University Physics I · unit UPI-07
Potential Energy & Conservation of 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.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
Conservative & Nonconservative Forces: infer an input with uncertainty
A force is conservative when its work on an object moving between two points is the same along every path joining them, which is the same as doing zero work round every closed path. Use the supplied topic model Y_out = -f*hypot(8,h), where Y_out is DRAG, STRAIGHT ROUTE (J). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; hypot(a,b) means √(a²+b²). h is Rise of B above A, h, measured in m. Keep d (Detour height above B, d) = 1.5 m; f (Drag force size, f) = 4 N fixed. For this exercise use only 0.425 m ≤ h ≤ 3.575 m 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 = -33.348118 J, with a bounded reading uncertainty ±0.19545 J. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown h by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000315 m, 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.
DRAG, STRAIGHT ROUTE against h
Predictions from the supplied Conservative & Nonconservative Forces 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): DRAG, STRAIGHT ROUTE / J
Horizontal axis (x): h: Rise of B above A, h / m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| h: Rise of B above A, h / m | DRAG, STRAIGHT ROUTE / J |
|---|---|
| 0.425 | -32.045124434147546 |
| 1.2125 | -32.36545226008745 |
| 2 | -32.984845004941285 |
| 2.7874999999999996 | -33.88690750127547 |
| 3.5749999999999997 | -35.04982168285596 |
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(h) = 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(0.425) = -32.0451 J and Y_out(3.575) = -35.0498 J. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives h_line = 0.425 + (-33.3481 − -32.0451)(3.575 − 0.425)/(-35.0498 − -32.0451) = 1.791 m. 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 h = 2.3465 m; substituting back gives Y_out = -33.348118 J.
- The interpolation error in the input is |h_line − h| = 0.555495 m. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -33.543568 J and -33.152668 J. Sort the roots even if the response decreases: 2.1665 m ≤ h ≤ 2.51459 m.
- The lower and upper input deviations are 0.180002 m and 0.168095 m. 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 h ≈ 2.3465 m; compatible interval [2.1665, 2.51459] m. Endpoint interpolation gives 1.791 m.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Near-Earth Gravitational Potential Energy: infer an input with uncertainty
Near-Earth gravitational potential energy is the energy stored in the Earth–object system by the object's height, U_g = mg(y − y₀), measured from a reference level y₀ you choose, so that only its changes ΔU_g = mgΔy carry physical meaning. Use the supplied topic model Y_out = 2*g*(yi-y0), where Y_out is U_i = mg(y_i − y₀) (J). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; g = 9.81 m s⁻². yi is Start height y_i, measured in m. Keep yf (End height y_f) = 0.8 m; y0 (Reference level y₀) = 0 m fixed. For this exercise use only 0.265625 m ≤ yi ≤ 2.23438 m 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 = 28.773956 J, with a bounded reading uncertainty ±2.2597 J. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown yi by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000196875 m, 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.
U_i = mg(y_i − y₀) against yi
Predictions from the supplied Near-Earth Gravitational 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): U_i = mg(y_i − y₀) / J
Horizontal axis (x): yi: Start height y_i / m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| yi: Start height y_i / m | U_i = mg(y_i − y₀) / J |
|---|---|
| 0.265625 | 5.2115625 |
| 0.7578125 | 14.86828125 |
| 1.25 | 24.525000000000002 |
| 1.7421875 | 34.18171875 |
| 2.234375 | 43.838437500000005 |
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(yi) = 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(0.265625) = 5.21156 J and Y_out(2.23438) = 43.8384 J. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives yi_line = 0.265625 + (28.774 − 5.21156)(2.23438 − 0.265625)/(43.8384 − 5.21156) = 1.46656 m. 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 yi = 1.46656 m; substituting back gives Y_out = 28.773956 J.
- The interpolation error in the input is |yi_line − yi| = 2.22045e-16 m. This is within the requested numerical tolerance.
- Invert both limits Y_out = 26.514256 J and 31.033656 J. Sort the roots even if the response decreases: 1.35139 m ≤ yi ≤ 1.58174 m.
- The lower and upper input deviations are 0.115173 m and 0.115173 m. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer yi ≈ 1.46656 m; compatible interval [1.35139, 1.58174] m. Endpoint interpolation gives 1.46656 m.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.
Force from Potential Energy: infer an input with uncertainty
For a conservative force acting along one axis, the force component at any position is the negative of the rate at which potential energy changes with position: Fₓ = −dU/dx. Use the supplied topic model Y_out = 16*cos(acos(3*sqrt(3)*f/8)/3-2*pi/3)^2-4, where Y_out is U″ AT HUMP (N/m). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; unmarked trigonometric functions use radians. f is Uniform applied force F₀, measured in N. Keep p (Bead position x) = 0.5 m fixed. For this exercise use only 0.07875 N ≤ f ≤ 1.18125 N 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 = -3.5424648 N/m, with a bounded reading uncertainty ±0.067932 N/m. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown f by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00011025 N, 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.
U″ AT HUMP against f
Predictions from the supplied Force from 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): U″ AT HUMP / N/m
Horizontal axis (x): f: Uniform applied force F₀ / N
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f: Uniform applied force F₀ / N | U″ AT HUMP / N/m |
|---|---|
| 0.07874999999999988 | -3.995345217658943 |
| 0.3543749999999999 | -3.9042932263154007 |
| 0.6299999999999999 | -3.686119597284891 |
| 0.9056249999999999 | -3.3071953928347106 |
| 1.18125 | -2.678420226121326 |
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(f) = 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(0.07875) = -3.99535 N/m and Y_out(1.18125) = -2.67842 N/m. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives f_line = 0.07875 + (-3.54246 − -3.99535)(1.18125 − 0.07875)/(-2.67842 − -3.99535) = 0.457891 N. 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 f = 0.751275 N; substituting back gives Y_out = -3.5424648 N/m.
- The interpolation error in the input is |f_line − f| = 0.293384 N. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -3.6103968 N/m and -3.4745328 N/m. Sort the roots even if the response decreases: 0.697343 N ≤ f ≤ 0.800379 N.
- The lower and upper input deviations are 0.053932 N and 0.0491045 N. 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 f ≈ 0.751275 N; compatible interval [0.697343, 0.800379] N. Endpoint interpolation gives 0.457891 N.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.