University Physics I · topic UPI-07.03
Elastic potential energy
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Elastic Potential Energy: predict a finite change
Elastic potential energy is the energy stored in a deformed spring or elastic body by the work done deforming it; for an ideal spring deformed by x from its natural length it is U = ½kx². Use the supplied topic model Y_out = k*(xa+dx)^2/20000, where Y_out is STORED AT x₂ ½kx₂² (J). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. xa is Starting stretch x₁, measured in cm. Keep k (Spring stiffness k) = 400 N m⁻¹; dx (Extra stretch Δx) = 4 cm fixed. For this exercise use only 0.85 cm ≤ xa ≤ 7.15 cm on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at xa = 2.236 cm and 3.118 cm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δxa across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
STORED AT x₂ ½kx₂² against xa
Predictions from the supplied Elastic Potential Energy model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): STORED AT x₂ ½kx₂² / J
Horizontal axis (x): xa: Starting stretch x₁ / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| xa: Starting stretch x₁ / cm | STORED AT x₂ ½kx₂² / J |
|---|---|
| 0.85 | 0.4704499999999999 |
| 2.425 | 0.8256125 |
| 4 | 1.28 |
| 5.574999999999999 | 1.8336125 |
| 7.1499999999999995 | 2.486449999999999 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of xa separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At xa = 2.236 cm, substitute into Y_out = k*(xa+dx)^2/20000: Y₀ = 0.777754 J.
- At xa = 3.118 cm, the same fixed-parameter model gives Y₁ = 1.01332 J.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.235565 J; the input change is Δxa = 0.882 cm.
- Average sensitivity is ΔY_out/Δxa = 0.26708 J per cm. Its positive sign means that this response increases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 0.777754 J, Y₁ = 1.01332 J; ΔY_out = 0.235565 J; average sensitivity = 0.26708 J per cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Elastic Potential Energy: infer an input with uncertainty
Elastic potential energy is the energy stored in a deformed spring or elastic body by the work done deforming it; for an ideal spring deformed by x from its natural length it is U = ½kx². Use the supplied topic model Y_out = k*(xa+dx)^2/20000, where Y_out is STORED AT x₂ ½kx₂² (J). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. xa is Starting stretch x₁, measured in cm. Keep k (Spring stiffness k) = 400 N m⁻¹; dx (Extra stretch Δx) = 4 cm fixed. For this exercise use only 0.85 cm ≤ xa ≤ 7.15 cm on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 1.511365 J, with a bounded reading uncertainty ±0.14626 J. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown xa by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00063 cm, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
STORED AT x₂ ½kx₂² against xa
Predictions from the supplied Elastic Potential Energy model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): STORED AT x₂ ½kx₂² / J
Horizontal axis (x): xa: Starting stretch x₁ / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| xa: Starting stretch x₁ / cm | STORED AT x₂ ½kx₂² / J |
|---|---|
| 0.85 | 0.4704499999999999 |
| 2.425 | 0.8256125 |
| 4 | 1.28 |
| 5.574999999999999 | 1.8336125 |
| 7.1499999999999995 | 2.486449999999999 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(xa) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(0.85) = 0.47045 J and Y_out(7.15) = 2.48645 J. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives xa_line = 0.85 + (1.51137 − 0.47045)(7.15 − 0.85)/(2.48645 − 0.47045) = 4.10286 cm. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives xa = 4.693 cm; substituting back gives Y_out = 1.511365 J.
- The interpolation error in the input is |xa_line − xa| = 0.590141 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.365105 J and 1.657625 J. Sort the roots even if the response decreases: 4.26167 cm ≤ xa ≤ 5.10391 cm.
- The lower and upper input deviations are 0.431326 cm and 0.410914 cm. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer xa ≈ 4.693 cm; compatible interval [4.26167, 5.10391] cm. Endpoint interpolation gives 4.10286 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.