Work, Energy and Power · Lesson 16 of 16
Chapter Summary and Practice
“Connect the chapter’s methods and practise choosing the correct energy and momentum balance.”
• Choose a method from the physical conditions. • Recall formulas with their limitations. • Solve and check mixed conceptual and numerical problems.
Choose the question before choosing a formula
If you need a speed after a known displacement, work–energy often avoids finding acceleration or time. If the work is conservative and other work vanishes, use mechanical energy. If time or rate is central, use power. During a short collision with negligible external impulse, use momentum and a stated collision condition. A multistage question may require different methods in different intervals. Begin with a sketch, a system boundary, an inertial frame and a consistent energy reference.
| Goal | Main relation | Check first |
|---|---|---|
| Work of a force | W = ∫F·dr | Which force and which displacement? |
| Endpoint speed | ΔK = Wnet | Have all works been counted? |
| Conservative transfer | K+U = constant | Does remaining work vanish? |
| Dissipative motion | Δ(K+U) = Wother | What energy leaves the mechanical account? |
| Rate or input rating | P = F·v; Pin = Puseful/η | Useful versus input, instantaneous versus average |
| Impact velocities | Vector momentum plus collision condition | Is external impulse negligible? |
Formula reference with conditions
Practice: pull, slope and an energy ledger
Problem
A 2 kg box moves 5 m horizontally from rest under a 10 N pull at 60° upward. Let μk = 0.2 and g = 10 m/s². Find final speed.
- 1.Vertical balance gives N = mg − F sin60° = 20 − 8.660 = 11.340 N, assuming contact remains.
- 2.Pull work is 10×5×cos60° = 25 J. Friction is 0.2N = 2.268 N, so its work is −11.340 J.
- 3.Weight and normal do zero work. Net work is 13.660 J = ½×2×v².
- 4.Final speed is √13.660 ≈ 3.70 m/s. Setting N=mg would miss the pull’s reduction of friction.
Problem
A 1 kg block slides 4 m down a 30° incline from rest with μk = 0.2. Use g = 10 m/s². Find its final speed.
- 1.Vertical drop is 4 sin30° = 2 m, so gravity supplies 20 J.
- 2.Normal force is mg cos30° = 8.660 N. Friction dissipates 0.2×8.660×4 = 6.928 J.
- 3.Kf = 20 − 6.928 = 13.072 J. Speed is √(2×13.072) ≈ 5.11 m/s.
- 4.The energy route and the force route a=g(sin30°−μcos30°) give the same result.
Practice: compare kinetic energies and particle speeds
Problem
An electron has K = 10 keV and a proton has K = 100 keV. Use me = 9.11×10⁻³¹ kg and mp = 1.67×10⁻²⁷ kg. Which is faster in the chapter’s classical approximation?
- 1.Since v = √(2K/m), ve/vp = √[(Ke/Kp)(mp/me)].
- 2.The ratio is √[0.1×(1.67×10⁻²⁷)/(9.11×10⁻³¹)] ≈ 13.54. The electron is faster despite its smaller kinetic energy.
- 3.Using 1 eV = 1.60×10⁻¹⁹ J gives ve ≈ 5.93×10⁷ m/s and vp ≈ 4.38×10⁶ m/s.
- 4.The electron speed is about 0.20c, so relativity produces a small correction. The requested textbook result uses the nonrelativistic approximation; it should not be extrapolated to much higher electron energies.
Practice: combine a collision with a spring
Problem
A 1 kg block at 4 m/s sticks to a resting 1 kg block. The joined blocks then compress a 100 N/m spring on a smooth surface. Find maximum compression.
- 1.During the impact, conserve momentum: 1×4 = 2v, so v = 2 m/s.
- 2.Initial K before impact is 8 J, but K after impact is ½×2×2² = 4 J. The other 4 J becomes internal energy during sticking.
- 3.During spring compression, mechanical energy is conserved: 4 = ½×100×x².
- 4.Maximum compression is √0.08 = 0.283 m. Conserving the original 8 J through the inelastic impact would overestimate compression.
Common reasoning errors to avoid
A force and its third-law partner do not necessarily do opposite work. Zero net work does not imply absence of forces or work by individual forces. Negative potential energy is allowed; negative classical kinetic energy is not. Friction does not destroy total energy. Constant power does not mean constant force. In a collision, system momentum may be conserved while each object’s momentum changes. In an elastic collision, endpoint kinetic energies match, but kinetic energy need not remain constant during compression. Potential energy is defined only for the conservative interactions included in the model.
A final self-check
Before accepting a numerical answer, convert units, state the reference level, verify that K≥0, check that contact constraints such as N≥0 or T≥0 still hold, and substitute the result into the original energy equation. At a collision, verify signed momentum and the appropriate energy condition separately. An energy equation usually gives speed rather than direction, so recover direction from the physical geometry or additional dynamics. The theorem can be extended to an accelerating frame only with the appropriate inertial-force work; use an inertial frame by default.
Check your understanding
Quiz
A sticking collision followed by spring compression should use
Explanation: The inelastic impact changes mechanical energy, while the later ideal spring stage conserves it.
A graph shows E−U<0 at an endpoint. The endpoint is
Explanation: Kinetic energy cannot be negative.
A 4 kW device operating 30 min transfers
Explanation: Energy=4×0.5=2 kWh.
Zero instantaneous velocity at a turning point implies
Explanation: Force and potential energy can remain nonzero.
For a box pulled upward at an angle on a floor, before using μN you should
Explanation: The pull’s vertical component can reduce the normal force and friction.
Key Takeaways
• Work, energy, power and momentum answer related but different questions. • State the model, frame, force and system for every equation. • Check units, signs, contact conditions and conservation balances.
Previous · Lesson 15
Oblique Collisions and Multi-Stage Energy Problems
Next
End of chapter