2.8 Spring Forces — Worked Solutions
Work through Topic 2.8
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
A horizontal spring has k = 100 N/m and signed extension x = −0.02 m. What is F_s,x?
- −2 N
- +2 N
- Zero
Step 1. Measure x from relaxed length and write F_s,x = −kx.
Why this step? A restoring spring force points opposite the signed deformation, so the minus sign belongs to the force law.
Step 2. Substitute F_s,x = −100(−0.02) = +2 N.
Why this step? The compressed position has negative x; the double negative produces a force in the positive direction toward relaxed length.
Step 3. Choose +2 N rather than −2 N or zero.
Why this step? The deformation is nonzero, and the restoring direction must oppose compression rather than push the mass farther into it.
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Question 2
A mass hangs at equilibrium from a vertical spring. Which statement is correct?
- The spring force is zero
- The spring force balances gravity
- Its extension must be measured from the equilibrium position to use Hooke’s law
Step 1. The hanging mass has gravity downward and spring force upward. At equilibrium their sum is zero.
Why this step? Static equilibrium requires zero acceleration, not zero individual interactions.
Step 2. Thus the upward spring force has magnitude mg and the spring is stretched from relaxed length.
Why this step? The weight has to be supported; a spring with zero deformation and zero force could not balance a nonzero mg.
Step 3. Use deformation from relaxed length for Hooke’s law, even though the equilibrium position is lower.
Why this step? Gravity shifts the balance position; it does not move the spring’s zero-force length or turn its spring force into zero there.
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Question 3
A signed spring-force graph F_s,x versus x has slope −250 N/m. What is k?
- −250 N/m
- 250 N/m
- 500 N/m
Step 1. From F_s,x = −kx, the signed force-versus-x slope is −k.
Why this step? Opposite restoring direction makes the force decrease as positive extension increases; the graph is not the external holding-force graph.
Step 2. Set −k = −250 N/m to obtain k = +250 N/m.
Why this step? Spring stiffness is the positive force-per-deformation magnitude; the negative graph slope encodes direction rather than negative stiffness.
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Question 4
A spring of relaxed length 0.15 m is held at length 0.21 m by a 9 N force. Find k, then predict the holding-force magnitude for length 0.18 m.
Step 1. Initial extension is x = 0.21 − 0.15 = 0.06 m.
Why this step? The 0.15 m relaxed length contributes no Hooke-law force; the spring’s full held length is not its deformation.
Step 2. At rest the holding force balances the spring magnitude, so k = F_hold/x = 9/0.06 = 150 N/m.
Why this step? Force divided by deformation determines stiffness because the spring is assumed linear in this range.
Step 3. At length 0.18 m, extension is 0.18 − 0.15 = 0.03 m.
Why this step? Both states must use the same relaxed-length reference, rather than taking the second length relative to the first.
Step 4. The new holding-force magnitude is kx = 150·0.03 = 4.5 N.
Why this step? The deformation is half as large, so a linear ideal spring needs half the original 9 N supporting pull.
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Question 5
Derive the vertical equilibrium extension mg/k. For displacement y from that equilibrium, show why the net force is −ky even though the spring force isn’t −ky.
Step 1. Take downward deformation x from relaxed length as positive. The total force is mg − kx.
Why this step? Gravity points along the positive axis while the stretched spring pulls upward; both interactions must be counted separately.
Step 2. At equilibrium, mg − kx_eq = 0, giving x_eq = mg/k.
Why this step? Zero net force fixes the extension where spring support equals weight; it does not make the spring force itself zero.
Step 3. If y is displacement below that equilibrium, the actual spring deformation is x = x_eq + y.
Why this step? Changing coordinate origin subtracts the equilibrium offset from position, but Hooke’s law still needs the relaxed-length deformation.
Step 4. Substitute: F_net = mg − k(x_eq + y) = mg − kx_eq − ky = −ky.
Why this step? The equilibrium weight and spring-offset terms cancel because kx_eq = mg, leaving a restoring net force about the shifted position.
Step 5. The spring force alone is −k(x_eq + y), not −ky.
Why this step? Only the combined spring-plus-gravity equation loses the constant offset; omitting it from the spring force would incorrectly predict no support at equilibrium.
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Question 6
Experimental pairs are (extension, force): (0.01 m, 2 N), (0.02 m, 4 N), and (0.03 m, 6 N). Sketch the graph, infer k with units, and state one measurement that would test the limit of the linear model.
Step 1. Plot extension on x and holding-force magnitude on y; the three points lie on one straight line.
Why this step? The stated force rises by 2 N for every 0.01 m deformation, so constant slope is testable across more than one load.
Step 2. Use k = ΔF/Δx = (6 − 2)/(0.03 − 0.01) = 200 N/m.
Why this step? This holding-force graph has positive slope k; dividing force change by length change gives the correct stiffness units.
Step 3. The extrapolated intercept is 2 − 200·0.01 = 0 N.
Why this step? A zero intercept agrees with measuring extension from a zero-force relaxed length rather than from the spring’s total length.
Step 4. Measure additional larger extensions without damaging the spring, then unload it and check its relaxed length.
Why this step? Curvature would show a changing stiffness, while failure to recover relaxed length would show that the assumed linear elastic model no longer applies.