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Figure 20 shows a child on a playground toy - AQA - GCSE Physics - Question 9 - 2021 - Paper 1

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Question 9

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Figure 20 shows a child on a playground toy. The springs have been elastically deformed. Explain what is meant by ‘elastically deformed’. Question 9 continues on ... show full transcript

Worked Solution & Example Answer:Figure 20 shows a child on a playground toy - AQA - GCSE Physics - Question 9 - 2021 - Paper 1

Step 1

Explain what is meant by ‘elastically deformed’

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Answer

The term 'elastically deformed' refers to a alteration in the shape or length of an object, such that it can return to its original form once the force causing the deformation is removed. In this case, the springs on the playground toy are compressed under the weight of the child and will return to their original state when the child gets off.

Step 2

Describe a method the student could use to obtain the results given in Figure 21.

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Answer

To investigate the properties of the spring:

  1. Set up the apparatus: Secure a clamp stand to a table and attach the spring to it using a clamp.
  2. Add weights: Hang weights on the spring and record the force applied to the spring.
  3. Measure extension: For each weight added, measure the extension of the spring from its original length.
  4. Repeat measurements: Gradually increase the weight in increments and measure the corresponding extension up to the maximum weight.
  5. Risk assessment: Ensure the clamp is secure to prevent it from falling. Use safety goggles to avoid injuries from falling weights.

Step 3

Which equation links extension (e), force (F) and spring constant (k)

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Answer

The correct equation that links extension (e), force (F), and spring constant (k) is:

F=kimeseF = k imes e

Step 4

Determine the spring constant of the spring.

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Answer

To determine the spring constant (k) from the data in Figure 21, use the formula:

k=Fek = \frac{F}{e}

If we assume the values from the graph, for example, a force (F) of 5 N and extension (e) of 0.125 m, then:

k=50.125=40N/mk = \frac{5}{0.125} = 40 \, N/m

Step 5

Describe how Figure 21 supports the student’s conclusion.

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Answer

Figure 21 supports the student's conclusion by demonstrating a linear relationship between the force applied to the spring and its extension. As more weight is added, the spring stretches proportionally, indicating that the extension directly correlates with the applied force, hence confirming the observation about proportionality.

Step 6

Calculate the elastic potential energy of the spring when the extension of the spring was 20 cm.

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Answer

To calculate the elastic potential energy (E_e) stored in the spring, we can use the formula:

Ee=12ke2E_e = \frac{1}{2} k e^2

Given:

  • Spring constant, k=13N/mk = 13 \, N/m
  • Extension, e=20cm=0.20me = 20 \, cm = 0.20 \, m, we substitute these values:

Ee=12imes13imes(0.20)2=12×13×0.04=0.26JE_e = \frac{1}{2} imes 13 imes (0.20)^2 = \frac{1}{2} \times 13 \times 0.04 = 0.26 \, J

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