Home / Calculators / Work–Energy Theorem Dynamics
Find speed or travel distance using work and kinetic energy for a particle, including constant forces, gravity, springs, and kinetic friction.
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These are the solver's own reference problems — the answers come from published sources or independent hand computation, never from the solver itself. The solution below is the live solver output for each.
Example 1
pushed block: m=2 kg, v1=3 m/s, F=10 N over 5 m Given
mass 2 kg
initial velocity 3 m/s
solve for finalVelocity work terms 1. kind constant-force force 10 N distance 5 m angle deg 0 Assumptions The body is treated as a particle; work terms are the TOTAL work done on it between states 1 and 2; g = 9.80665 m/s². Solution steps Set up the work–energy theorem
The change in kinetic energy equals the total work done on the body.
T1 + ΣU = T2 with T = m·v²/2
Total work ΣU = 50 J
Solve for the final speed
Rearrange the theorem for the unknown kinetic energy, then take the square root.
v = √(v_known² ± 2·ΣU/m)
T1 = 9 J; ΣU = 50 J; T2 = 59 J; v = 7.681 m/s
= 7.681 m/s
Where this answer was checked source T1 + ΣU = T2 (NCEES FE Reference Handbook work-energy) — single constant force along the motion verified by hand-recomputed derivation T1 = 0.5·2·9 = 9 J; U = 10·5 = 50 J; T2 = 59 J; v2 = √(2·59/2) = √59 = 7.6811 m/s. Example 2
spring launcher: k=200 N/m compressed 0.1 m, m=0.5 kg from rest Given
mass 0.5 kg
initial velocity 0 m/s
solve for finalVelocity work terms 1. kind spring stiffness 200 N/m initial deformation 0.1 m final deformation 0 m Assumptions The body is treated as a particle; work terms are the TOTAL work done on it between states 1 and 2; g = 9.80665 m/s². Solution steps Set up the work–energy theorem
The change in kinetic energy equals the total work done on the body.
T1 + ΣU = T2 with T = m·v²/2
Total work ΣU = 1 J
Solve for the final speed
Rearrange the theorem for the unknown kinetic energy, then take the square root.
v = √(v_known² ± 2·ΣU/m)
T1 = 0 J; ΣU = 1 J; T2 = 1 J; v = 2 m/s
= 2 m/s
Where this answer was checked source Spring work = ½k(x1² − x2²) (NCEES FE Reference Handbook) — released to natural length verified by hand-recomputed derivation U = 0.5·200·(0.01 − 0) = 1 J; v = √(2·1/0.5) = 2 m/s exactly. Example 3
US friction stop: 2-slug block at 30 ft/s, μ=0.5 — how far does it slide? Given
mass 2 slug
initial velocity 30 ft/s
final velocity 0 ft/s
solve for distance work terms 1. kind friction friction coefficient 0.5 normal force 64.348 lbf Assumptions The body is treated as a particle; work terms are the TOTAL work done on it between states 1 and 2; g = 9.80665 m/s². Solution steps Set up the work–energy theorem
The change in kinetic energy equals the total work done on the body.
T1 + ΣU = T2 with T = m·v²/2
Fixed work ΣU = 0 ft·lbf plus -32.17 lbf per unit distance
Solve for the distance
The unknown distance appears linearly in the work sum.
d = (ΔT − W_fixed)/c
ΔT = -900 ft·lbf; d = 27.97 ft
= 27.97 ft
Results Change in kinetic energy ΔT
-900ft·lbf
Where this answer was checked source Cross-check closed form d = v0²/(2·μ·g) — friction-only stop, solve for distance verified by hand-recomputed derivation N = m·g = 2·32.174 = 64.348 lbf. ΔT = 0 − 0.5·2·900 = −900 ft·lbf; friction = −0.5·64.348·d. d = 900/32.174 = 27.973 ft. Independent check: v0²/(2·μ·g) = 900/(2·0.5·32.174) = 27.973 ✓.