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.

GCSE Edexcel Physics Combined Science Power: A cyclist is riding a bicycle at

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 find the kinetic energy (KE) of the cyclist and bicycle, we can use the equation:

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

Substituting the values:

Mass (m) = 68 kg
Velocity (v) = 12 m/s

Thus,

$KE = \frac{1}{2} \times 68 \times (12)^2$

Calculating this gives:

$KE = \frac{1}{2} \times 68 \times 144 = 4896 \text{ J}$

So, the kinetic energy is approximately 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 uses the brakes to stop, the kinetic energy of the cyclist and bicycle is transformed into thermal energy. The cyclist applies the brakes, which creates friction between the brake pads and the wheels. This friction causes the kinetic energy to decrease while increasing the thermal energy of the brakes and surrounding air, causing them to heat up.

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 calculate power, we use the formula:

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

For session 1:

Energy = 45.2 kJ = 45200 J
Time = 300 s

Thus,

$\text{Power}_1 = \frac{45200}{300} = 150.67 \text{ W} \approx 0.15 \text{ kW}$

For session 2:

Energy = 37.9 kJ = 37900 J
Time = 300 s

So,

$\text{Power}_2 = \frac{37900}{300} = 126.33 \text{ W} \approx 0.13 \text{ kW}$

In conclusion, the athlete produced more power during session 1 compared to 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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