University Physics I · unit UPI-05
Applications of Newton's Laws
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.
Inclined planes: infer an input with uncertainty
An inclined-plane analysis is Newton's second law written along and perpendicular to a tilted surface: the perpendicular equation fixes the normal force, and the along-slope equation, with friction, decides whether the object holds or slides and with what acceleration. Use the supplied topic model Y_out = tand(t)/s, where Y_out is DEMAND ÷ CEILING tan θ/μₛ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. s is Static coefficient μₛ, dimensionless. Keep t (Ramp angle θ) = 20 °; k (Kinetic coefficient μₖ) = 0.3 fixed. For this exercise use only 0.419063 ≤ s ≤ 0.930937 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 = 0.49769879, with a bounded reading uncertainty ±0.016009. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown s by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000511875, 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.
DEMAND ÷ CEILING tan θ/μₛ against s
Predictions from the supplied Inclined planes 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): DEMAND ÷ CEILING tan θ/μₛ
Horizontal axis (x): s: Static coefficient μₛ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Static coefficient μₛ | DEMAND ÷ CEILING tan θ/μₛ |
|---|---|
| 0.4190625 | 0.8685344889275521 |
| 0.54703125 | 0.6653554696668652 |
| 0.675 | 0.5392151618758553 |
| 0.80296875 | 0.45328069649809205 |
| 0.9309375 | 0.3909717185806806 |
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(s) = 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(0.419063) = 0.868534 and Y_out(0.930937) = 0.390972. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives s_line = 0.419063 + (0.497699 − 0.868534)(0.930937 − 0.419063)/(0.390972 − 0.868534) = 0.816542. 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 s = 0.731306; substituting back gives Y_out = 0.49769879.
- The interpolation error in the input is |s_line − s| = 0.085236. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.48168979 and 0.51370779. Sort the roots even if the response decreases: 0.708516 ≤ s ≤ 0.755611.
- The lower and upper input deviations are 0.0227902 and 0.024305. 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 s ≈ 0.731306; compatible interval [0.708516, 0.755611]. Endpoint interpolation gives 0.816542.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Connected Objects: infer an input with uncertainty
Connected objects are bodies joined by a taut string, a rigid bar or direct contact, so that they share one acceleration and can be treated either as a single composite system or one at a time to reveal the forces between them. Use the supplied topic model Y_out = (pull-0.4*5*g-grip*m2*g)/(5+m2), where Y_out is SHARED a IF STRING TAUT (m s⁻²). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; g = 9.81 m s⁻². m2 is Towed block mass m₂, measured in kg. Keep pull (Pull on the front block F) = 40 N; grip (Towed block friction μ₂) = 0.4 fixed. For this exercise use only 1.6375 kg ≤ m2 ≤ 6.3625 kg 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 = 0.27779101 m s⁻², with a bounded reading uncertainty ±0.10222 m s⁻². Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown m2 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0004725 kg, 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.
SHARED a IF STRING TAUT against m2
Predictions from the supplied Connected Objects 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): SHARED a IF STRING TAUT / m s⁻²
Horizontal axis (x): m2: Towed block mass m₂ / kg
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m2: Towed block mass m₂ / kg | SHARED a IF STRING TAUT / m s⁻² |
|---|---|
| 1.6375 | 2.102365348399246 |
| 2.8187499999999996 | 1.1919072741806556 |
| 4 | 0.5204444444444442 |
| 5.1812499999999995 | 0.004790669122160749 |
| 6.3625 | -0.40364796479647985 |
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(m2) = 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(1.6375) = 2.10237 m s⁻² and Y_out(6.3625) = -0.403648 m s⁻². The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives m2_line = 1.6375 + (0.277791 − 2.10237)(6.3625 − 1.6375)/(-0.403648 − 2.10237) = 5.07767 kg. 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 m2 = 4.51975 kg; substituting back gives Y_out = 0.27779101 m s⁻².
- The interpolation error in the input is |m2_line − m2| = 0.557921 kg. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.17557101 m s⁻² and 0.38001101 m s⁻². Sort the roots even if the response decreases: 4.29366 kg ≤ m2 ≤ 4.75712 kg.
- The lower and upper input deviations are 0.226093 kg and 0.237368 kg. 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 m2 ≈ 4.51975 kg; compatible interval [4.29366, 4.75712] kg. Endpoint interpolation gives 5.07767 kg.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Pulley and Atwood-type systems: infer an input with uncertainty
An Atwood-type system is a set of bodies joined by ideal ropes over pulleys, solved by writing Newton's second law for each body and closing the equations with one length constraint per rope. Use the supplied topic model Y_out = mc*ml*g*(1+n)/(ml+n*n*mc), where Y_out is TENSION PER SEGMENT T (N). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; g = 9.81 m s⁻². ml is Load mass M, measured in kg. Keep n (Segments holding the load n) = 2; mc (Counterweight mass m) = 4 kg fixed. For this exercise use only 6.125 kg ≤ ml ≤ 21.875 kg 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 = 58.363819 N, with a bounded reading uncertainty ±1.4439 N. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown ml by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.001575 kg, 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.
TENSION PER SEGMENT T against ml
Predictions from the supplied Pulley and Atwood-type systems 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): TENSION PER SEGMENT T / N
Horizontal axis (x): ml: Load mass M / kg
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| ml: Load mass M / kg | TENSION PER SEGMENT T / N |
|---|---|
| 6.125 | 32.58915254237288 |
| 10.0625 | 45.450647482014396 |
| 14 | 54.936 |
| 17.9375 | 62.220331491712706 |
| 21.875 | 67.99009900990099 |
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(ml) = 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(6.125) = 32.5892 N and Y_out(21.875) = 67.9901 N. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives ml_line = 6.125 + (58.3638 − 32.5892)(21.875 − 6.125)/(67.9901 − 32.5892) = 17.5922 kg. 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 ml = 15.7325 kg; substituting back gives Y_out = 58.363819 N.
- The interpolation error in the input is |ml_line − ml| = 1.85974 kg. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 56.919919 N and 59.807719 N. Sort the roots even if the response decreases: 14.9789 kg ≤ ml ≤ 16.5237 kg.
- The lower and upper input deviations are 0.753594 kg and 0.791172 kg. 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 ml ≈ 15.7325 kg; compatible interval [14.9789, 16.5237] kg. Endpoint interpolation gives 17.5922 kg.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Elevators & Apparent Weight: infer an input with uncertainty
Apparent weight is the normal force a support exerts on a body, N = m(g + a_y) with upward positive; it is the quantity a scale actually reads, and it equals the true weight mg only while the vertical acceleration is zero. Use the supplied topic model Y_out = m*max(g+a,0), where Y_out is NORMAL FORCE N (N). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; max(a,b) selects the larger value; g = 9.81 m s⁻². m is Rider mass m, measured in kg. Keep a (Lift acceleration a_y (up positive)) = 1.5 m s⁻²; v (Lift velocity v_y (up positive)) = 2 m s⁻¹ fixed. For this exercise use only 48.5 kg ≤ m ≤ 111.5 kg 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 = 983.1783 N, with a bounded reading uncertainty ±41.683 N. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown m by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0063 kg, 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.
NORMAL FORCE N against m
Predictions from the supplied Elevators & Apparent Weight 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): NORMAL FORCE N / N
Horizontal axis (x): m: Rider mass m / kg
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| m: Rider mass m / kg | NORMAL FORCE N / N |
|---|---|
| 48.5 | 548.535 |
| 64.25 | 726.6675 |
| 80 | 904.8000000000001 |
| 95.75 | 1082.9325000000001 |
| 111.5 | 1261.065 |
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(m) = 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(48.5) = 548.535 N and Y_out(111.5) = 1261.07 N. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives m_line = 48.5 + (983.178 − 548.535)(111.5 − 48.5)/(1261.07 − 548.535) = 86.93 kg. 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 m = 86.93 kg; substituting back gives Y_out = 983.1783 N.
- The interpolation error in the input is |m_line − m| = 0 kg. This is within the requested numerical tolerance.
- Invert both limits Y_out = 941.4953 N and 1024.8613 N. Sort the roots even if the response decreases: 83.2445 kg ≤ m ≤ 90.6155 kg.
- The lower and upper input deviations are 3.6855 kg and 3.6855 kg. 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 m ≈ 86.93 kg; compatible interval [83.2445, 90.6155] kg. Endpoint interpolation gives 86.93 kg.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.