3.3 Potential Energy — Worked Solutions
Work through Topic 3.3
These are the full worked solutions to the lesson questions. Try the questions first, then check each step and the reason for it.
Question 1
You shift the potential-energy zero upward. What changes in a fixed lift?
- The energy values but not ΔU
- The required energy difference
- The gravitational force
Step 1. Moving the zero-height reference to y₀ gives U′ = mg(y − y₀) = U − mgy₀.
Why this step? The same reference shift applies to every position in this fixed gravitational model, so it adds one constant offset.
Step 2. Then ΔU′ = (U_f − mgy₀) − (U_i − mgy₀) = U_f − U_i = ΔU.
Why this step? The reference offsets cancel when subtracting the two states; the fixed lift’s energy transfer cannot depend on where we label zero.
Step 3. The individual values change, but the energy difference does not.
Why this step? The interaction and height change stayed physical constants of the lift; changing their labels cannot change the required work.
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Question 2
A spring is compressed and stretched by equal magnitudes from relaxed length. Compare U_s.
- Compression has negative energy
- The elastic energies are equal
- Stretching has twice the energy
Step 1. Write the two relaxed-length deformations as x = +d and x = −d.
Why this step? Equal stretch and compression magnitudes differ by direction, while the spring model uses the same k and relaxed reference.
Step 2. Both energies are ½k(±d)² = ½kd².
Why this step? Squaring removes the deformation sign, so compressing does not create negative elastic energy.
Step 3. Choose equal elastic energies.
Why this step? The stored energy depends on deformation magnitude squared rather than on whether the restoring force points left or right.
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Question 3
For U_g = −GMm/r with zero at infinity, what happens as r increases?
- U becomes more negative
- U increases toward zero
- U stays constant
Step 1. At finite r, U = −GMm/r is negative; at 2r it is −GMm/(2r).
Why this step? The specified infinity zero makes a bound finite-separation configuration lower in energy than infinite separation.
Step 2. The second value is less negative, so U increases as separation grows.
Why this step? Multiplying a positive reciprocal distance by the negative constant reverses the comparison; halving its magnitude moves the value upward.
Step 3. As r becomes arbitrarily large, U approaches zero from below.
Why this step? The reciprocal term tends toward zero without changing sign, so the curve does not become more negative or stay constant.
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Question 4
A 300 N/m spring is stretched 0.10 m from relaxed length. Find its energy and the energy after doubling that stretch.
Step 1. Use x = 0.10 m from relaxed length in U_s = ½kx².
Why this step? The given displacement is the spring deformation relevant to elastic storage, not a shift from a gravity-modified equilibrium.
Step 2. Calculate U_s = ½(300)(0.10²) = 1.5 J.
Why this step? Squaring the 0.10 m deformation before multiplying gives energy units N·m rather than the force units of kx.
Step 3. At doubled stretch x = 0.20 m, U_s = ½(300)(0.20²) = 6.0 J.
Why this step? The coefficient k is held fixed; the doubled deformation enters twice through the square.
Step 4. The energy ratio is 6.0/1.5 = 4.
Why this step? This check confirms a quadratic deformation dependence and rejects the tempting linear doubling to 3 J.
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Question 5
Three equal masses m occupy the corners of an equilateral triangle of side r. Write the total gravitational potential energy with zero at infinite separation. Explain the number of terms.
Step 1. List the distinct interacting pairs: AB, AC, and BC.
Why this step? Potential energy belongs to each two-mass interaction; exchanging the names of one pair does not create a second interaction.
Step 2. Each pair contributes −Gm²/r because both masses equal m and every separation is r.
Why this step? The equilateral geometry supplies the same center separation for all three pairs, with infinity as the stated zero.
Step 3. Add the three contributions: U_total = −3Gm²/r.
Why this step? Pairwise gravitational energies add once per distinct pair; summing AB and BA separately would double-count the same stored interaction.
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Question 6
An object moves from center distance R to 2R from a planet. Find ΔU symbolically and describe how the general U–r graph represents that change.
Step 1. The initial and final values are U_i = −GMm/R and U_f = −GMm/(2R).
Why this step? The question gives center distances, so they can be substituted directly into the general potential rather than a near-surface mgh approximation.
Step 2. Subtract final minus initial: ΔU = −GMm/(2R) + GMm/R = +GMm/(2R).
Why this step? The initial negative value is subtracted; keeping that sign is what makes the outward move a positive energy change.
Step 3. On U versus r, the point at 2R lies higher and closer to zero than the point at R.
Why this step? Moving against attraction increases the system’s potential energy even though both plotted absolute values remain negative.