University Physics I · unit UPI-06
Work & Kinetic 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.
Work by a Constant Force: infer an input with uncertainty
The work done by a constant force is the dot product of the force with the displacement of the point where it acts, W = F⋅Δr: a signed scalar, measured in joules, that counts only the part of the force along the motion. Use the supplied topic model Y_out = -5*mu*(15*g-pull*sind(th)), where Y_out is WORK BY FRICTION W_f (J). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees; g = 9.81 m s⁻². th is Rope angle θ above the motion, measured in °. Keep pull (Rope pull F) = 40 N; mu (Kinetic friction coefficient μₖ) = 0.2 fixed. For this exercise use only 94.725 ° ≤ th ≤ 160.875 ° 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 = -118.90352 J, with a bounded reading uncertainty ±1.7426 J. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown th by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.006615 °, 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.
WORK BY FRICTION W_f against th
Predictions from the supplied Work by a Constant Force 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): WORK BY FRICTION W_f / J
Horizontal axis (x): th: Rope angle θ above the motion / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th: Rope angle θ above the motion / ° | WORK BY FRICTION W_f / J |
|---|---|
| 94.725 | -107.28593841890674 |
| 111.26249999999999 | -109.87284898441996 |
| 127.8 | -115.54379950497238 |
| 144.3375 | -123.82961684066598 |
| 160.875 | -134.04479281753225 |
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(th) = 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(94.725) = -107.286 J and Y_out(160.875) = -134.045 J. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives th_line = 94.725 + (-118.904 − -107.286)(160.875 − 94.725)/(-134.045 − -107.286) = 123.445 °. 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 th = 135.077 °; substituting back gives Y_out = -118.90352 J.
- The interpolation error in the input is |th_line − th| = 11.6319 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -120.64612 J and -117.16092 J. Sort the roots even if the response decreases: 131.433 ° ≤ th ≤ 138.502 °.
- The lower and upper input deviations are 3.64324 ° and 3.42526 °. 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 th ≈ 135.077 °; compatible interval [131.433, 138.502] °. Endpoint interpolation gives 123.445 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Work by a Variable Force: infer an input with uncertainty
The work done by a force whose component F(x) along a line changes with position is W = ∫ F(x) dx from the start position to the end position, which is the signed area under the force-position graph between them. Use the supplied topic model Y_out = -200*(xb-xa)*(xa+(xb-xa)*(n-1)/(2*n)), where Y_out is STRIP SUM Σ F Δx (J). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. xa is Start position x₁, measured in m. Keep xb (End position x₂) = 0.3 m; n (Number of strips N) = 4 fixed. For this exercise use only -0.23625 m ≤ xa ≤ 0.04725 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 = -5.2991763 J, with a bounded reading uncertainty ±0.28208 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 0.00002835 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.
STRIP SUM Σ F Δx against xa
Predictions from the supplied Work by a Variable Force 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): STRIP SUM Σ F Δx / J
Horizontal axis (x): xa: Start position x₁ / m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| xa: Start position x₁ / m | STRIP SUM Σ F Δx / J |
|---|---|
| -0.23625 | 3.770507812499997 |
| -0.165375 | -0.8507636718750012 |
| -0.0945 | -4.2162187499999995 |
| -0.023625000000000007 | -6.325857421874998 |
| 0.047249999999999986 | -7.1796796875 |
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(-0.23625) = 3.77051 J and Y_out(0.04725) = -7.17968 J. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives xa_line = -0.23625 + (-5.29918 − 3.77051)(0.04725 − -0.23625)/(-7.17968 − 3.77051) = -0.00143617 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 xa = -0.063315 m; substituting back gives Y_out = -5.2991763 J.
- The interpolation error in the input is |xa_line − xa| = 0.0618788 m. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -5.5812563 J and -5.0170963 J. Sort the roots even if the response decreases: -0.0721485 m ≤ xa ≤ -0.0537978 m.
- The lower and upper input deviations are 0.00883351 m and 0.00951716 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 xa ≈ -0.063315 m; compatible interval [-0.0721485, -0.0537978] m. Endpoint interpolation gives -0.00143617 m.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Work–Kinetic Energy Theorem: infer an input with uncertainty
The work–kinetic energy theorem says that the net work done on a particle by all the forces acting on it equals the change in its kinetic energy: W_net = ½mv_f² − ½mv_i². Use the supplied topic model Y_out = p*min(vi^2/max(19.62*mu-p,0.001),3), where Y_out is WORK BY PUSH (J). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; min(a,b) selects the smaller value; max(a,b) selects the larger value. mu is Friction coefficient μₖ, dimensionless. Keep p (Push F) = 8 N; vi (Initial speed v₀) = 1 m s⁻¹ fixed. For this exercise use only 0.42875 ≤ mu ≤ 0.468125 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 = 9.056711 J, with a bounded reading uncertainty ±0.34552 J. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown mu by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000039375, 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.
WORK BY PUSH against mu
Predictions from the supplied Work–Kinetic Energy Theorem 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): WORK BY PUSH / J
Horizontal axis (x): mu: Friction coefficient μₖ
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| mu: Friction coefficient μₖ | WORK BY PUSH / J |
|---|---|
| 0.42874999999999996 | 19.413941636837965 |
| 0.43859374999999995 | 13.218565888871117 |
| 0.44843749999999993 | 10.020746075860192 |
| 0.45828125 | 8.068760972411768 |
| 0.46812499999999996 | 6.7532631978811635 |
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(mu) = 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(0.42875) = 19.4139 J and Y_out(0.468125) = 6.75326 J. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives mu_line = 0.42875 + (9.05671 − 19.4139)(0.468125 − 0.42875)/(6.75326 − 19.4139) = 0.460961. 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 mu = 0.452769; substituting back gives Y_out = 9.056711 J.
- The interpolation error in the input is |mu_line − mu| = 0.00819247. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 8.711191 J and 9.402231 J. Sort the roots even if the response decreases: 0.451114 ≤ mu ≤ 0.454554.
- The lower and upper input deviations are 0.00165448 and 0.00178573. 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 mu ≈ 0.452769; compatible interval [0.451114, 0.454554]. Endpoint interpolation gives 0.460961.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Kinetic energy: infer an input with uncertainty
Kinetic energy is the energy a body has because it is moving, K = ½mv²: a scalar set by its mass and its speed in a chosen frame, equal to the work needed to bring it to that speed from rest. Use the supplied topic model Y_out = 0.5*2*(v2^2 - v1^2), where Y_out is ΔK, GROUND FRAME (J). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. v1 is Velocity before, v₁, measured in m/s. Keep v2 (Velocity after, v₂) = 4 m/s; u (Observer frame velocity u) = 2 m/s fixed. For this exercise use only 0.2125 m/s ≤ v1 ≤ 3.3625 m/s 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 = 11.446044 J, with a bounded reading uncertainty ±0.67632 J. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown v1 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/s, 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.
ΔK, GROUND FRAME against v1
Predictions from the supplied Kinetic 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): ΔK, GROUND FRAME / J
Horizontal axis (x): v1: Velocity before, v₁ / m/s
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| v1: Velocity before, v₁ / m/s | ΔK, GROUND FRAME / J |
|---|---|
| 0.21250000000000036 | 15.95484375 |
| 1.0000000000000002 | 15 |
| 1.7875 | 12.80484375 |
| 2.575 | 9.369374999999998 |
| 3.3625 | 4.693593750000002 |
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(v1) = 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(0.2125) = 15.9548 J and Y_out(3.3625) = 4.69359 J. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives v1_line = 0.2125 + (11.446 − 15.9548)(3.3625 − 0.2125)/(4.69359 − 15.9548) = 1.4737 m/s. 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 v1 = 2.134 m/s; substituting back gives Y_out = 11.446044 J.
- The interpolation error in the input is |v1_line − v1| = 0.660297 m/s. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 10.769724 J and 12.122364 J. Sort the roots even if the response decreases: 1.96917 m/s ≤ v1 ≤ 2.28698 m/s.
- The lower and upper input deviations are 0.164829 m/s and 0.15298 m/s. 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 v1 ≈ 2.134 m/s; compatible interval [1.96917, 2.28698] m/s. Endpoint interpolation gives 1.4737 m/s.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.