Dynamics

Work–Energy Theorem

Applies T1 + ΣU = T2 to a particle: kinetic energy plus the work of constant forces, gravity (height change), springs, and kinetic friction. Solves for the final speed, the initial speed, or an unknown shared distance (e.g. 'how far does it slide before stopping'). Use for: blocks on ramps with friction, spring launchers, 'speed at the bottom' problems. Not for: rigid-body rotation or momentum/impact questions (impulse-momentum).

Direct solving is free and needs no account. Have a word problem instead? Submit it as text.

Inputs

Units

Load a sample problem:

massMass of the body [dimension: mass]initial velocity (optional)Speed at state 1 (omit when solving for it; 0 = starts at rest) [dimension: velocity]final velocity (optional)Speed at state 2 (omit when solving for it) [dimension: velocity]solve forThe unknown. 'distance' solves for a distance shared by every constant-force/friction term that omits its own distance

work termsWork terms acting on the body between states 1 and 2

work terms 1type

forceApplied force magnitude [dimension: force]distance (optional)Distance moved along the path (omit when solving for distance) [dimension: length]angle deg (optional)Angle between the force and the motion direction, degrees (0 = pulling forward, 180 = opposing)

Worked examples

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-forceforce 10 Ndistance 5 mangle 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

Results

Final speed v2

7.681m/s

Kinetic energy T1

0.009kJ

Total work ΣU

0.05kJ

Kinetic energy T2

0.059kJ

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 springstiffness 200 N/minitial deformation 0.1 mfinal 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

Results

Final speed v2

2m/s

Kinetic energy T1

0kJ

Total work ΣU

0.001kJ

Kinetic energy T2

0.001kJ

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 frictionfriction coefficient 0.5normal 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

Solved distance

27.97ft

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 ✓.