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7(a) Figure 16 shows part of the inside of a pen - Edexcel - GCSE Physics - Question 7 - 2023 - Paper 2

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7(a) Figure 16 shows part of the inside of a pen. The pen contains a spring that can be compressed. tip inside body ... show full transcript

Worked Solution & Example Answer:7(a) Figure 16 shows part of the inside of a pen - Edexcel - GCSE Physics - Question 7 - 2023 - Paper 2

Step 1

Calculate the force needed to compress the spring by the amount shown in Figure 16.

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Answer

To calculate the force required to compress the spring by the distance shown in Figure 16 (6.2 cm or 0.062 m), we use Hooke's Law, which states that the force applied to a spring is proportional to the distance it is compressed or stretched. The formula is given by:

F=kimesxF = k imes x

where:

  • FF is the force (in N),
  • kk is the spring constant (260 N/m),
  • xx is the compression distance (0.062 m).

Substituting the values, we find:

F=260imes0.062=16.12extNF = 260 imes 0.062 = 16.12 ext{ N}

Rounding to two significant figures, the answer is therefore:

16 N.

Step 2

Describe how the student can determine the change in length of the spring.

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Answer

To determine the change in length of the spring, the student can follow these steps:

  1. Measure the initial length: The student should use the ruler to measure the length of the spring before any force is applied.

  2. Apply force to the spring: The student can gently press down on the spring using a finger, ensuring that the force applied is consistent.

  3. Measure the compressed length: While pressing down, the student should again measure the length of the spring.

  4. Calculate the change in length: Finally, the student can subtract the compressed length from the initial length to find the change in length:

    extChangeinlength=extInitiallengthextCompressedlength ext{Change in length} = ext{Initial length} - ext{Compressed length}

This method provides a clear and quantitative way to measure how much the spring has changed length.

Step 3

Describe one way that the student could improve the procedure.

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Answer

To improve the accuracy of the measurement, the student could use a pointer or a set square to more precisely locate the top of the spring before and after the force is applied. This would minimize parallax error and ensure consistent readings on the ruler. Additionally, using a weight to compress the spring rather than manual pressure would provide a more consistent and repeatable measurement.

Step 4

The force exerted on the top end of the spring by the hook is

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Answer

To analyze the forces acting on the spring, we recognize that the weight of the block (5 N down) and the weight of the spring (1 N down) exert forces on the spring itself. The force exerted by the hook on the spring must therefore counterbalance these forces.

The total weight acting downwards is:

extTotalWeight=extWeightofblock+extWeightofspring=5extN+1extN=6extN ext{Total Weight} = ext{Weight of block} + ext{Weight of spring} = 5 ext{ N} + 1 ext{ N} = 6 ext{ N}

Thus, the correct answer is: D 6 N up.

Step 5

Complete the diagram in Figure 19 to show the size and direction of the resultant force, R, on point X.

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Answer

To complete the diagram in Figure 19, we must first resolve the two forces acting at point X: P (30 N down) and Q (40 N right). The resultant force, R, can be found using vector addition.

R is the diagonal of the right triangle formed by P and Q. It can be calculated as follows:

R=exthypotenuse=extsqrt(P2+Q2)=extsqrt((30)2+(40)2)=extsqrt(900+1600)=extsqrt(2500)=50extNR = ext{hypotenuse} = ext{sqrt}(P^2 + Q^2) = ext{sqrt}((30)^2 + (40)^2) = ext{sqrt}(900 + 1600) = ext{sqrt}(2500) = 50 ext{ N}

Therefore, draw an arrow from point X to represent R, measuring about 50 N, directed diagonally down and to the right.

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