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91/3334–6FE Reference Handbook 10.4 · Dynamics · Kinetic energy
1. Constant Acceleration2. Newton’s Second Law3. Work–Energy Principle4. No-Time Acceleration Equation5. Position–Time Equation6. Impulse–Momentum Principle7. Translational Kinetic Energy8. Normal (Centripetal) Acceleration9. Level-Ground Projectile Range10. Power11. Angular Form of Newton’s Second Law12. Relative Velocity13. Coefficient of Restitution14. Elastic Spring Energy15. Rigid-Body Rotation: Tangential and Normal Accel.16. Mass Moment of Inertia17. Conservation of Mechanical Energy18. Angular Impulse–Momentum19. Rolling Without Slip20. Polar / Path Normal–Tangential Acceleration21. Instantaneous Center of Zero Velocity22. Work of a Variable Force

Handbook formula

Translational Kinetic Energy

Scalar energy of a particle due to speed. Work of all forces equals ΔT. Doubling speed quadruples kinetic energy; the unit is the joule (N·m).

(J)
Kinetic energy
(kg)
Mass
(m/s)
Speed

Step-by-step solved example

Kinetic energy of a 1200 kg vehicle traveling 25 m/s?

SSDprtbraking
KE = ½mv² along the path — braking work removes it.

  1. 1. Square the speed

    v² = 625 m²/s².

  2. 2. Apply T = ½mv²

    T = ½×1200×625 = 375000 J = 375 kJ.

Answer: 375 kJ

10 practice questions

0/10 correct

1.KE of 4 kg at 5 m/s is

2.2 kg has T = 16 J. v =

3.Doubling v multiplies T by

4.SI unit of kinetic energy is

5.800 kg at 10 m/s: T =

6.Kinetic energy is a

7.Same T, mass ×4. The new speed is

8.The work–energy principle states that net work equals

9.0.5 kg at 20 m/s: T =

10.If v = 0, translational T is