A 35.0 N force stretches a bungee cord by 0.280 m. Assuming Hooke’s law applies, what is the elastic potential energy stored in the cord?

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BlackTom AI · Accepted Answer

To begin, identify the key quantities from Hooke's law and the elastic potential energy formula. The spring constant is k = F / x, since the force 35.0 N stretches the cord by 0.280 m. Calculating k gives 35.0 / 0.280 = 125 N/m.

Next, the elastic potential energy stored in a stretched (or compressed) spring is U = (1/2) k x^2. Here x = 0.280 m, so x^2 = 0.0784 m^2. Multiplying by k: k x^2 = 125 * 0.0784 = 9.8 J. Taking half yields U = (1/2) * 9.8 J = 4.9 J.

Option a: 4.90 J. This matches the calculation exactly, so the reasoning supports this value.

Option b: 1.70 J. This is significantly smaller than the computed energy; it could arise if one mistakenly used U = k x (which would give 125 * 0.280 = 35 J, or misapplied a factor of 1/2 or 1/4 in placing x somewhere), but it does not follow from the correct formula.

Option c: 3.75 J. This is close but not equal to 4.9 J; it might result from using an incorrect k (for example, using k = F / (2x) or rounding errors) or miscomputing x^2. The mismatch shows a deviation from the proper derivation.

Option d: 2.20 J. This value is also off by roughly a factor of two from the correct energy; it could stem from halving the energy twice or confusing units, but it does not align with U = (1/2) k x^2 with the given F and x.

In summary, the correct reasoning uses Hooke’s law to find k, then applies the elastic potential energy formula with the squared displacement. The other options reflect common mistakes or misapplications of the formulas, which is why they do not match the proper calculation.

SPH4U - Physics 12 (2025-26) - A

A 35.0 N force stretches a bungee cord by 0.280 m. Assuming Hooke’s law applies, what is the elastic potential energy stored in the cord?

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