Physics · Professor Newton

Forces and motion, Year 11: Hooke's law and elastic energy:全部卡片

整套按顺序列出——孩子看到之前,您可以先通读一遍。

1 GCSE-PHYS-FOR-0015

A change of shape that is completely reversed when the forces are removed, so the object goes back to its original length

Elastic deformation

提示A rubber band shows it every time you let go.

为什么If 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.

2 GCSE-PHYS-FOR-0016

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

提示Picture pulling one end while nobody holds the other.

为什么To 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.

3 GCSE-PHYS-FOR-0017

force = ____ × extension

spring constant

提示It is measured in N/m and says how stiff the thing being stretched is.

为什么In 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.

4 GCSE-PHYS-FOR-0018

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

提示Put the extension into metres, then divide the force by it.

为什么F = 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.

5 GCSE-PHYS-FOR-0019

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

提示Extension means how much longer it has become.

为什么The 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.

6 GCSE-PHYS-FOR-0020

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

提示Think about how a reading seems to shift when you look at it from above or below.

为什么Looking 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.

7 GCSE-PHYS-FOR-0021

elastic potential energy = 0.5 × spring constant × (____)²

extension

提示It is how much longer the spring has become, not how long it now is.

为什么In 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.

8 GCSE-PHYS-FOR-0022

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

提示Change the centimetres into the unit that matches N/m before you square anything.

为什么The 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.

9 GCSE-PHYS-FOR-0023

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

提示Look at what the equation does to the one quantity that has changed.

为什么The 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.

10 GCSE-PHYS-FOR-0024

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

提示Rearrange the energy equation so that k stands alone; you will divide by a square.

为什么Start 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.

11 GCSE-PHYS-FOR-0025

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

提示Ask where the energy from the pupil's pull has gone.

为什么A 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.

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