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Figure 9 shows a spring standing on a table - Edexcel - GCSE Physics Combined Science - Question 4 - 2020 - Paper 1

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Figure 9 shows a spring standing on a table. Weights are added to the spring as shown in Figures 9b and 9c. a) (i) Estimate the original length of the spring as sh... show full transcript

Worked Solution & Example Answer:Figure 9 shows a spring standing on a table - Edexcel - GCSE Physics Combined Science - Question 4 - 2020 - Paper 1

Step 1

Estimate the original length of the spring as shown in Figure 9a.

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Answer

The original length of the spring can be estimated by directly measuring it using a ruler or a similar device. Based on the provided figures, the estimated range for the original length of the spring is approximately 50.0 to 55.0 mm.

Step 2

Describe how the reduction in the length of the spring when weights are added can be determined.

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Answer

To determine the reduction in length when weights are added, first measure the original length of the spring without any weights. Then, progressively add weights and measure the new lengths after each addition. Finally, subtract the new length from the original length to find the reduction.

Step 3

State two ways that the experimental procedure could be improved.

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Answer

  1. Use a ruler with smaller divisions for more accurate measurements of the spring length.
  2. Ensure that the measurements are taken at eye level to avoid parallax error.

Step 4

Give one reason why the reduction in length eventually reaches a limit as more weights are added.

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Answer

The reduction in length reaches a limit because the coils of the spring become fully compressed when the maximum load is applied, preventing any further extension.

Step 5

Calculate the spring constant k in N/m.

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Answer

Given the force of 0.5 N results in an extension of 13 mm (0.013 m), the spring constant k is calculated using Hooke's law:

k=Fx=0.5N0.013m38.46N/mk = \frac{F}{x} = \frac{0.5 \, N}{0.013 \, m} \approx 38.46 \, N/m

Step 6

Calculate the extension of the spring.

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Answer

Using the work-energy principle, the work done on a spring is given by:

W=12kx2W = \frac{1}{2} k x^2

Substituting the given values:

0.14=12175x20.14 = \frac{1}{2} \cdot 175 \cdot x^2

Rearranging gives:

x2=0.1421750.0016x^2 = \frac{0.14 \cdot 2}{175} \\ \approx 0.0016

Thus, evaluating gives:

x0.04mx \approx 0.04 \, m

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