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Figure 14 shows an athlete using a fitness device - Edexcel - GCSE Physics - Question 8 - 2019 - Paper 1

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Figure 14 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 ... show full transcript

Worked Solution & Example Answer:Figure 14 shows an athlete using a fitness device - Edexcel - GCSE Physics - Question 8 - 2019 - Paper 1

Step 1

(i) Calculate the useful power output of the athlete when stretching the.

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Answer

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

P=work donetimeP = \frac{\text{work done}}{\text{time}}

Given that the work done is 45 J and the time taken is 0.6 s, we can substitute the values into the equation:

P=45 J0.6 sP = \frac{45 \text{ J}}{0.6 \text{ s}}

Calculating this gives:

P=75 WP = 75 \text{ W}

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

Step 2

(ii) Calculate the extension of the spring.

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Answer

To find the extension of the spring, we can use Hooke's Law, which states:

F=kxF = k \cdot x

Where:

  • FF is the force (in Newtons),
  • kk is the spring constant (in N/m), and
  • xx is the extension (in meters).

The work done on the spring can also be expressed as:

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

We can rearrange this formula to find xx:

  1. Substitute W=45W = 45 J and k=140k = 140 N/m: 45=12(140)x245 = \frac{1}{2} (140) x^2
  2. Rearranging gives: x2=45×2140x^2 = \frac{45 \times 2}{140}
  3. Calculating: x2=90140=914x^2 = \frac{90}{140} = \frac{9}{14}
  4. Finally, taking the square root to find xx: x=9140.75 mx = \sqrt{\frac{9}{14}} \approx 0.75 \text{ m}

Therefore, the extension of the spring is approximately 0.75 m.

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