University Physics I · topic UPI-05.02
Connected objects
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.
Connected Objects: predict a finite change
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. Calculate Y_out at m2 = 2.677 kg and 3.3385 kg. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δm2 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
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
Substitute each value of m2 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At m2 = 2.677 kg, substitute into Y_out = (pull-0.4*5*g-grip*m2*g)/(5+m2): Y₀ = 1.28637 m s⁻².
- At m2 = 3.3385 kg, the same fixed-parameter model gives Y₁ = 0.873026 m s⁻².
- Signed change: ΔY_out = Y₁ − Y₀ = -0.413343 m s⁻²; the input change is Δm2 = 0.6615 kg.
- Average sensitivity is ΔY_out/Δm2 = -0.624857 m s⁻² per kg. Its negative sign means that this response decreases 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 = 1.28637 m s⁻², Y₁ = 0.873026 m s⁻²; ΔY_out = -0.413343 m s⁻²; average sensitivity = -0.624857 m s⁻² per kg.
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.
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.