Work-energy and impulse-momentum methods

Work of forces, kinetic and potential energy, the work-energy principle, linear and angular impulse-momentum, power, and direct impact with the coefficient of restitution.

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Why it matters

Newton's second law gives accelerations, but many machine problems ask for a speed after a distance (brakes, springs, lifts, cams) or a velocity after a short, violent event (impacts, forging hammers, pile drivers, couplings). The work-energy and impulse-momentum methods answer these directly, without integrating accelerations, and they are the basis of energy audits and impact design.

Key ideas

Work of a force. U = ∫ F · dr: only the component of force along the displacement does work. A force perpendicular to the path (normal reaction on a fixed surface, tension in a pendulum string) does no work. Work is a scalar in joules (N·m).

  • Constant force along a straight path: U = F s cos θ.
  • Weight: U = −m g Δh (negative when the body rises).
  • Linear spring: U = −½ k (x2² − x1²), where x is the deflection from the free length.
  • Kinetic friction: U = −μk N s, always negative; the energy becomes heat.

Principle of work and energy. The net work of all forces on a particle or rigid body equals the change in its kinetic energy: T1 + U1→2 = T2. Kinetic energy of a rigid body in plane motion: T = ½ m v_G² + ½ I_G ω².

Conservative forces and potential energy. Gravity and spring forces do work that depends only on the end points, so they can be written as potential energies V. If only conservative forces do work, T1 + V1 = T2 + V2 (conservation of mechanical energy). Friction and drag are non-conservative; include their work separately.

Power and efficiency. Power is the rate of doing work, P = F v (or P = T ω for a shaft). Efficiency = output power / input power.

Linear impulse and momentum. m v1 + ∫ F dt = m v2. Momentum and impulse are vectors, so write the equation in components. The impulse of a variable force is the area under its force–time curve. For a system of particles, internal impulses cancel; if the external impulse is zero in a direction, momentum is conserved in that direction.

Angular impulse and momentum. For a rigid body rotating about a fixed axis: I ω1 + ∫ M dt = I ω2. If no external moment acts about an axis, angular momentum about it is conserved (a skater pulling in her arms speeds up).

Impact. In a short collision, impulsive contact forces are very large and finite forces such as weight are negligible during the impact. Total momentum along the line of impact is conserved. The coefficient of restitution e = (v2' − v1') / (v1 − v2) (separation speed / approach speed) ranges from 0 (perfectly plastic, bodies move together) to 1 (perfectly elastic, no kinetic energy lost). For 0 ≤ e < 1 kinetic energy is lost, so you must not use energy conservation across an impact.

Choosing a method. Force and distance known → work-energy. Force and time known, or impact → impulse-momentum. Need the force at an instant → Newton's second law.

Formulas

U = F s cos θ (constant force F at angle θ to a straight displacement s); U in J, F in N, s in m.

T = ½ m v² (particle); T = ½ m v_G² + ½ I_G ω² (rigid body in plane motion); T in J.

T1 + U1→2 = T2 (work-energy); T1 + V1 = T2 + V2 (only conservative forces work).

V_g = m g h, V_e = ½ k x²; k = spring stiffness (N/m), x = deflection from free length (m).

P = F v = T ω; P in W, T = torque (N·m), ω in rad/s.

m v1 + Σ∫F dt = m v2 (linear impulse-momentum; impulse in N·s).

I_O ω1 + Σ∫M_O dt = I_O ω2 (angular impulse-momentum about a fixed axis; angular momentum in kg·m²/s).

m1 u1 + m2 u2 = m1 v1 + m2 v2 and e = (v2 − v1)/(u1 − u2) (direct central impact; u before, v after).

ΔT_loss = ½ · (m1 m2 / (m1 + m2)) · (1 − e²) · (u1 − u2)² (kinetic energy lost in direct central impact).

Worked examples

Example 1 (standard, work-energy). A 5 kg block starts from rest and slides 4 m down a 30° incline with μk = 0.2. Find its speed at the bottom. g = 9.81 m/s².

  1. Work of gravity: U_g = m g s sin θ = 5 × 9.81 × 4 × 0.5 = 98.1 J.
  2. Normal force: N = m g cos θ = 5 × 9.81 × 0.866 = 42.48 N; work of friction: U_f = −μk N s = −0.2 × 42.48 × 4 = −33.98 J.
  3. Normal force does no work (perpendicular to motion).
  4. T2 = 0 + 98.1 − 33.98 = 64.12 J = ½ × 5 × v², so v = √(2 × 64.12 / 5) = 5.06 m/s.

Answer: v ≈ 5.06 m/s.

Example 2 (GATE level, impact). Ball A (2 kg) moving at 6 m/s strikes ball B (3 kg) at rest, head-on; e = 0.5. Find the velocities after impact, the kinetic energy lost and the impulse on B.

  1. Momentum: 2 × 6 + 3 × 0 = 2 v_A + 3 v_B, so 2 v_A + 3 v_B = 12.
  2. Restitution: v_B − v_A = e (u_A − u_B) = 0.5 × 6 = 3.
  3. Solve: from (2) v_B = v_A + 3; then 2 v_A + 3 v_A + 9 = 12, so v_A = 0.6 m/s and v_B = 3.6 m/s.
  4. Energy before: ½ × 2 × 6² = 36 J. After: ½ × 2 × 0.6² + ½ × 3 × 3.6² = 0.36 + 19.44 = 19.8 J. Loss = 16.2 J.
  5. Check with the formula: ½ × (2 × 3/5) × (1 − 0.25) × 6² = ½ × 1.2 × 0.75 × 36 = 16.2 J. ✓
  6. Impulse on B: 3 × 3.6 − 0 = 10.8 N·s (equal and opposite on A: 2 × (0.6 − 6) = −10.8 N·s).

Answer: v_A = 0.6 m/s, v_B = 3.6 m/s (both forward); energy lost 16.2 J; impulse 10.8 N·s.

Example 3 (spring). A 2 kg block sliding at 3 m/s on a smooth floor hits a spring of stiffness 800 N/m. Maximum compression: ½ m v² = ½ k x², so x = v √(m/k) = 3 × √(2/800) = 3 × 0.05 = 0.15 m. Answer: 0.15 m.

Common mistakes

  • Using energy conservation across an impact (energy is lost unless e = 1); use momentum and e instead.
  • Forgetting the rotational term ½ I ω² for rolling or rotating bodies.
  • Treating momentum or impulse as scalars in two-dimensional problems.
  • Measuring spring deflection from the wrong reference; it is from the free length.
  • Counting the work of the normal force, or of static friction on a body rolling without slipping (neither does work).
  • Sign errors in e: use separation speed over approach speed, both along the same positive direction.

For GATE PI

Typical questions: speed after sliding on rough inclines, spring–mass energy problems, collisions with a given coefficient of restitution and energy loss, impulse from force–time graphs, angular momentum conservation, and power of machines. Practise picking the right method within a minute of reading the question.

Quick check

  1. A 2 kg body speeds up from 3 m/s to 5 m/s. What net work was done on it?
  2. A constant 10 N force acts on a 4 kg body for 3 s along its motion. What is the change in velocity?
  3. What is e for a perfectly plastic impact?
  4. Does the normal reaction of a fixed smooth surface do work on a sliding block?

Answers: 1. 16 J. 2. 7.5 m/s. 3. Zero. 4. No, it is perpendicular to the motion.

Interview questions

All Engineering Mechanics and Strength of Materials interview questions

Try answering each one aloud before you open it.

  1. 1.What is the work-energy principle in engineering mechanics?Concept

    The net work done by all forces acting on a particle or rigid body between two positions equals the change in its kinetic energy: T1 + U1→2 = T2. For a rigid body T includes both ½ m v_G² and ½ I_G ω². If only conservative forces (gravity, springs) do work, it reduces to conservation of mechanical energy, T + V = constant; friction work must be added separately as a loss. It is the quickest route when forces and distances are known and a speed is wanted.

  2. 2.Explain the impulse-momentum principle.Concept

    The impulse-momentum principle states that the change in momentum of an object is equal to the impulse applied to it. Impulse is the product of force and the time duration over which it acts, and momentum is the product of mass and velocity. Mathematically, it is expressed as F·Δt = m·Δv, where F is the force, Δt is the time duration, m is the mass, and Δv is the change in velocity.

  3. 3.How is the work-energy method used to solve problems in engineering mechanics?Application

    The work-energy method is used to solve problems by calculating the work done by forces and equating it to the change in kinetic energy. This method is particularly useful when dealing with problems involving variable forces or when the path of motion is complex. It simplifies the analysis by focusing on energy changes rather than forces and accelerations.

  4. 4.Why is the impulse-momentum method preferred in collision analysis?Application

    The impulse-momentum method is preferred in collision analysis because it directly relates the forces during the collision to the change in momentum, which is often more straightforward to calculate than forces and accelerations. It is particularly useful in short-duration events where forces are not constant and can be difficult to measure.

  5. 5.What happens to the kinetic energy of a system if only conservative forces are acting on it?Application

    If only conservative forces are acting on a system, the total mechanical energy (sum of kinetic and potential energy) remains constant. This means that any change in kinetic energy is exactly balanced by a change in potential energy, and vice versa. Therefore, the kinetic energy can change, but the total energy remains constant.

  6. 6.How does the presence of non-conservative forces affect the work-energy principle?Application

    Non-conservative forces, such as friction, cause energy to be dissipated as heat or other forms of energy, which means that the total mechanical energy is not conserved. In the work-energy principle, the work done by non-conservative forces must be accounted for separately, as it results in a loss of mechanical energy from the system.

  7. 7.A 5 kg object is moving with a velocity of 10 m/s. Calculate the impulse required to stop the object.Numerical

    To stop the object, its final velocity must be 0 m/s. The initial momentum is m·v = 5 kg × 10 m/s = 50 kg·m/s. The change in momentum required is 50 kg·m/s. Therefore, the impulse required is also 50 N·s.

  8. 8.A force of 20 N is applied to a 2 kg object for 3 seconds. What is the change in velocity of the object?Numerical

    Impulse is given by F·Δt = 20 N × 3 s = 60 N·s. The change in momentum is equal to the impulse, so m·Δv = 60 kg·m/s. Solving for Δv gives Δv = 60 kg·m/s / 2 kg = 30 m/s.

  9. 9.Explain how energy conservation is used in analyzing projectile motion.Application

    In projectile motion, energy conservation is used by equating the initial total mechanical energy (kinetic plus potential) to the total mechanical energy at any other point in the motion. This allows for the calculation of velocities and heights without directly solving the equations of motion, simplifying the analysis.

  10. 10.What is the significance of the coefficient of restitution in collision problems?Concept

    The coefficient of restitution is a measure of how elastic a collision is. It is defined as the ratio of relative velocity after the collision to the relative velocity before the collision. A value of 1 indicates a perfectly elastic collision, while a value of 0 indicates a perfectly inelastic collision. It helps in determining the final velocities of colliding bodies.

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