04 // Interactive Formula Compendium // PHYSICSW = F · d · cos(θ) = ΔK = ½ m v_f² − ½ m v_i²
Work-Energy Theorem & Mechanical Power Formula Sheet
The Work-Energy Theorem states that the net mechanical work W = ∫ F · dr (or F d cos θ for constant force at angle θ to displacement d) performed on a rigid particle equals its change in kinetic energy ΔK = ½mv_f² − ½mv_i².
Formula Reference
Main Formula
W = F · d · cos(θ) = ΔK = ½ m v_f² − ½ m v_i²
Rearranged Forms
W = ∫_{r₁}^{r₂} F(r) · dr (Line Integral for Variable Force)
P_avg = W / Δt, P_inst = F · v · cos(θ) (Mechanical Power in Watts)
1 hp (mechanical) = 745.7 W, 1 kWh = 3.6 × 10⁶ J
Variables & Units
| Symbol | Name | Unit |
|---|---|---|
| W | Mechanical Work Done (F·d·cos θ) | J |
| F | Applied Force Magnitude (F) | N |
| d | Displacement Distance (d) | m |
| theta_deg | Angle Between F and d (degrees) | deg |
Inputs & Parameters
[N]
[m]
[deg]
Calculated Result (Mechanical Work Done (F·d·cos θ))
W =2,598.076J
Formula Notes & Background
- Start with Newton's second law along the path s: F_t = m a_t = m (dv/dt) = m (dv/ds)(ds/dt) = m v (dv/ds).
- Multiply by ds and integrate from s₁ to s₂: W = ∫ F_t ds = ∫_{v_i}^{v_f} m v dv.
- Evaluating the definite integral gives W = ½mv_f² − ½mv_i² = ΔK.
Step-by-Step Examples
#1: A worker pulls a crate d = 12 m across a floor with tension F = 250 N via a rope angled at θ = 30° above horizontal.
F = 250 Nd = 12 mθ = 30° (cos 30° ≈ 0.86603)
W = 250 × 12 × cos(30°) = 3,000 × 0.86603
→ W = 2,598.08 J (2.60 kJ)
#2: Calculate the braking work required to bring a 2,000 kg car from 30 m/s to rest.
m = 2,000 kgv_i = 30 m/sv_f = 0 m/s
W = ΔK = 0 − ½(2000)(30²) = −1000 × 900
→ W = −900,000 J (−900 kJ dissipated as brake heat)
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Frequently Asked Questions
Why does a centripetal force do zero work on an object in circular motion?+
Because the centripetal force points radially inward perpendicular (θ = 90°) to the tangential velocity vector, cos(90°) = 0 and W = 0 J.
What does negative work (W < 0) signify physically?+
Negative work occurs when a force component opposes the direction of motion (90° < θ ≤ 180°), removing kinetic energy from the moving body.
Whydoes doubling your driving speed quadruple your braking distance?+
Kinetic energy scales as v² (½mv² = μ m g d_stop), so 2× speed carries 4× the kinetic energy and requires 4× the stopping distance.