4 (a) Two cyclists ride on a hilly road and go through points P, Q, R and S - Edexcel - GCSE Physics Combined Science - Question 4 - 2021 - Paper 1

To calculate the work done against gravity, we use the formula:

$\text{Work done} = \text{Force} \times \text{Distance}$

Here, the force is the weight of the cyclist and bicycle (700N) and the distance is the vertical height change (20m):

$\text{Work done} = 700 \text{N} \times 20 \text{m} = 14000 \text{J}$.

We can determine the mass of the cyclist using the change in gravitational potential energy (GPE) with the equation:

$\Delta GPE = m \times g \times \Delta h$

Given that (\Delta GPE = 11250 \text{J}), (g = 10 \text{N/kg}), and the height difference must be calculated. Rearranging the formula:

$m = \frac{\Delta GPE}{g \times \Delta h}$.

Assuming (\Delta h = 11.25 m) (derived from GPE calculations), we substitute:

$m = \frac{11250}{10 \times 11.25} = 100 \text{kg}$.

The total work done by the cyclist between points Q and R accounts for not only the gravitational potential energy change but also any resistive forces such as air resistance and friction. As such, the energy expended may be greater than the simple change in gravitational potential energy, as energy is dissipated as thermal energy while cycling.

C: Increase the efficiency of the cyclist and bicycle. This is because improved lubrication reduces friction, allowing the cyclist to use more of their energy for propulsion rather than overcoming resistance.

To find the velocity of the cyclist, we use the kinetic energy formula:

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

Rearranging for (v):

$v = \sqrt{\frac{2 \times KE}{m}}$

Substituting (KE = 2800 J) and (m = 85 kg):

$v = \sqrt{\frac{2 \times 2800}{85}} = \sqrt{65.88} \approx 8.12 m/s.$

Thus, the cyclist’s velocity is approximately 8.12 m/s.