Every card in Forces and motion, Year 11: Hooke's law and elastic energy
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- A change of shape that is completely reversed when the forces are removed, so the object goes back to its original length
Elastic deformation
HintA rubber band shows it every time you let go.
WhyIf the object stays stretched or bent after the forces are removed, the change is called inelastic instead. A spring that has been pulled too far shows this: take the load off and it is permanently longer than it was.
- A spring lies at rest on a bench. Why can it not be stretched by applying just one force to it?
One force alone would just move the whole spring
HintPicture pulling one end while nobody holds the other.
WhyTo stretch, squash or bend a stationary object, at least two forces must act on it — for a spring, one at each end pulling in opposite directions. In the school experiment the clamp holding the top of the spring supplies the second force.
- force = ____ × extension
spring constant
HintIt is measured in N/m and says how stiff the thing being stretched is.
WhyIn symbols, F = k e, with F in newtons, k in newtons per metre and e in metres. It holds only up to the limit of proportionality, and works for squashing too, with e as the compression. AQA expects this equation to be recalled.
- A spring extends by 6 cm when a 3 N weight hangs from it, within its limit of proportionality. Calculate its spring constant in N/m. (number only)
50
HintPut the extension into metres, then divide the force by it.
WhyF = k e rearranges to k = F ÷ e. The extension must be in metres for an answer in N/m: 6 cm = 0.06 m, so k = 3 ÷ 0.06. A stiffer spring has a bigger spring constant because it needs more force for each metre of stretch.
- In the required practical on force and extension, a spring hangs from a clamp beside a ruler and weights are added one at a time. The ruler gives the length of the spring. How is the extension found from each reading?
Subtract the spring's original, unloaded length
HintExtension means how much longer it has become.
WhyThe force (the weight added) is the independent variable and the extension is the dependent variable; the spring itself is kept the same. The length with no load is measured first, and each extension is the loaded length minus that starting length. Plotting force against extension then gives a straight line through the origin up to the limit of proportionality.
- In the force and extension practical, why should your eye be level with the bottom of the spring (or its pointer) each time you read the ruler?
To avoid a parallax error
HintThink about how a reading seems to shift when you look at it from above or below.
WhyLooking from an angle makes the pointer line up with the wrong mark on the ruler, so every length is slightly out. Reading at eye level, and fixing a pointer to the bottom of the spring, makes each measurement more accurate. Waiting for the spring to stop bouncing before reading helps too.
- elastic potential energy = 0.5 × spring constant × (____)²
extension
HintIt is how much longer the spring has become, not how long it now is.
WhyIn symbols, Ee = ½ k e², with Ee in joules, k in newtons per metre and e in metres. Only the e is squared. AQA gives this equation on the Physics equation sheet, so pupils must be able to select and apply it rather than recall it, and it holds only while the spring has not passed its limit of proportionality.
- A spring with a spring constant of 800 N/m is stretched by 5 cm, staying within its limit of proportionality. Calculate the elastic potential energy stored in it, in joules. (number only)
1
HintChange the centimetres into the unit that matches N/m before you square anything.
WhyThe equation, Ee = ½ k e², is on the equation sheet and needs e in metres: 5 cm = 0.05 m. Then Ee = 0.5 × 800 × 0.05² = 0.5 × 800 × 0.0025. Square the extension first, then multiply by k and by a half.
- A spring is stretched, and then stretched further until its extension has doubled. It stays within its limit of proportionality. What happens to the elastic potential energy stored in it?
It becomes four times as large
HintLook at what the equation does to the one quantity that has changed.
WhyThe energy stored depends on extension squared (Ee = ½ k e²), so doubling the extension multiplies the energy by 2². The second half of the stretch is harder than the first, because the force needed keeps growing as the spring gets longer.
- A spring stores 4 J of elastic potential energy when it is stretched by 0.2 m, within its limit of proportionality. Calculate its spring constant.
200 N/m
HintRearrange the energy equation so that k stands alone; you will divide by a square.
WhyStart from Ee = ½ k e² (on the equation sheet). Doubling both sides gives 2 Ee = k e², so k = 2 Ee ÷ e² = (2 × 4) ÷ 0.2² = 8 ÷ 0.04. Check: 0.5 × 200 × 0.04 gives back the 4 J.
- A pupil does 3 J of work stretching a spring. The spring is not permanently deformed. How much elastic potential energy is now stored in the spring?
3 J
HintAsk where the energy from the pupil's pull has gone.
WhyA force that stretches or squashes a spring does work on it, and that energy is stored as elastic potential energy. As long as the spring has not been inelastically deformed, the two are equal, so the stored energy can be found by working out the work done. This is why AQA repeats the equation in the Forces topic.
1★ GCSE-PHYS-FOR-0015
2★ GCSE-PHYS-FOR-0016
3★ GCSE-PHYS-FOR-0017
4★ GCSE-PHYS-FOR-0018
5★ GCSE-PHYS-FOR-0019
6★ GCSE-PHYS-FOR-0020
7★ GCSE-PHYS-FOR-0021
8★ GCSE-PHYS-FOR-0022
9★ GCSE-PHYS-FOR-0023
10★ GCSE-PHYS-FOR-0024
11★ GCSE-PHYS-FOR-0025
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