7 - Power and Energy Applications

A progressive guide to analyzing energy transfers, calculating power, evaluating efficiency, and solving mechanical applications involving lifting, friction, and power graphs.

1. Set Up the Energy Model

Energy methods begin by deciding what belongs to the system and comparing its initial and final states. The system could be a single object, an object together with Earth, or an object plus a machine and its surroundings.

A useful accounting equation is

ΔK+ΔU+ΔEth=Wext.\Delta K+\Delta U+\Delta E_{\text{th}}=W_{\text{ext}}.

Here, KK is , UU represents stored potential energy, ΔEth\Delta E_{\text{th}} is the increase in , and WextW_{\text{ext}} is work done by forces outside the system. For an isolated system, Wext=0W_{\text{ext}}=0, so total energy is conserved.

For motion near Earth’s surface, is Ug=mgyU_g=mgy. For a spring, elastic potential energy is Us=12kx2U_s=\frac{1}{2}kx^2. The zero level for can be selected freely; only changes in that energy matter.

Takeaway: Define the system before choosing an equation, and represent friction as a transfer into rather than as destroyed energy.

2. Relate Energy and

describes how quickly energy is transferred or converted, whereas energy describes the amount transferred. The average rate is

Pavg=ΔEΔt=WΔt.P_{\text{avg}}=\frac{\Delta E}{\Delta t}=\frac{W}{\Delta t}.

The SI unit is the watt:

1 W=1 Js.1\ \text{W}=1\ \frac{\text{J}}{\text{s}}.

Thus, a device rated at 500 W500\ \text{W} transfers or converts energy at a rate of 500 J/s500\ \text{J/s} under the stated conditions. For constant , the transferred energy is

ΔE=PΔt.\Delta E=P\Delta t.

When changes with time, total energy is the area under the -versus-time graph:

ΔE=∫P dt.\Delta E=\int P\,dt.

Takeaway: Keep energy and conceptually separate: joules measure an amount, while watts measure an amount per unit time.

3. Calculate from Force and Motion

The delivered by a force depends on the component of the force along the velocity:

P=F⃗⋅v⃗=Fvcos⁡θ.P=\vec F\cdot\vec v=Fv\cos\theta.

Equivalently, if F∥F_{\parallel} is the force component parallel to the velocity, then

P=F∥v.P=F_{\parallel}v.

Important cases are:

  • If the force is parallel to the velocity, P=FvP=Fv.

  • If the force is opposite the velocity, P=−FvP=-Fv, so the force removes mechanical energy.

  • If the force is perpendicular to the velocity, P=0P=0, so it does no instantaneous work.

For an object lifted upward at constant speed, the lifting force approximately balances the weight. The mechanical delivered to the object is therefore

P=Fv≈mgv.P=Fv\approx mgv.

This is the useful mechanical . A motor with less than perfect requires greater input .

Takeaway: Use P=FvP=Fv only when the force is parallel to the velocity; otherwise include the angle or use the parallel component.

4. Evaluate

compares the desired output with the total input:

η=Euseful,outEin\eta=\frac{E_{\text{useful,out}}}{E_{\text{in}}}

or, during steady operation,

η=Puseful,outPin.\eta=\frac{P_{\text{useful,out}}}{P_{\text{in}}}.

As a percentage,

η%=100η.\eta_{\%}=100\eta.

For an ordinary machine, 0<η<10<\eta<1. The input can be found from

Pin=Puseful,outη,P_{\text{in}}=\frac{P_{\text{useful,out}}}{\eta},

and useful output energy can be found from

Euseful,out=ηEin.E_{\text{useful,out}}=\eta E_{\text{in}}.

The part of the input that is not useful output may become , sound, vibration, or another unintended form. It has not disappeared.

Example: If a motor provides 3.68 kW3.68\ \text{kW} of useful while receiving 5.0 kW5.0\ \text{kW}, then

η=3.685.0=0.736≈74%.\eta=\frac{3.68}{5.0}=0.736\approx74\%.

Takeaway: An inefficient machine must receive more input than the useful it delivers.

5. Apply to Lifting and Friction

For a lifting problem, the useful output is often the increase in . A 250 kg250\ \text{kg} load lifted through 12 m12\ \text{m} in 8.0 s8.0\ \text{s} has useful output

Puseful=mghΔt=(250)(9.8)(12)8.0≈3.68×103 W.P_{\text{useful}}=\frac{mgh}{\Delta t}=\frac{(250)(9.8)(12)}{8.0}\approx3.68\times10^3\ \text{W}.

Therefore, Puseful≈3.7 kWP_{\text{useful}}\approx3.7\ \text{kW}. The average upward speed is v=h/Δt=1.5 m/sv=h/\Delta t=1.5\ \text{m/s}, which gives the same result through P=mgvP=mgv.

For a car traveling at constant speed, the engine can still transfer energy even though the net force is zero. If a car moves at 20 m/s20\ \text{m/s} against a resistive force of 600 N600\ \text{N}, the engine supplies

P=Fv=(600)(20)=1.20×104 W=12 kW.P=Fv=(600)(20)=1.20\times10^4\ \text{W}=12\ \text{kW}.

The supplied energy is transferred primarily into in the tires, road, air, and engine components.

Takeaway: Constant speed means zero net force and zero net work on the object, not zero from the engine.

6. Use Energy Accounting with Friction

An is especially useful when friction is present. Consider a 2.0 kg2.0\ \text{kg} block that starts from rest at a height of 3.0 m3.0\ \text{m}, with friction transferring 12 J12\ \text{J} into . Choose the block–Earth system and take the bottom as y=0y=0. Since there is no external work,

ΔK+ΔUg+ΔEth=0.\Delta K+\Delta U_g+\Delta E_{\text{th}}=0.

The initial is zero, so

Kf−mgh+12=0.K_f-mgh+12=0.

Thus,

12mv2=mgh−12.\frac{1}{2}mv^2=mgh-12.

Substituting the values gives

12(2.0)v2=(2.0)(9.8)(3.0)−12=46.8 J,\frac{1}{2}(2.0)v^2=(2.0)(9.8)(3.0)-12=46.8\ \text{J},

so

v=46.8≈6.8 m/s.v=\sqrt{46.8}\approx6.8\ \text{m/s}.

Without friction, more would become and the final speed would be higher.

Takeaway: Include frictional energy explicitly as an increase in ; do not simply remove energy without representing where it goes.

7. Read –Time Graphs

A -versus-time graph represents a rate, so its area represents transferred energy. For a machine operating at 800 W800\ \text{W} for 5.0 s5.0\ \text{s} and then at 300 W300\ \text{W} for 10 s10\ \text{s}, the total input energy is

ΔE=(800)(5.0)+(300)(10)=7.0×103 J.\Delta E=(800)(5.0)+(300)(10)=7.0\times10^3\ \text{J}.

The total time is 15 s15\ \text{s}, so the is

Pavg=7.0×10315≈4.67×102 W.P_{\text{avg}}=\frac{7.0\times10^3}{15}\approx4.67\times10^2\ \text{W}.

This average is not the same as the during either individual interval. For a graph with several segments, calculate the area of each segment and add the results before dividing by the full time interval.

Takeaway: Find total energy from graph area, then find by dividing that total energy by total time.

8. Solve and Check Applications

Use the following sequence for unfamiliar problems:

  1. Define the system. Decide which objects and energy stores are included.

  2. Identify the states. Record initial and final speeds, heights, spring compression, and time intervals.

  3. List transfers. Include external work, , potential energy, and .

  4. Select the equation. Use P=ΔE/ΔtP=\Delta E/\Delta t, P=Fvcos⁡θP=Fv\cos\theta, an relation, or an .

  5. Convert units. Use seconds for time, meters for distance, watts for , and consistent energy units.

  6. Check the result. should have units of watts, should be dimensionless or a percentage, and an inefficient machine should require more input than useful output .

Common checks:

  • Do not confuse with energy.

  • Do not use P=FvP=Fv when the force is not parallel to the velocity.

  • Do not infer that constant speed means no energy transfer.

  • Do not accept an greater than 100%100\% without finding a system, input, or unit error.

  • Do not omit the thermal-energy increase caused by friction.

Final takeaway: A clear system boundary, a complete energy account, and careful attention to units connect , energy transfer, friction, and into one consistent method.