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:

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

Given that the cyclist's weight is 700N and the distance from P to Q is 20m:

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

Thus, the total amount of work done against gravity is 14,000 J.

Using the equation for gravitational potential energy:

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

we know:\n- (\Delta GPE = 11250 J)\n- (g = 10 N/kg)\nLet's assume (\Delta h) (change in height) from Q to R can be derived if necessary. Reorganizing the formula for mass gives:

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

Substituting for (m):

• if (\Delta h = (height at Q - height at R)):
• find (\Delta h) and continue.

With complete data, substitute to find (m).

The total amount of work done by a cyclist includes not only the work done against gravity but also the work needed to overcome other forces such as friction and air resistance. In contrast, the change in gravitational potential energy (GPE) only considers the work done against gravity as the cyclist moves between points Q and R.

C increase the efficiency of the cyclist and bicycle. Lubrication reduces friction, allowing for smoother motion and less energy wastage.

Using the kinetic energy formula:

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

To find velocity, rearranging gives:

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

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

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

Therefore, the velocity of the cyclist is approximately 8.12 m/s.