The image you uploaded contains a physics problem involving energy conservation and drag force. Let's break it down:

Problem Overview: A 65 kg bicyclist is initially at rest at the top of a hill, coasting down and reaches a speed of 16.9 m/s at the bottom. The height of the hill is 19 m, and the horizontal distance is 80 m. We're also asked to determine the strength of drag.

Steps to solve:

1. Initial Energy (at the top of the hill):

   • The initial energy is all in the form of gravitational potential energy (GPE) since the cyclist starts at rest. $GPE = mgh$ where:
   • $m = 65 \, \text{kg}$
   • $g = 9.8 \, \text{m/s}^2$
   • $h = 19 \, \text{m}$

   So: $GPE = 65 \times 9.8 \times 19 = 12,350 \, \text{J}$

2. Final Energy (at the bottom of the hill):

   • At the bottom, all the gravitational potential energy would have been converted into kinetic energy (KE) if there were no losses (such as drag or friction).
   • The cyclist reaches a speed of $v = 16.9 \, \text{m/s}$. $KE = \frac{1}{2} m v^2$ $KE = \frac{1}{2} \times 65 \times (16.9)^2$ $KE \approx 18,677.85 \, \text{J}$

3. Energy Loss (due to drag):

   • The energy initially available as GPE has been partially converted into KE, but not completely. The difference between the initial GPE and the final KE is the energy lost due to drag (resistance). $\text{Loss} = GPE - KE$ $\text{Loss} = 12,350 - 18,677.85 \approx -6,327.85 \, \text{J}$

4. Calculating the drag force:

   • The drag force is responsible for the energy lost. We can calculate the work done by drag (which is negative because it's opposing the motion) using: $W = F \times d$ where $d = 80 \, \text{m}$ (the distance traveled horizontally).

   Thus: $F = \frac{\text{Loss}}{d} = \frac{-6,327.85}{80} \approx -79.1 \, \text{N}$

The strength of the drag force is approximately 79.1 N.

Final Answer:

The drag force is approximately 79.1 N.

Would you like a more detailed explanation or further clarification on any steps? Here are some related questions that might help expand on this topic:

1. How does friction affect the total energy conversion in a similar scenario?
2. How would the drag force change if the cyclist were moving faster?
3. What role does the coefficient of friction play in energy loss calculations?
4. How do you calculate the work done by non-conservative forces (like drag)?
5. What would happen to the final kinetic energy if there were no drag at all?

Tip: When calculating energy loss due to drag or friction, always consider that the total mechanical energy is not conserved when non-conservative forces are present.