https://file.aiquickdraw.com/mathbot/file/17292602181142m3xfn5w.jpg

To address the question in the image, we need to analyze the change in velocity and determine the work done on the puck in each situation. 

Work \( W \) is related to the change in kinetic energy, and the formula is:

\[
W = \Delta KE = \frac{1}{2} m (v_f^2 - v_i^2)
\]

Where:
- \( v_i \) is the initial velocity,
- \( v_f \) is the final velocity,
- \( m \) is the mass of the puck (which can be ignored since we are only comparing changes).

### Situation (a)
- Initial velocity: \( v_i = 6 \, \text{m/s} \) along the \( x \)-axis.
- Final velocity: \( v_f = 5 \, \text{m/s} \) along the \( y \)-axis.
  
The change in velocity is from 6 m/s in the \( x \)-direction to 5 m/s in the \( y \)-direction, indicating the puck undergoes a directional change with a reduced speed in total.

The work done will be:

\[
W_a = \frac{1}{2} m (5^2 - 6^2) = \frac{1}{2} m (25 - 36) = \frac{1}{2} m (-11) = -5.5m
\]

### Situation (b)
- Initial velocity: \( v_i = 4 \, \text{m/s} \) along the \( x \)-axis.
- Final velocity: \( v_f = 3 \, \text{m/s} \) along the \( y \)-axis.
  
The puck also changes direction here but with a larger decrease in speed.

\[
W_b = \frac{1}{2} m (3^2 - 4^2) = \frac{1}{2} m (9 - 16) = \frac{1}{2} m (-7) = -3.5m
\]

### Situation (c)
- Initial velocity: \( v_i = 4 \, \text{m/s} \) at an angle.
- Final velocity: \( v_f = 2 \, \text{m/s} \) in a different direction.
  
In this case, both speed and direction change significantly, with a greater reduction in velocity.

\[
W_c = \frac{1}{2} m (2^2 - 4^2) = \frac{1}{2} m (4 - 16) = \frac{1}{2} m (-12) = -6m
\]

### Ranking the Work Done:
- Most positive to most negative:
  1. Situation (b): \( W_b = -3.5m \)
  2. Situation (a): \( W_a = -5.5m \)
  3. Situation (c): \( W_c = -6m \)

### Conclusion:
The correct ranking for the work done, from most positive to most negative, is:
\( b > a > c \).

Would you like any further details on these calculations?

Here are some questions that expand on this information:
1. How does the direction of velocity affect the work done in a system?
2. What would happen if the surface were not frictionless?
3. Can we relate this concept to other real-world scenarios involving kinetic energy?
4. How would the work ranking change if the initial velocities were different?
5. What additional factors could affect the final velocities besides the applied force?

**Tip**: Work is only done by a force if there is a displacement in the direction of the force.

In three situations, a briefly applied horizontal force changes the velocity of a hockey puck that slides over frictionless ice. The overhead views of the figure indicate, for each situation, the puck’s initial speed v_i, its final speed v_f, and the directions of the corresponding velocity vectors. Rank the situations according to the work done on the puck by the applied force, most positive first and most negative last.

Analyze the changes in velocity of a hockey puck due to applied force on frictionless ice. Learn how to rank the work done using the work-energy theorem and kinetic energy formula. Suitable for high school physics students.

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