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A student applies different forces to a spring and measures the extension of the spring each time - OCR Gateway - GCSE Physics - Question 19 - 2023 - Paper 1

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A student applies different forces to a spring and measures the extension of the spring each time. The force–extension graph shows their results. A, B and C are poin... show full transcript

Worked Solution & Example Answer:A student applies different forces to a spring and measures the extension of the spring each time - OCR Gateway - GCSE Physics - Question 19 - 2023 - Paper 1

Step 1

Which letter represents where the spring has elastic deformation?

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Answer

The letter A represents where the spring has elastic deformation, as this is the region where the spring returns to its original length when the force is removed.

Step 2

Which letter represents where the spring obeys Hooke’s Law?

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Answer

The letter P represents where the spring obeys Hooke's Law. This is the linear region of the graph where the force is proportional to the extension.

Step 3

Which letter represents the elastic limit of the spring?

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Answer

The letter B represents the elastic limit of the spring. Beyond this point, the spring will not return to its original shape.

Step 4

Which letter represents where the graph is non-linear?

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Answer

The letter Q represents where the graph is non-linear. This indicates that the spring has reached a permanent deformation.

Step 5

State the minimum number of forces that need to be applied to the spring in order to stretch it.

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Answer

At least one force must be applied to stretch the spring.

Step 6

Calculate the force exerted by the spring when it is extended by 0.15 m.

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Answer

Using the formula:

extForce=extSpringConstantimesextExtension ext{Force} = ext{Spring Constant} imes ext{Extension}

we have: extForce=28extN/mimes0.15extm=4.2extN ext{Force} = 28 ext{ N/m} imes 0.15 ext{ m} = 4.2 ext{ N}.

Step 7

Calculate the energy transferred when the spring is extended by 0.15 m.

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Answer

The energy transferred can be calculated using the formula:

ext{Energy} = rac{1}{2} imes ext{Spring Constant} imes ( ext{Extension})^2

Thus, substituting the values:

= rac{1}{2} imes 28 imes 0.0225 \ = 0.315 ext{ J}.$$

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