A cyclist is riding a bicycle at a steady velocity of 12 m/s - Edexcel - GCSE Physics - Question 3 - 2018 - Paper 1

When the cyclist uses the brakes, the kinetic energy of the cyclist and the bicycle decreases. This kinetic energy is then transformed into thermal energy due to friction between the brake pads and the wheels. As a result, the thermal energy of the brakes increases, causing them to heat up. This process essentially converts mechanical energy into thermal energy.

To find the average force exerted by the cyclist, we use the formula:
$ext{Force} = \frac{\text{Work}}{\text{Distance}}$
In this case,

• Work = 1600 J
• Distance = 28 m
  Substituting the values:
  $\text{Force} = \frac{1600}{28}$
  $\text{Force} \approx 57.14 N$
  Thus, the average force exerted by the cyclist is approximately 57 N.

To calculate the average power for each session, we use the formula:
$\text{Power} = \frac{\text{Energy}}{\text{Time}}$
For Session 1:

• Energy = 45.2 kJ = 45200 J
• Time = 300 s
  Thus,
  $\text{Power}_{1} = \frac{45200}{300} \approx 150.67 W$
  For Session 2:
• Energy = 37.9 kJ = 37900 J
• Time = 300 s
  Thus,
  $\text{Power}_{2} = \frac{37900}{300} \approx 126.33 W$
  In summary, the athlete developed more power in Session 1 (approximately 150.67 W) compared to Session 2 (approximately 126.33 W).