A cyclist is riding a bicycle at a steady velocity of 12 m/s. The cyclist and bicycle have a total mass of 68 kg. (a) Calculate the kinetic energy of the cyclist and bicycle. Use the equation $$KE = \frac{1}{2} \times m \times v^2$$ kinetic energy = _____________ J (b) Describe the energy transfer that happens when the cyclist uses the brakes to stop. (d) An athlete uses a training machine in a gym. The display on the machine shows the time spent on the machine and the amount of energy transferred during a training session. Figure 3 shows the displays for two different sessions by the same athlete. Energy 45.2 kJ Time 300 s session 1 Energy 37.9 kJ Time 300 s session 2 Explain what the displays show about the average power of the athlete in each of these two sessions.

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A cyclist is riding a bicycle at a steady velocity of 12 m/s - Edexcel - GCSE Physics Combined Science - Question 2 - 2018 - Paper 1

Question 2

A cyclist is riding a bicycle at a steady velocity of 12 m/s. The cyclist and bicycle have a total mass of 68 kg. (a) Calculate the kinetic energy of the cyclist a... show full transcript

Worked Solution & Example Answer:A cyclist is riding a bicycle at a steady velocity of 12 m/s - Edexcel - GCSE Physics Combined Science - Question 2 - 2018 - Paper 1

Step 1

Calculate the kinetic energy of the cyclist and bicycle.

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Answer

To calculate the kinetic energy (KE), we use the formula:

$KE = \frac{1}{2} \times m \times v^2$

Where:

m = 68 kg (mass of the cyclist and bicycle)
v = 12 m/s (velocity)

Substituting these values:

$KE = \frac{1}{2} \times 68 \times (12)^2$ $KE = \frac{1}{2} \times 68 \times 144$ $KE = 4900 \text{ J}$

Thus, the kinetic energy of the cyclist and bicycle is 4900 J.

Step 2

Describe the energy transfer that happens when the cyclist uses the brakes to stop.

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Answer

When the cyclist applies the brakes to stop, kinetic energy is transformed into thermal energy due to friction between the brake pads and the wheel. As the brakes are applied, the kinetic energy of the cyclist and bicycle decreases, while the thermal energy of the brake system (and surroundings) increases. This energy transformation leads to a reduction in the speed of the cyclist and an increase in temperature of the brake system, ensuring the cyclist comes to a stop.

Step 3

Explain what the displays show about the average power of the athlete in each of these two sessions.

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Answer

To compare the average power of the athlete in the two sessions, we can use the formula for power:

$\text{Power} = \frac{\text{Work}}{\text{Time}}$

For session 1:

Energy = 45.2 kJ = 45200 J
Time = 300 s

Average power in session 1 is: $\text{Power} = \frac{45200 ext{ J}}{300 ext{ s}} = 150.67 ext{ W} \approx 0.15 ext{ kW}$

For session 2:

Energy = 37.9 kJ = 37900 J
Time = 300 s

Average power in session 2 is: $\text{Power} = \frac{37900 ext{ J}}{300 ext{ s}} = 126.33 ext{ W} \approx 0.13 ext{ kW}$

Therefore, the athlete has more power output in session 1 than in session 2.

A cyclist is riding a bicycle at a steady velocity of 12 m/s - Edexcel - GCSE Physics Combined Science - Question 2 - 2018 - Paper 1

Worked Solution & Example Answer:A cyclist is riding a bicycle at a steady velocity of 12 m/s - Edexcel - GCSE Physics Combined Science - Question 2 - 2018 - Paper 1

Calculate the kinetic energy of the cyclist and bicycle.

Describe the energy transfer that happens when the cyclist uses the brakes to stop.

Explain what the displays show about the average power of the athlete in each of these two sessions.

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