Work–Energy Calculator

Solve the most common work–energy questions fast: compute work from a force and displacement (W = F·d·cos(θ)) or use the work–energy theorem (W_net = ΔK). Includes quick picks, steps, and clear interpretation.

Background

Work measures how much energy is transferred when a force acts over a distance. If the force points in the direction of motion, work is positive. If it points opposite the motion (like friction), work is negative. Net work changes kinetic energy: W_net = ΔK = ½m(v_f² − v_i²).

Enter values

Calculator mode

Work from force — W = F·d·cos(θ) Work–energy theorem — W_net = ½m(v_f² − v_i²)

Tip: If you’re given speeds, use Work–energy. If you’re given force + distance, use Work from force.

Work from force

Solve for

Angle θ is between the force vector and the displacement direction.

Force (F)

Select units to auto-convert.

Distance (d)

Select units to auto-convert.

Angle (θ)

Units: degrees

Work (W)

Select units to auto-convert.

Friction add-on (optional)

Include friction

Work–energy theorem

Solve for

Uses W_net = ½m(v_f² − v_i²). Speeds are magnitudes (non-negative).

Mass (m)

Units: kg

Net work / ΔK (Wnet)

Select units to auto-convert.

Initial speed (vᶦ)

Units: m/s

Final speed (vᶠ)

Units: m/s

Friction helper (optional)

Show friction work

Options

Show step-by-step

Show interpretation

Display rounding

Rounding affects display only.

Quick picks

Chips prefill and calculate immediately.

Result

No results yet. Enter values and click Calculate.

How to use this calculator

Choose Work from force for W = F·d·cos(θ).
Choose Work–energy theorem for W_net = ½m(v_f² − v_i²).
Pick what you want to solve for, enter the other values, then click Calculate.

How this calculator works

Work from force: W = F·d·cos(θ)
Work–energy theorem: W_net = ΔK = ½m(v_f² − v_i²)
Friction work: W_f = −μ_k·N·d
Sign intuition: positive work speeds you up, negative work slows you down.

Formula & Equations Used

Work (constant force): W = F·d·cos(θ)

Work–energy theorem: W_net = ΔK

Kinetic energy change: ΔK = ½m(v_f² − v_i²)

Friction work: W_f = −μ_k·N·d

Example Problem & Step-by-Step Solution

Example 1 — Work from force

A force of 50 N pushes a box 8 m in the same direction (θ = 0°). Find the work.

W = F·d·cos(θ)
W = 50·8·cos(0°) = 400·1
W = 400 J

Example 2 — Work–energy (find net work)

A 2.0 kg cart speeds up from 3 m/s to 9 m/s. Find W_net.

W_net = ½m(v_f² − v_i²)
W_net = ½·2(9² − 3²) = 1(81 − 9)
W_net = 72 J

Example 3 — Work–energy (solve vᶠ)

A 1.5 kg object has vᶦ = 2 m/s and net work W_net = 60 J. Find vᶠ.

W_net = ½m(v_f² − v_i²)
v_f² = v_i² + 2W_net/m = 2² + 2·60/1.5
v_f² = 4 + 80 = 84
v_f = √84 ≈ 9.17 m/s

Frequently Asked Questions

Q: When is work negative?

Work is negative when the force points opposite the displacement (like friction slowing an object down), meaning energy is removed from kinetic energy.

Q: Does the normal force do work?

Usually no—if the normal force is perpendicular to the motion, then θ = 90° and W = 0.

Q: What does net work tell me?

Net work equals the change in kinetic energy. Positive net work increases speed; negative net work decreases speed.

Q: What if I get a negative value under the square root when solving for speed?

That means the inputs are physically inconsistent (not enough energy/work for that speed change). Double-check signs, units, and given values.