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Figure 9 shows a 10 N weight hanging from a spring - Edexcel - GCSE Physics - Question 5 - 2019 - Paper 1

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Figure 9 shows a 10 N weight hanging from a spring. One of the forces acting to stretch the spring is shown in Figure 9. Complete Figure 9 by adding an arrow to sho... show full transcript

Worked Solution & Example Answer:Figure 9 shows a 10 N weight hanging from a spring - Edexcel - GCSE Physics - Question 5 - 2019 - Paper 1

Step 1

Complete Figure 9 by adding an arrow to show the other force acting to stretch the spring.

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Answer

In Figure 9, add a downward arrow below the support to indicate the weight of the spring, which acts downwards due to gravity. The arrow should be of the same length as the 10 N upward force to show that it balances the weight.

Step 2

Calculate the spring constant, k, of the spring.

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Answer

To calculate the spring constant, use Hooke's Law, which states that the force, FF, is equal to the spring constant, kk, times the extension, xx:

F=k×xF = k \times x

Given that the weight (force) F=4.0extNF = 4.0 ext{ N} and the extension x=0.06extmx = 0.06 ext{ m}, we can rearrange the formula:

k=Fx=4.0extN0.06extm=66.67extN/mk = \frac{F}{x} = \frac{4.0 ext{ N}}{0.06 ext{ m}} = 66.67 ext{ N/m}

Step 3

State what measurements should be made to determine the extension of the spring produced by the 4.0 N weight.

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Answer

To determine the extension of the spring caused by the 4.0 N weight, measure the original length of the spring before applying the weight and then measure the new length after the weight has been applied. The extension is the difference between these two lengths.

Step 4

Calculate the work done in stretching the spring by 0.30 m.

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Answer

To calculate the work done, use the following equation:

E=12×k×x2E = \frac{1}{2} \times k \times x^2

Substituting the given data:

  • k=250 N/mk = 250 \text{ N/m}
  • x=0.30 mx = 0.30 \text{ m}

The calculation is as follows:

E=12×250 N/m×(0.30extm)2E = \frac{1}{2} \times 250 \text{ N/m} \times (0.30 ext{ m})^2
E=12×250×0.09=11.25extJE = \frac{1}{2} \times 250 \times 0.09 = 11.25 ext{ J}

Thus, the work done in stretching the spring by 0.30 m is 11.25 J, and the unit is Joules (J).

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