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 Energy 37.9 kJ Time 300 s Time 300 s Figure 3 Explain what the displays show about the average power of the athlete in each of these two sessions.

Photo AI

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.

96%

114 rated

Answer

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

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

Substituting the values:

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

The calculation becomes:

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

First, calculate ((12)^2 = 144):

Then,

$KE = \frac{1}{2} \times 68 \times 144$ $KE = 34 \times 144 = 4896 \text{ J}$

Thus, the kinetic energy is approximately 4900 J.

Step 2

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

99%

104 rated

Answer

When the cyclist uses the brakes, the kinetic energy of the cyclist and the bicycle is converted into thermal energy due to friction between the brake pads and the wheels. As the brakes are applied, energy is transformed from the kinetic energy (energy of motion) to thermal energy, which causes the brake components to heat up. Therefore, the kinetic energy of the cyclist and bicycle decreases as energy is transferred to the surroundings in the form of heat.

Step 3

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

96%

101 rated

Answer

To find the average power output of the athlete during the two training sessions, we can use the formula for power:

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

Session 1:

Energy transferred = 45.2 kJ = 45200 J
Time = 300 s

Average Power in Session 1: $\text{Power} = \frac{45200}{300} \approx 150.67 \text{ W} \text{ (or 0.15 kW)}$

Session 2:

Energy transferred = 37.9 kJ = 37900 J
Time = 300 s

Average Power in Session 2: $\text{Power} = \frac{37900}{300} \approx 126.33 \text{ W} \text{ (or 0.13 kW)}$

Thus, the displays show that the athlete produced more power in Session 1 compared to Session 2, indicating a higher energy output over the same time period.

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.

Session 1:

Session 2:

Join the GCSE students using SimpleStudy...