided its elastic limit has not been exceeded."},{"text":"It remains in its deformed shape.","correct":false,"explanation":"Remaining deformed describes inelastic deformation, not elastic behaviour. The correct answer is that the object returns to its original shape after the force is removed."},{"text":"It returns to its new shape.","correct":false,"explanation":"Elastic deformation does not create a permanent new shape. The correct answer is that the object returns to its original shape once the deforming force is removed."},{"text":"It changes into its original material.","correct":false,"explanation":"Removing a force changes the object's shape, not its material identity. The correct answer is that elastic deformation allows it to return to its original shape."}]},{"id":"0dc6e92a-c2cf-44f3-b364-3ff8d805b846","cardType":"basic","points":1,"front":"What happens to an object after the deforming force is removed if its deformation is inelastic?","back":"It does not return completely to its original shape.","mcqOptions":[{"text":"It does not return completely to its original shape.","correct":true,"explanation":"Inelastic deformation is not fully reversible. After the deforming force is removed, the object keeps some permanent change and does not return completely to its original shape."},{"text":"It does return completely to its original shape.","correct":false,"explanation":"Complete recovery describes elastic deformation, whereas inelastic deformation leaves a permanent change. The correct answer is that the object does not return completely to its original shape."},{"text":"It does not return temporarily to its original shape.","correct":false,"explanation":"A temporary delay is not the defining feature of inelastic deformation. The correct answer is that some permanent deformation remains, so the object does not return completely."},{"text":"It returns partly to a completely new shape.","correct":false,"explanation":"Inelastic deformation may leave a permanent change, but the key point is incomplete return towards the original shape. The correct answer states that it does not return completely."}]},{"id":"846ee2e8-637c-4f05-a1b7-977948f891d3","cardType":"basic","points":1,"front":"How is extension calculated?","back":"$$\\text{extension} = \\text{stretched length} - \\text{original length}$","mcqOptions":[{"text":"$$\\text{extension} = \\text{stretched length} - \\text{original length}$","correct":true,"explanation":"Extension is the increase in length, so subtracting the original length from the stretched length gives the amount by which the object has extended."},{"text":"$$\\text{extension} = \\text{original length} - \\text{stretched length}$","correct":false,"explanation":"Reversing the subtraction would give a negative value for a stretched object. Extension is found by subtracting the original length from the stretched length."},{"text":"$$\\text{extension} = \\text{stretched length} + \\text{original length}$","correct":false,"explanation":"Adding the two lengths gives a total length rather than the increase in length. Extension is the stretched length minus the original length."},{"text":"$$\\text{extension} = \\dfrac{\\text{stretched length}}{\\text{original length}}$","correct":false,"explanation":"Dividing the lengths gives a ratio, not the change in length. Extension is calculated by subtracting the original length from the stretched length."}]},{"id":"0675f93a-2f2f-442e-ae1d-8d5ad3bb4556","cardType":"basic","points":2,"front":"State Hooke's law and its equation.","back":"For an elastic object below the limit of proportionality, extension is directly proportional to force: $F = ke$.","mcqOptions":[{"text":"For an elastic object below the limit of proportionality, extension is directly proportional to force: $F = ke$.","correct":true,"explanation":"Hooke’s law states that, below the limit of proportionality, extension increases in direct proportion to force. The equation $F = ke$ relates force, spring constant and extension."},{"text":"For an elastic object below the limit of proportionality, extension is inversely proportional to force: $F = \\dfrac{k}{e}$.","correct":false,"explanation":"Inverse proportionality would make extension decrease as force increases, which is not Hooke’s law. Below the limit of proportionality, extension is directly proportional to force and $F = ke$."},{"text":"For an elastic object below the limit of proportionality, extension is directly proportional to force: $F = \\dfrac{e}{k}$.","correct":false,"explanation":"The proportional relationship is described correctly, but the equation rearranges the variables incorrectly. Hooke’s law is $F = ke$, where $k$ is the spring constant."},{"text":"For an elastic object below the limit of proportionality, extension is directly proportional to force: $F = k + e$.","correct":false,"explanation":"Adding force-related quantities does not represent the proportional relationship between force and extension. Hooke’s law uses multiplication: $F = ke$, below the limit of proportionality."}]},{"id":"6fe16992-17d9-4dcd-b1f7-9cd535bcccd7","cardType":"basic","points":2,"front":"What does a larger spring constant mean?","back":"The spring is stiffer; more force is needed for the same extension.","mcqOptions":[{"text":"The spring is stiffer; more force is needed for the same extension.","correct":true,"explanation":"A larger spring constant means the spring needs more force to produce a given extension, so it is stiffer than a spring with a smaller spring constant."},{"text":"The spring is less stiff; less force is needed for the same extension.","correct":false,"explanation":"A smaller spring constant indicates a less stiff spring. A larger spring constant means more force is needed for the same extension, so the spring is stiffer."},{"text":"The spring is stiffer; less force is needed for the same extension.","correct":false,"explanation":"This reverses the force requirement. A stiffer spring has a larger spring constant and therefore requires more force to produce the same extension."},{"text":"The spring has a greater maximum extension; more force is needed for the same extension.","correct":false,"explanation":"The spring constant describes stiffness, not directly the maximum extension before breaking. A larger value means the spring is stiffer and needs more force for the same extension."}]},{"id":"33261c71-29bb-4062-9d8c-0fa1437b79c9","cardType":"basic","points":1,"front":"What does the gradient of an extension-force graph represent?","back":"The gradient is $\\dfrac{1}{k}$, where $k$ is the spring constant.","mcqOptions":[{"text":"The gradient is $\\dfrac{1}{k}$, where $k$ is the spring constant.","correct":true,"explanation":"For an extension-force graph, extension is plotted against force. Rearranging $F = ke$ gives $e = \\dfrac{F}{k}$, so the gradient is $\\dfrac{1}{k}$."},{"text":"The gradient is $k$, where $k$ is the spring constant.","correct":false,"explanation":"This is the gradient of a force-extension graph, where force is plotted against extension. For an extension-force graph, rearranging Hooke’s law gives a gradient of $\\dfrac{1}{k}$."},{"text":"The gradient is $\\dfrac{1}{e}$, where $e$ is the extension.","correct":false,"explanation":"The gradient depends on the graph’s axes and the spring constant, not the reciprocal of one particular extension. An extension-force graph has gradient $\\dfrac{1}{k}$."},{"text":"The gradient is $ke$, where $k$ is the spring constant.","correct":false,"explanation":"$$ke$ represents the force in Hooke’s law, not the gradient of an extension-force graph. Since $e = \\dfrac{F}{k}$, the graph’s gradient is $\\dfrac{1}{k}$."}]},{"id":"04391d74-25e6-45a1-8c8b-ebcc04bee0bf","cardType":"basic","points":2,"front":"What does a straight line through the origin on a force-extension graph show?","back":"Force and extension are directly proportional, so Hooke's law applies.","mcqOptions":[{"text":"Force and extension are directly proportional, so Hooke's law applies.","correct":true,"explanation":"A straight line through the origin shows that the ratio of force to extension is constant. This is direct proportionality, so Hooke’s law applies in that region."},{"text":"Force and extension are inversely proportional, so Hooke's law applies.","correct":false,"explanation":"An inverse relationship would not produce a straight line through the origin. The graph shows direct proportionality, which is the condition required for Hooke’s law."},{"text":"Force and extension are directly proportional, so the spring has reached its limit of proportionality.","correct":false,"explanation":"The limit of proportionality is where the graph stops being straight through the origin. A straight line through the origin shows that Hooke’s law still applies."},{"text":"Force and extension are not proportional, so Hooke's law applies.","correct":false,"explanation":"A straight line through the origin is the defining graph pattern for direct proportionality. If the quantities were not proportional, Hooke’s law would not apply."}]},{"id":"e4b01db9-3757-4558-a4a2-ad776672c68d","cardType":"basic","points":2,"front":"What does a curved section of a force-extension graph show?","back":"The relationship is non-linear; doubling force no longer doubles extension, so $F = ke$ no longer holds with constant $k$.","mcqOptions":[{"text":"The relationship is non-linear; doubling force no longer doubles extension, so $F = ke$ no longer holds with constant $k$.","correct":true,"explanation":"A curved graph means force and extension are no longer directly proportional. The spring has passed the limit of proportionality, so a constant spring constant cannot describe the relationship."},{"text":"The relationship is linear; doubling force doubles extension, so $F = ke$ still holds with constant $k$.","correct":false,"explanation":"A linear section would be a straight line through the origin. Curvature shows that doubling force no longer doubles extension, so $F = ke$ does not hold with constant $k$."},{"text":"The relationship is non-linear; doubling force doubles extension, so $F = ke$ still holds with constant $k$.","correct":false,"explanation":"The term non-linear is appropriate, but a non-linear graph does not show doubling extension for doubled force. The constant-$k$ form of Hooke’s law no longer applies."},{"text":"The relationship is non-linear; doubling force no longer doubles extension, so $F = ke$ holds with a larger constant $k$.","correct":false,"explanation":"Once the graph is curved, the ratio of force to extension changes, so there is no single larger constant $k$ that makes $F = ke$ valid throughout the section."}]},{"id":"03942c62-ef80-4a06-b422-21d579023ad3","cardType":"basic","points":2,"front":"In a spring investigation, which variable is independent and which is dependent?","back":"

Independent: force.
Dependent: extension.

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

Independent: force.
Dependent: extension.

","correct":true,"explanation":"The force is deliberately changed by the investigator, making it the independent variable. The resulting extension is measured, making it the dependent variable."},{"text":"

Independent: extension.
Dependent: force.

","correct":false,"explanation":"This reverses the variables. In the investigation, force is changed deliberately and extension is measured in response, so force is independent and extension is dependent."},{"text":"

Independent: spring constant.
Dependent: force.

","correct":false,"explanation":"The spring constant is a property of the spring and is not normally the variable deliberately changed in this investigation. Force is changed and extension is measured."},{"text":"

Independent: extension.
Dependent: spring constant.

","correct":false,"explanation":"Neither variable assignment matches the investigation. Force is the controlled input, while extension is the measured outcome; the spring constant is calculated or treated as a property."}]},{"id":"bdde44a6-3451-4878-b2d3-d743d8a6a647","cardType":"basic","points":1,"front":"Why are spring readings taken at eye level?","back":"To avoid parallax error when reading the ruler.","mcqOptions":[{"text":"To avoid parallax error when reading the ruler.","correct":true,"explanation":"Viewing the ruler at eye level keeps the pointer and scale aligned, preventing the apparent position from shifting when viewed from above or below."},{"text":"To avoid zero error when reading the ruler.","correct":false,"explanation":"Zero error comes from an instrument not starting at zero and is checked before measuring. Eye-level viewing prevents parallax error caused by the viewing angle."},{"text":"To avoid random error when reading the ruler.","correct":false,"explanation":"Random error causes readings to vary unpredictably and is reduced through repeated measurements. Eye-level positioning specifically prevents parallax error from the viewing angle."},{"text":"To avoid reaction-time error when reading the ruler.","correct":false,"explanation":"Reaction-time error affects measurements involving timing, such as using a stopwatch. Reading a ruler at eye level avoids parallax error instead."}]},{"id":"b1a1e87f-5c5b-45b9-a5ea-1b6e040c70c3","cardType":"basic","points":2,"front":"Why should anomalous spring readings be repeated rather than included in the mean?","back":"Repeating them checks whether they are random errors before calculating the mean.","mcqOptions":[{"text":"Repeating them checks whether they are random errors before calculating the mean.","correct":true,"explanation":"Repeating an anomalous measurement shows whether the unusual result occurs again. If it does not, it is likely a random error and should be excluded from the mean."},{"text":"Repeating them checks whether they are systematic errors before calculating the mean.","correct":false,"explanation":"A systematic error affects measurements consistently and will not usually be identified by repeating one anomalous reading. Repetition helps identify random errors, which should be excluded from the mean."},{"text":"Repeating them checks whether they are zero errors before calculating the mean.","correct":false,"explanation":"A zero error shifts every reading by a similar amount, so repeating an anomalous result does not diagnose it. Repetition checks whether the anomaly is a random error."},{"text":"Repeating them checks whether they are parallax errors before calculating the mean.","correct":false,"explanation":"Parallax is a reading error caused by viewing a scale from the wrong angle and can affect repeated readings. The purpose here is to test whether the anomaly is random."}]},{"id":"150802d8-217f-44ce-a83f-a99a1ff2f81b","cardType":"basic","points":2,"front":"A spring extends by $0.040\\ \\text{m}$ under a force of $6.0\\ \\text{N}$. Calculate its spring constant.","back":"$$k = \\dfrac{F}{e} = \\dfrac{6.0}{0.040} = \\mathbf{150\\ \\text{N/m}}$","mcqOptions":[{"text":"$$k = \\dfrac{F}{e} = \\dfrac{6.0}{0.040} = \\mathbf{150\\ \\text{N/m}}$","correct":true,"explanation":"The spring constant is force divided by extension: $k=F/e$. Substituting $6.0\\ \\text{N}$ and $0.040\\ \\text{m}$ gives $150\\ \\text{N/m}$."},{"text":"$$k = \\dfrac{F}{e} = \\dfrac{6.0}{0.040} = \\mathbf{15\\ \\text{N/m}}$","correct":false,"explanation":"The division is miscalculated by a factor of ten. Dividing $6.0$ by $0.040$ gives $150$, so the spring constant is $150\\ \\text{N/m}$."},{"text":"$$k = \\dfrac{F}{e} = \\dfrac{6.0}{0.040} = \\mathbf{0.15\\ \\text{N/m}}$","correct":false,"explanation":"This treats the extension as though its decimal conversion introduced an extra factor of $1000$. Using metres correctly gives $6.0/0.040=150\\ \\text{N/m}$."},{"text":"$$k = \\dfrac{F}{e} = \\dfrac{6.0}{0.040} = \\mathbf{240\\ \\text{N/m}}$","correct":false,"explanation":"This value can result from multiplying force and extension instead of dividing, or from an unrelated arithmetic slip. The defining equation gives $k=6.0/0.040=150\\ \\text{N/m}$."}]},{"id":"db061c16-6039-47c6-b91f-34afb806697c","cardType":"basic","points":1,"front":"What is elastic potential energy?","back":"Energy stored in a stretched or compressed elastic object because work is done to deform it.","mcqOptions":[{"text":"Energy stored in a stretched or compressed elastic object because work is done to deform it.","correct":true,"explanation":"Work done when an elastic object is stretched or compressed is transferred into stored elastic potential energy, provided the object can return to its original shape."},{"text":"Energy stored in a stretched or compressed rigid object because work is done to deform it.","correct":false,"explanation":"A rigid object does not significantly deform elastically, so this description does not identify elastic potential energy. The energy is stored in an elastic object that deforms and can recover."},{"text":"Energy transferred by a moving object because work is done to accelerate it.","correct":false,"explanation":"Work done to accelerate a moving object changes its kinetic energy, not its elastic potential energy. Elastic potential energy specifically results from reversible stretching or compression."},{"text":"Energy stored in a stretched or compressed elastic object because heat is transferred to it.","correct":false,"explanation":"Heating transfers thermal energy and may cause expansion, but it is not the defining mechanism here. Elastic potential energy is stored when work is done to deform the object."}]},{"id":"3470ebb5-c646-4c30-914e-528cd8c70152","cardType":"basic","points":1,"front":"State the equation for elastic potential energy stored up to the limit of proportionality.","back":"$$E_{\\text{e}} = \\tfrac{1}{2}ke^2$","mcqOptions":[{"text":"$$E_{\\text{e}} = \\tfrac{1}{2}ke^2$","correct":true,"explanation":"Elastic potential energy equals half the spring constant multiplied by the square of the extension. This equation applies while the force-extension relationship remains proportional."},{"text":"$$E_{\\text{e}} = ke^2$","correct":false,"explanation":"The factor of one-half is missing. Since the force increases as the spring extends, the energy is the triangular area under the force-extension graph, giving $E_{\\text{e}}=\\tfrac{1}{2}ke^2$."},{"text":"$$E_{\\text{e}} = \\tfrac{1}{2}ke$","correct":false,"explanation":"The extension must be squared because energy depends on the area under the graph and force is proportional to extension. The correct relationship contains $e^2$, not just $e$."},{"text":"$$E_{\\text{e}} = \\dfrac{2e^2}{k}$","correct":false,"explanation":"This reverses the role of the spring constant and places it in the denominator. The graph area gives half of $ke^2$, so the correct equation is $\\tfrac{1}{2}ke^2$."}]},{"id":"c7addfe7-310f-44e6-942a-736bd979f0cd","cardType":"cloze","points":null,"front":"A stationary object can be deformed by balanced forces if they act at [ ].","back":"A stationary object can be deformed by balanced forces if they act at different positions.","mcqOptions":[{"text":"different positions","correct":true,"explanation":"Balanced forces can still produce a turning effect when their lines of action are separated. Equal opposite forces acting at different positions can therefore deform a stationary object."},{"text":"the same position","correct":false,"explanation":"Forces acting at the same position have no separation to create a turning effect. Deformation by balanced forces requires equal opposite forces acting at different positions."},{"text":"the centre of mass","correct":false,"explanation":"Forces through the centre of mass do not necessarily create the separated turning effects needed for this deformation. The relevant condition is that balanced forces act at different positions."},{"text":"different times","correct":false,"explanation":"Forces acting at different times do not provide simultaneous balanced forces that can deform the object in this way. They must act at different positions to create a turning effect."}]},{"id":"f8f00186-a240-4bbf-a504-e30a44c0846b","cardType":"cloze","points":null,"front":"Elastic deformation is reversed when the deforming forces are [ ].","back":"Elastic deformation is reversed when the deforming forces are removed.","mcqOptions":[{"text":"removed","correct":true,"explanation":"Elastic deformation is reversible: when the deforming forces are removed, the object returns to its original shape, provided its elastic limit has not been exceeded."},{"text":"increased","correct":false,"explanation":"Increasing the deforming forces generally increases the deformation rather than reversing it. Reversal occurs when the forces are removed and the elastic limit has not been exceeded."},{"text":"balanced","correct":false,"explanation":"Balanced forces can maintain a stationary situation but do not necessarily remove the deformation. Elastic deformation reverses when the deforming forces are removed."},{"text":"reversed","correct":false,"explanation":"Reversing the direction of the forces may change the deformation but does not state the condition for recovery. The object returns to its shape when the forces are removed."}]},{"id":"a70ceb30-ec11-44d7-85d8-40ac3bc82e40","cardType":"cloze","points":null,"front":"The limit of proportionality is reached where a force-extension graph begins to [ ].","back":"The limit of proportionality is reached where a force-extension graph begins to curve.","mcqOptions":[{"text":"curve","correct":true,"explanation":"Before the limit of proportionality, extension is directly proportional to force and the graph is straight. Beginning to curve shows that proportionality has been lost."},{"text":"straighten","correct":false,"explanation":"A straight graph indicates proportional behaviour rather than the point where proportionality ends. The limit is reached when the force-extension graph begins to curve."},{"text":"flatten","correct":false,"explanation":"A flattening graph describes a particular change in gradient but is not the defining criterion for the limit of proportionality. The key observation is that the graph begins to curve."},{"text":"reverse","correct":false,"explanation":"The graph does not need to reverse direction when proportionality is lost. The limit is identified by the graph changing from a straight line and beginning to curve."}]},{"id":"2d56a1c6-3297-407c-b1dc-1968df1be012","cardType":"cloze","points":null,"front":"In the practical, the spring's extension is plotted on the [ ] against force on the x-axis.","back":"In the practical, the spring's extension is plotted on the y-axis against force on the x-axis.","mcqOptions":[{"text":"y-axis","correct":true,"explanation":"In this practical, force is the independent variable placed on the x-axis, while extension is the dependent variable measured in response and plotted on the y-axis."},{"text":"x-axis","correct":false,"explanation":"Putting extension on the x-axis reverses the chosen variables for this graph. Force is plotted on the x-axis, and the resulting extension is plotted on the y-axis."},{"text":"origin","correct":false,"explanation":"The origin is the point where both axes meet, not an axis on which data are plotted. Extension is plotted on the y-axis against force on the x-axis."},{"text":"gradient","correct":false,"explanation":"The gradient is a calculated feature of a graph, not one of its axes. The extension values belong on the y-axis, with force on the x-axis."}]},{"id":"d2f45b16-c612-449c-9102-92edef61ab41","cardType":"cloze","points":null,"front":"If the spring is not inelastically deformed, the work done equals the [ ].","back":"If the spring is not inelastically deformed, the work done equals the elastic potential energy stored.","mcqOptions":[{"text":"elastic potential energy stored","correct":true,"explanation":"When deformation remains elastic, the spring returns to its original shape, so the work done in stretching or compressing it is stored as elastic potential energy."},{"text":"thermal energy dissipated","correct":false,"explanation":"Thermal energy is produced when work is dissipated through inelastic deformation or friction. For an elastic spring with no inelastic deformation, the work is stored elastically."},{"text":"kinetic energy gained","correct":false,"explanation":"Kinetic energy is associated with motion, whereas a stationary deformed spring stores energy due to its deformation. The work done becomes elastic potential energy."},{"text":"gravitational potential energy gained","correct":false,"explanation":"Gravitational potential energy changes when an object moves vertically in a gravitational field. Stretching an elastic object stores the work done as elastic potential energy."}]}]}],"examCountdown":{"examDate":"2027-05-25","paperLabel":"Paper 1"},"heading":"5.3 Forces and elasticity","title":"5.3 Forces and elasticity Flashcards","topbar":{"backHref":"/gcse/physics/aqa/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"f786d035-4565-5b85-9a90-dea455b1cd4e","walled":"$L38"}],"$L39","$L3a"]