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A teacher stretches an elastic band by increasing the force applied and measures the extension during loading - OCR Gateway - GCSE Physics - Question 18 - 2023 - Paper 3

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A teacher stretches an elastic band by increasing the force applied and measures the extension during loading. The teacher then reduces the force applied and measure... show full transcript

Worked Solution & Example Answer:A teacher stretches an elastic band by increasing the force applied and measures the extension during loading - OCR Gateway - GCSE Physics - Question 18 - 2023 - Paper 3

Step 1

(i) The elastic band obeys Hooke's Law.

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Answer

The statement is incorrect. An elastic band does not obey Hooke's Law because, beyond its elastic limit, it does not return to its original length. Hooke's Law applies only to materials that obey the linear elastic behavior.

Step 2

(ii) The elastic band undergoes plastic deformation.

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Answer

The statement is correct. An elastic band can experience plastic deformation if it is stretched beyond its elastic limit, affecting its ability to return to its original shape.

Step 3

(iii) There is a linear relationship between force and extension for the elastic band.

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Answer

The statement is incorrect. While there may be a linear relationship in the initial phase of stretching, once the elastic limit is exceeded, the relationship becomes non-linear.

Step 4

Calculate the work done stretching this spring by 0.20 m.

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Answer

Work done (W) can be calculated using the formula:

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

Where:

  • k=28 N/mk = 28 \text{ N/m} (spring constant)
  • x=0.20 mx = 0.20 \text{ m} (extension)

Substituting the values gives:

W=12×28×(0.20)2=12×28×0.04=0.56 JW = \frac{1}{2} \times 28 \times (0.20)^2 = \frac{1}{2} \times 28 \times 0.04 = 0.56 \text{ J}

Step 5

Determine the magnitude of the resultant force.

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Answer

To calculate the resultant force when two forces act at right angles, we use the Pythagorean theorem:

R=F12+F22R = \sqrt{F_1^2 + F_2^2}

Where:

  • F1=2.0 NF_1 = 2.0 \text{ N}
  • F2=3.0 NF_2 = 3.0 \text{ N}

Substituting the values gives:

R=(2.0)2+(3.0)2=4+9=133.6 NR = \sqrt{(2.0)^2 + (3.0)^2} = \sqrt{4 + 9} = \sqrt{13} \approx 3.6 \text{ N}

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