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Engineering & Technology · Dynamics, Vibration & Mechanical Response

Work-energy and momentum methods

Two force histories can give a carriage the same final velocity while using different forces, durations and distances. Momentum and energy methods describe different aspects of that change. Their units help prevent a force-time area from being mistaken for work.

Use Forces, Energy & Simple Systems and Engineering Sketches, Models & Technical Communication. Be comfortable with signed quantities, ratios, squares and graph axes. Radians, frequency and each new symbol are introduced here. Sinusoidal response formulas are supplied and interpreted without requiring a differential-equation derivation. All numerical records and model parameters are original constructed cases, not equipment limits, operating instructions or dynamic qualification.

Back to the chapter

After reading, you can

Choose a force-distance or force-time account according to the question and preserve its system boundary.

01

Force over time changes momentum

Linear momentum is p=mv for a particle of constant mass; its direction follows velocity. Net external impulse is the accumulated force-time area. For constant net force, J=FΔt; for a varying force, the area can be found from a suitable record. The impulse-momentum relation says J=pf−pi.

N·s and kg·m/s are equivalent momentum units. Impulse is not a stock of energy and does not require displacement to appear in its equation. If external impulse is negligible for a chosen system, its total momentum is approximately conserved. For one carriage alone, a drive force is external; enlarging the boundary changes which interactions are internal.

02

Force along displacement changes kinetic energy

For our translating particle, K=mv²/2. Net work equals the change in K. A constant force component parallel to displacement contributes FΔx; an opposing component gives negative work. Force perpendicular to the instantaneous displacement does no work in this particle model, though it can change the direction of velocity.

Joules are N·m, not N·s. Work-energy reasoning often gives speed after a known distance without first finding duration. It does not by itself supply the direction after a complicated path or the peak force during a short event. Momentum reasoning can supply a velocity change from a time record while leaving travel distance to a motion calculation.

03

Use the net quantity or account for other destinations

A drive’s work need not all become carriage translational energy. Resistance can transfer energy to the surroundings; rotation and internal deformation can store some. If the calculation uses net work of all forces on the particle, it already combines driving and opposing contributions. Do not subtract a resistance transfer twice.

Similarly, an impulse determines the momentum change but not the detailed force waveform. The same impulse can have a sharp peak or a lower force over longer time. A mean force is not automatically a maximum or a permissible contact load. Real impact and stopping behaviour require compliance, geometry, material and time-dependent evidence beyond this simple account.

Equal force-time areas need not use equal force or distanceTwo separate constant net-force histories each deliver six newton seconds to a two-kilogram particle initially at rest.Separate cases, same initial rest and 2 kg mass00.511.52036A: solid, 6 N for 1 sB: dashed, 3 N for 2 sNet force / NTime / s; each enclosed area is 6 N·s
Case A net 6N for 1s; case B net 3N for 2s. Both start with 2kg at rest, impulse 6N·s and final velocity 3m/s. Constant accelerations 3 and 1.5m/s² give distances 1.5 and 3m respectively; net work 6×1.5=3×3=9J. Curves describe two alternatives, not simultaneous added forces.

Compare two equal-impulse histories

Take a 2 kg particle initially at rest. Case A supplies constant net force 6 N for 1 s: impulse 6 N·s, velocity change 6/2=3 m/s and acceleration 3 m/s². Its distance is½×3×1²=1.5 m, so net work 6×1.5=9 J. The final kinetic energy is½×2×3²=9 J.

Case B instead supplies 3 N for 2 s. Impulse is again 6 N·s and final velocity 3 m/s. Acceleration is 1.5 m/s²; distance½×1.5×2²=3 m. Work is 3×3=9 J. Equal final energy and momentum do not imply equal force, duration or displacement. The two plotted alternatives should not be added as simultaneous forces.

Try the reasoning

  1. What changes if the same net impulse acts on twice the mass initially at rest?

    Show the answer and reasoning

    The velocity change halves, because Δv=J/m.

  2. Why does 6 N ×1 s not equal 6 J?

    Show the answer and reasoning

    It is 6 N·s, an impulse. Work requires the force component along a displacement.

  3. Does an average stopping force identify the largest instantaneous force?

    Show the answer and reasoning

    No. Many force-time waveforms can have the same area or mean.

What to carry forward

Use impulse for momentum change and net work for kinetic-energy change; neither replaces the missing event details.

References

Teaching resources and research supporting this lesson’s explanations and examples.

  1. OpenStax · University Physics Volume 13 sources
    1. 5.3 Newton’s Second LawOpenStax · University Physics Volume 1Teaching resourceopenstax.orgRead source
    2. 7.3 Work-Energy TheoremOpenStax · University Physics Volume 1Teaching resourceopenstax.orgRead source
    3. 9.2 Impulse and CollisionsOpenStax · University Physics Volume 1Teaching resourceopenstax.orgRead source