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Power

Reference data and engineering information about power for dynamics applications.

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Overview

Engineering reference data for Power in dynamics.

Key Formulas

Newton's Second Law

F=maF = ma

Force = mass × acceleration.

Kinetic Energy

Ek=12mv2E_k = \frac{1}{2}mv^2

Energy of motion.

Momentum

p=mvp = mv

Mass × velocity.

Work

W=Fdcos⁡θW = Fd\cos\theta

Force × displacement × cos(angle).

Variables

Symbol Description Unit
FF Force N
mm Mass kg
aa Acceleration m/s²
vv Velocity m/s
EkE_k Kinetic energy J

Comparative Power Outputs

Power Unit Conversions

Power can be expressed in various units. Key conversion factors include:

  • 1 kW = 1.341 hp (UK/US, mechanical)
  • 1 hp (mechanical) = 0.7457 kW = 550 ft·lbf/s = 2545 Btu/h
  • 1 metric horsepower (PS, Pferdestärke) ≈ 0.7355 kW
  • 1 Btu/h ≈ 0.000293071 kW

Common equivalences for 1 kW:

  • 1000 W
  • 3412 Btu/h
  • 737.6 ft·lbf/s
  • 1.3596 metric horsepower (PS)
  • 1.341 mechanical horsepower (bhp)
  • 101.97 kgf·m/s

Worked Examples

Example 1: Power Required to Lift a Mass

Calculate the power needed to lift a 1000 kg mass 10 m in 10 seconds.

Step 1: Calculate work done. W=F⋅s=m⋅g⋅s=(1000 kg)×(9.81 m/s2)×(10 m)=98100 JW = F \cdot s = m \cdot g \cdot s = (1000\ \text{kg}) \times (9.81\ \text{m/s}^2) \times (10\ \text{m}) = 98100\ \text{J}

Step 2: Calculate power. P=Wdt=98100 J10 s=9810 W=9.81 kWP = \frac{W}{dt} = \frac{98100\ \text{J}}{10\ \text{s}} = 9810\ \text{W} = 9.81\ \text{kW}

Example 2: Work Done by an Electric Motor

Calculate the work done by a 1 kW electric motor running for 1 hour.

Rearrange the power formula: W=P⋅dt=(1 kW)×(1000 W/kW)×(1 h×3600 s/h)=3.6×106 J=3600 kJW = P \cdot dt = (1\ \text{kW}) \times (1000\ \text{W/kW}) \times (1\ \text{h} \times 3600\ \text{s/h}) = 3.6 \times 10^6\ \text{J} = 3600\ \text{kJ}

Example 3: Electric Winch Lifting a Mass

An electric winch with a 500 W motor lifts a 100 kg mass 10 m.

Step 1: Calculate the force (weight) on the mass. F=m⋅g=(100 kg)×(9.81 m/s2)=981 NF = m \cdot g = (100\ \text{kg}) \times (9.81\ \text{m/s}^2) = 981\ \text{N}

Step 2: Calculate work done. W=F⋅s=(981 N)×(10 m)=9810 JW = F \cdot s = (981\ \text{N}) \times (10\ \text{m}) = 9810\ \text{J}

Step 3: Calculate time required. dt=WP=9810 J500 W=19.62 sdt = \frac{W}{P} = \frac{9810\ \text{J}}{500\ \text{W}} = 19.62\ \text{s}

References