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Figure 1 shows a stretched spring - AQA - GCSE Physics Combined Science - Question 1 - 2021 - Paper 2

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Figure 1 shows a stretched spring. The spring is elastically deformed. **What is meant by 'elastically deformed'?** Tick (✓) one box. - As the force on the spring... show full transcript

Worked Solution & Example Answer:Figure 1 shows a stretched spring - AQA - GCSE Physics Combined Science - Question 1 - 2021 - Paper 2

Step 1

What is meant by 'elastically deformed'?

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Answer

The spring will return to its original length when the force is removed.

Step 2

Describe a method to determine the extension of the spring.

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Answer

Measure the original length of the spring and the extended length of the spring (with the metre rule). The extension is calculated by subtracting the original length from the extended length.

Step 3

Calculate the elastic potential energy of the spring.

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Answer

The extension e is given as 0.080 m. Using the formula for elastic potential energy, we have:

Ee=12ke2E_e = \frac{1}{2} k e^2 Substituting the values: Ee=12×40×(0.080)2=0.128 JE_e = \frac{1}{2} \times 40 \times (0.080)^2 = 0.128 \text{ J}

Step 4

Write down the equation which links extension (e), force (F) and spring constant (k).

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Answer

The equation is: F=keF = k \cdot e

Step 5

Calculate the spring constant of the spring.

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Answer

Given a force of 300 N and an extension of 0.40 m, we can rearrange the equation from the previous part:

k=Fek = \frac{F}{e} Substituting the values: k=3000.40=750 N/mk = \frac{300}{0.40} = 750 \text{ N/m}

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