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 140 N/m. The athlete does 45 J of work to extend the spring. The athlete takes 0.6 s to expand the spring. (i) Calculate the useful power output of the athlete when stretching the spring. (ii) Calculate the extension of the spring. Use an equation selected from the list of equations at the end of this paper.

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

Question 5

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

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

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Answer

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

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

where:

$P$ is the power output,
$E$ is the energy (work done), and
$t$ is the time taken.

Substituting the given values:

$P = \frac{45 \text{ J}}{0.6 \text{ s}} \approx 75 \text{ W}$

Therefore, 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 calculate the extension of the spring, we can use the formula for the elastic potential energy of a spring:

$E = \frac{1}{2} k x^2$

where:

$E$ is the work done (45 J),
$k$ is the spring constant (140 N/m), and
$x$ is the extension.

Rearranging the formula to find $x$ gives us:

$45 = \frac{1}{2} \times 140 \times x^2$

Solving for $x$ :

$x^2 = \frac{2 \times 45}{140} \implies x^2 = 0.64 \implies x \approx 0.8 \text{ m}$

Thus, the extension of the spring is 0.8 m.

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