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Figure 7 shows an athlete using a fitness device - Edexcel - GCSE Physics Combined Science - Question 5 - 2019 - Paper 1

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Figure 7 shows an athlete using a fitness device. The athlete stretches the spring in the device by pulling the handles apart. The spring constant of the spring is... show full transcript

Worked Solution & Example Answer:Figure 7 shows an athlete using a fitness device - Edexcel - GCSE Physics Combined Science - Question 5 - 2019 - Paper 1

Step 1

Calculate the useful power output of the athlete when stretching the spring.

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Answer

To calculate the useful power output, we use the formula:

P=EtP = \frac{E}{t}

where:

  • PP is the power,
  • EE is the work done (in joules),
  • tt is the time taken (in seconds).

Substituting the given values:

P=450.6=75 WP = \frac{45}{0.6} = 75 \text{ W}

Thus, the useful power output of the athlete is 75 W.

Step 2

Calculate the extension of the spring.

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Answer

To find the extension of the spring, we use Hooke's Law, represented as:

E=12kx2E = \frac{1}{2} k x^2

where:

  • EE is the energy stored in the spring (45 J),
  • kk is the spring constant (140 N/m),
  • xx is the extension of the spring in meters.

Rearranging the equation to solve for xx:

45=12×140×x245 = \frac{1}{2} \times 140 \times x^2

This simplifies to:

x2=2×45140x2=90140x2=0.642857x=0.6428570.8 mx^2 = \frac{2 \times 45}{140} \Rightarrow x^2 = \frac{90}{140} \Rightarrow x^2 = 0.642857 \Rightarrow x = \sqrt{0.642857} \approx 0.8 \text{ m}

Therefore, the extension of the spring is approximately 0.8 m.

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