MechanicsTopic #6 of 35

Power

Rate of energy transfer, mechanical power, and efficiency.

Overview

Power is the rate at which work is done or energy is transferred. It measures how quickly energy is converted from one form to another.

Definition

Average Power

Pavg=WΔt=ΔEΔtP_{\text{avg}} = \frac{W}{\Delta t} = \frac{\Delta E}{\Delta t}

Instantaneous Power

P=dWdt=dEdtP = \frac{dW}{dt} = \frac{dE}{dt}

Units

UnitSymbolEquivalent
WattWJ/s = kg·m²/s³
KilowattkW1000 W
Horsepowerhp746 W
MegawattMW10610^6 W

Power in Terms of Force and Velocity

For a constant force:

P=Fv=FvcosθP = \vec{F} \cdot \vec{v} = Fv\cos\theta

When force is parallel to velocity:

P=FvP = Fv

Energy-Power Relationship

Energy transferred over time:

E=P×tE = P \times t

Common unit: kilowatt-hour (kWh)

1 kWh=1000 W×3600 s=3.6×106 J=3.6 MJ1 \text{ kWh} = 1000 \text{ W} \times 3600 \text{ s} = 3.6 \times 10^6 \text{ J} = 3.6 \text{ MJ}

Mechanical Power

Rotational Power

P=τωP = \tau\omega

Where τ\tau is torque and ω\omega is angular velocity

Power and Constant Velocity

When an object moves at constant velocity against friction:

P=fk×v=μkmgvP = f_k \times v = \mu_k mgv

Power to Climb

Power needed to climb at velocity vv:

P=mgvP = mgv

Efficiency

The ratio of useful output power to input power:

η=PoutPin×100%\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \times 100\%

η=EoutEin×100%\eta = \frac{E_{\text{out}}}{E_{\text{in}}} \times 100\%

Energy Losses

Ploss=PinPout=Pin(1η)P_{\text{loss}} = P_{\text{in}} - P_{\text{out}} = P_{\text{in}}(1 - \eta)

Human Power Output

ActivityTypical Power
At rest~80 W
Walking~200 W
Cycling~400 W
Sprinting~500-2000 W
Maximum sustained~300-400 W

Examples

Example 1: Motor Power

A motor lifts a 500 kg load 20 m in 10 seconds. Find power.

W=mgh=500×9.8×20=98,000 JW = mgh = 500 \times 9.8 \times 20 = 98{,}000 \text{ J}

P=Wt=98,00010=9,800 W=9.8 kWP = \frac{W}{t} = \frac{98{,}000}{10} = 9{,}800 \text{ W} = 9.8 \text{ kW}

Example 2: Car Engine

A car moves at 30 m/s against 1000 N of friction. Find power needed.

P=Fv=1000×30=30,000 W=30 kW=40.2 hpP = Fv = 1000 \times 30 = 30{,}000 \text{ W} = 30 \text{ kW} = 40.2 \text{ hp}

Example 3: Climbing Stairs

A 70 kg person climbs 15 m of stairs in 20 s. Find average power.

P=mght=70×9.8×1520=515 WP = \frac{mgh}{t} = \frac{70 \times 9.8 \times 15}{20} = 515 \text{ W}

Example 4: Electric Bill

A 100 W light bulb runs for 5 hours. Find energy used in kWh.

E=Pt=100 W×5 h=500 Wh=0.5 kWhE = Pt = 100 \text{ W} \times 5 \text{ h} = 500 \text{ Wh} = 0.5 \text{ kWh}

Example 5: Efficiency

An engine produces 60 kW from fuel supplying 200 kW. Find efficiency.

η=60200×100%=30%\eta = \frac{60}{200} \times 100\% = 30\%

Power in Transportation

Terminal Velocity

At terminal velocity, driving force equals drag:

P=Fdrive×v=Fdrag×vP = F_{\text{drive}} \times v = F_{\text{drag}} \times v

Drag Power (at high speed)

Air resistance increases with v2v^2, so power to overcome it:

Pdragv3P_{\text{drag}} \propto v^3

This is why fuel consumption increases rapidly with speed.

Acceleration Power

Power needed to accelerate:

P=Fv=mavP = Fv = mav

Note: Power required increases with velocity even at constant acceleration.

Key Relationships Summary

QuantityFormula
Average powerP=W/tP = W/t
Instantaneous powerP=FvP = Fv
Rotational powerP=τωP = \tau\omega
Energy from powerE=PtE = Pt
Efficiencyη=Pout/Pin\eta = P_{\text{out}}/P_{\text{in}}