Questions: Total Mechanical Energy and Energy Conservation

Question 1

A particle with total mechanical energy E = 10 J moves in a potential U(x). At a certain position, U(x) = 10 J. What must be true at that position?

• A. The particle is at maximum kinetic energy
• B. The particle has zero velocity — it is at a turning point
• C. The particle cannot exist there because it requires negative kinetic energy
• D. The particle is accelerating away from this position

Answer: (B)

When U = E, kinetic energy K = E − U = 0, meaning the particle has zero velocity. This is a turning point: the particle arrives, momentarily stops, and reverses direction. It does not 'cannot exist there' — the particle can reach this exact point (with K = 0). It would be forbidden only if U > E, which would require K < 0.

Question 2

A particle moves in a potential well with U_min = 2 J and barriers of height U_barrier = 8 J on either side. Its total mechanical energy is E = 5 J. What can you conclude about its motion without solving any equations?

• A. The particle oscillates forever between the two turning points where U = 5 J
• B. The particle eventually escapes over the 8 J barrier due to accumulated kinetic energy
• C. The particle can only be found at the single point where U = 2 J
• D. The particle moves freely since E > U_min

Answer: (A)

The particle is confined to the region where U(x) ≤ E = 5 J. Since the barriers reach 8 J > 5 J, the particle cannot pass over them. It bounces between the two turning points (where U = 5 J) indefinitely, converting kinetic and potential energy back and forth. This entire analysis comes from reading the potential energy graph — no equations of motion needed.

Question 3

A particle with total mechanical energy E can never be found at a location where the potential energy U(x) is greater than E.

• T. True
• F. False

Answer: True

Since E = K + U and kinetic energy K = ½mv² ≥ 0 always, it follows that U = E − K ≤ E. A location where U > E would require K = E − U < 0, which is physically impossible — you cannot have negative kinetic energy. These classically forbidden regions can be read directly off a plot of U(x) by seeing where U exceeds the horizontal line at E.

Question 4

The energy conservation method E = K + U = constant can mainly be applied at turning points, where the particle is momentarily at rest.

• T. True
• F. False

Answer: False

Energy conservation holds at every instant along the trajectory, not just at turning points. E = K + U is a constant of the motion. You can apply it at any point: if you know E and U at some location, you immediately get K = E − U and thus the speed at that point. Turning points are just a special case where K = 0, but the method is valid and useful throughout the entire trajectory.

Question 5

Explain why the non-negativity of kinetic energy (K ≥ 0) is the central tool for analyzing particle motion under a conservative potential, and what it lets you determine from a graph alone.

Model answer: Since K = ½mv² ≥ 0, the particle can only exist in regions where U(x) ≤ E. On a graph of U(x) with a horizontal line drawn at height E, the particle is confined to where the curve lies below the line. The intersections (U = E, K = 0) are turning points — the particle reverses direction there. The minimum of U gives maximum K and maximum speed. You can identify confinement regions, oscillation boundaries, and escape conditions purely geometrically, without integrating the equation of motion. This transforms dynamics (motion in time) into a spatial geometry problem.

This is the core power of energy methods in mechanics: they trade the full complexity of Newton's second law (a differential equation requiring initial conditions) for a simple geometric condition. The approach extends to multi-dimensional problems via effective potential, where angular momentum adds a centrifugal barrier term and the full orbit problem reduces to a 1D energy problem.