Question

Potential energy (V) versus distance (x) is given by the graph. Rank various regions as per the magnitudes of the force (F) acting on a particle from high to low.

• A. \( F_{CD}>F_{AB}>F_{BC}>F_{DE} \)
• B. \( F_{CD}>F_{DE}>F_{AB}>F_{BC} \)
• C. \( F_{BC}>F_{AB}>F_{DE}>F_{CD} \)
• D. \( F_{BC}>F_{CD}>F_{DE}>F_{AB} \)

Hint:

Remember that the force is related to the \textbf{slope} of the potential energy graph, not the value of the potential energy itself. A steeper slope (either positive or negative) means a larger force magnitude. A horizontal line (zero slope) means zero force.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks us to rank the magnitudes of the force in different regions (AB, BC, CD, DE) based on a given potential energy (V) versus position (x) graph.
Step 2: Key Formula or Approach:
The relationship between force (F) and potential energy (V) is given by the negative gradient of the potential energy. For one dimension, this is:
\[ F = - \frac{dV}{dx} \] The magnitude of the force is the absolute value of the slope of the V-x graph:
\[ |F| = \left| - \frac{dV}{dx} \right| = \left| \text{slope of V-x graph} \right| \] We need to calculate the magnitude of the slope for each region.
Step 3: Detailed Explanation:
From the graph, we can determine the coordinates of the points:
A = (0, 0), B = (1, 1), C = (2, 3), D = (4, 3), E = (6, 1).
Now, we calculate the slope for each region:
\begin{itemize} \item Region AB: The slope is \( m_{AB} = \frac{V_B - V_A}{x_B - x_A} = \frac{1 - 0}{1 - 0} = 1 \).
Therefore, the magnitude of the force is \( |F_{AB}| = |1| = 1 \).
\item Region BC: The slope is \( m_{BC} = \frac{V_C - V_B}{x_C - x_B} = \frac{3 - 1}{2 - 1} = 2 \).
Therefore, the magnitude of the force is \( |F_{BC}| = |2| = 2 \).
\item Region CD: The slope is \( m_{CD} = \frac{V_D - V_C}{x_D - x_C} = \frac{3 - 3}{4 - 2} = 0 \).
Therefore, the magnitude of the force is \( |F_{CD}| = |0| = 0 \).
\item Region DE: The slope is \( m_{DE} = \frac{V_E - V_D}{x_E - x_D} = \frac{1 - 3}{6 - 4} = \frac{-2}{2} = -1 \).
Therefore, the magnitude of the force is \( |F_{DE}| = |-1| = 1 \).
\end{itemize} Step 4: Final Answer:
Ranking the magnitudes of the forces from high to low:
\( |F_{BC}| = 2 \)
\( |F_{AB}| = 1 \)
\( |F_{DE}| = 1 \)
\( |F_{CD}| = 0 \)
So, the ranking is \( F_{BC}>F_{AB} = F_{DE}>F_{CD} \).
Looking at the options, option (C) fits this ranking.