Lesson 10 The Efficiency Formula
The Efficiency Formula
To calculate the percent efficiency of any system or device, what do you need to know?
C10.2 moving metronome
To calculate the efficiency of a device, you must identify the useful output energy and the total input energy. You can then analyze the system or machine.
Calculating Efficiency
Efficiency is a ratio, so it has no units. The output and input energies must be in the same units so that they will cancel out.
Calculating Efficiency
Efficiency is a ratio, so it has no units. The output and input energies must be in the same units so that they will cancel out.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathcolor=¨#FFFFFF¨»percent«/mi»«mo mathcolor=¨#FFFFFF¨»§#160;«/mo»«mi mathcolor=¨#FFFFFF¨»efficiency«/mi»«mo mathcolor=¨#FFFFFF¨»=«/mo»«mfrac mathcolor=¨#FFFFFF¨»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«/mfrac»«mo mathcolor=¨#FFFFFF¨»§#215;«/mo»«mn mathcolor=¨#FFFFFF¨»100«/mn»«mo mathcolor=¨#FFFFFF¨»%«/mo»«/math»
Examples
- When a 100.0 W light bulb is on for 2.4 h, it uses 864 kJ of electrical energy. During that time, the light bulb emits 45.6 kJ of light. What is the efficiency of the light bulb in transforming electrical energy into light energy?
Step 1: List the variables.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mn»864«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mn»45«/mn»«mo».«/mo»«mn»6«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mtd»«/mtr»«/mtable»«/math»
Note: The information of 100.0 W and 2.4 h is not used in the calculation.
Step 2: Identify the correct formula, and rearrange if necessary.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathcolor=¨#191919¨»percent«/mi»«mo mathcolor=¨#191919¨»§#160;«/mo»«mi mathcolor=¨#191919¨»efficiency«/mi»«mo mathcolor=¨#191919¨»=«/mo»«mfrac mathcolor=¨#191919¨»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«/mfrac»«mo mathcolor=¨#191919¨»§#215;«/mo»«mn mathcolor=¨#191919¨»100«/mn»«mo mathcolor=¨#191919¨»%«/mo»«/math»
Step 3: Substitute the values into the formula.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«mo»=«/mo»«mfrac»«mrow»«mn»45«/mn»«mo».«/mo»«mn»6«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mrow»«mrow»«mn»864«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/math»
Step 4: Calculate the answer.
percent efficiency = 5.277...% = 5.28% (to three significant digits)
- A bobcat uses 346 kJ of chemical potential energy stored in the fuel to lift 2 725 kg of dirt 3.2 m straight up, to dump it into a truck. What was the efficiency of the bobcat in converting chemical potential energy into gravitational potential
energy of the dirt?
Step 1: List the variables.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mn»346«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mo»?«/mo»«mo»§#160;«/mo»«mi»kJ«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mo»§#160;«/mo»«mn»725«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»h«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»3«/mn»«mo».«/mo»«mn»2«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mtd»«/mtr»«/mtable»«/math»
Step 2: Identify the correct formula, and rearrange if necessary.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathcolor=¨#191919¨»percent«/mi»«mo mathcolor=¨#191919¨»§#160;«/mo»«mi mathcolor=¨#191919¨»efficiency«/mi»«mo mathcolor=¨#191919¨»=«/mo»«mfrac mathcolor=¨#191919¨»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«/mfrac»«mo mathcolor=¨#191919¨»§#215;«/mo»«mn mathcolor=¨#191919¨»100«/mn»«mo mathcolor=¨#191919¨»%«/mo»«/math»
Useful output energy must first be calculated. Enough information has been provided to determine gravitational potential energy.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»E«/mi»«mi»p«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»2«/mn»«mo»§#160;«/mo»«mn»725«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»g«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»9«/mn»«mo».«/mo»«mn»81«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd»«mi»h«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»3«/mn»«mo».«/mo»«mn»2«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«mtd»«/mtd»«mtd»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»E«/mi»«mi»p«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»m«/mi»«mi»g«/mi»«mi»h«/mi»«/mtd»«/mtr»«/mtable»«/math»
Step 3: Substitute the values into the formula.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨» «msub» «mi»E«/mi» «mi»p«/mi» «/msub» «mo»=«/mo» «mi»m«/mi» «mi»g«/mi» «mi»h«/mi» «mspace linebreak=¨newline¨»«/mspace» «msub» «mi»E«/mi» «mi»p«/mi» «/msub» «mo»=«/mo» «mfenced» «mrow» «mn»2«/mn» «mo»§#160;«/mo» «mn»725«/mn» «mo»§#160;«/mo» «mi»kg«/mi» «/mrow» «/mfenced» «mfenced» «mrow» «mn»9«/mn» «mo».«/mo» «mn»81«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»m«/mi» «mo»/«/mo» «msup» «mi mathvariant=¨normal¨»s«/mi» «mn»2«/mn» «/msup» «/mrow» «/mfenced» «mfenced» «mrow» «mn»3«/mn» «mo».«/mo» «mn»2«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»m«/mi» «/mrow» «/mfenced» «mspace linebreak=¨newline¨»«/mspace» «msub» «mi»E«/mi» «mi»p«/mi» «/msub» «mo»=«/mo» «mn»85«/mn» «mo»§#160;«/mo» «mn»543«/mn» «mo».«/mo» «mn»2«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»J«/mi» «/math»
Energy needs to be in the same units, so we convert Ep to kJ.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfenced»«mrow»«mn»85«/mn»«mo»§#160;«/mo»«mn»543«/mn»«mo».«/mo»«mn»2«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«/mfenced»«mfenced»«mfrac»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mtext»kJ«/mtext»«/mrow»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»000«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«/mfrac»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»85«/mn»«mo».«/mo»«mn»5432«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»85«/mn»«mo».«/mo»«mn»5432«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mrow»«mrow»«mn»346«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«mtd»«/mtd»«mtd»«/mtd»«/mtr»«/mtable»«/math»
Step 4: Calculate the answer.
percent efficiency = 24.723...% = 25% (to two significant digits)
- A pitching machine uses 185 J of electrical energy to pitch a 320 g baseball. If the speed of the ball is 24 m/s, what is the efficiency of the pitching machine in transforming electrical energy into kinetic energy of the baseball?
Step 1: List the variables.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mn»185«/mn»«mo»§#160;«/mo»«mi»kJ«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mo»?«/mo»«mo»§#160;«/mo»«mi»kJ«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»320«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»v«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»24«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/mtd»«/mtr»«/mtable»«/math»
Step 2: Identify the correct formula, and rearrange if necessary.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathcolor=¨#191919¨»percent«/mi»«mo mathcolor=¨#191919¨»§#160;«/mo»«mi mathcolor=¨#191919¨»efficiency«/mi»«mo mathcolor=¨#191919¨»=«/mo»«mfrac mathcolor=¨#191919¨»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«/mfrac»«mo mathcolor=¨#191919¨»§#215;«/mo»«mn mathcolor=¨#191919¨»100«/mn»«mo mathcolor=¨#191919¨»%«/mo»«/math»
Useful output energy must first be calculated. Enough information has been provided to determine kinetic energy. Remember, mass needs to be in kg when using the kinetic energy formula, so it needs to be converted.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»320«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mtd»«/mtr»«/mtable»«/math»
since mass needs to be in kg, the mass needs to be converted
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mfenced»«mrow»«mn»320«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»g«/mi»«/mrow»«/mfenced»«mfenced»«mfrac»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mrow»«mrow»«mn»1«/mn»«mo»§#160;«/mo»«mn»000«/mn»«mo»§#160;«/mo»«mi»g«/mi»«/mrow»«/mfrac»«/mfenced»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mo».«/mo»«mn»320«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»v«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»24«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«msup»«mi mathvariant=¨normal¨»v«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«/mtable»«/math»
Step 3: Substitute the values into the formula.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mfenced»«mrow»«mn»0«/mn»«mo».«/mo»«mn»320«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mrow»«/mfenced»«msup»«mfenced»«mrow»«mn»23«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/mrow»«/mfenced»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»84«/mn»«mo».«/mo»«mn»64«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»84«/mn»«mo».«/mo»«mn»64«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mn»185«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/mtd»«/mtr»«/mtable»«/math»
Step 4: Calculate the answer.
percent efficiency = 45.751...% = 46% (to two significant digits)
- A 3.7 kg steel ball is dropped on a spring, compressing it. As a result, the compressed spring stores 72 J of elastic potential energy. If the gravitational potential energy of the steel ball was converted into elastic potential energy in
the spring with an efficiency of 80.4%, from what height was the steel ball dropped?
Step 1: List the variables.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mn»80«/mn»«mo».«/mo»«mn»4«/mn»«mo»%«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mo»?«/mo»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mn»72«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»3«/mn»«mo».«/mo»«mn»7«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»g«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»9«/mn»«mo».«/mo»«mn»81«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd»«mi»h«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mtd»«/mtr»«/mtable»«/math»
Step 2: Identify the correct formula, and rearrange if necessary.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathcolor=¨#191919¨»percent«/mi»«mo mathcolor=¨#191919¨»§#160;«/mo»«mi mathcolor=¨#191919¨»efficiency«/mi»«mo mathcolor=¨#191919¨»=«/mo»«mfrac mathcolor=¨#191919¨»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«/mfrac»«mo mathcolor=¨#191919¨»§#215;«/mo»«mn mathcolor=¨#191919¨»100«/mn»«mo mathcolor=¨#191919¨»%«/mo»«/math»
Total input energy must first be calculated.
To isolate total input energy, you must multiple each side by total input energy. To move total input energy to the other side, you must use the opposite operation. Multiplication is opposite to division.
Then you would have the equation
(total input energy)(percent efficiency) = useful output energy × 100%
To isolate total input energy, you must divide each side by percent efficiency. To move percent efficiency to the other side, you must use the opposite operation. Division is opposite to multiplication.
Then you would have the equation
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»=«/mo»«mfrac»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/math»
Step 3: Substitute the values into the formula.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»72«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mn»80«/mn»«mo».«/mo»«mn»4«/mn»«mo»%«/mo»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»89«/mn»«mo».«/mo»«mn»552«/mn»«mo»§#160;«/mo»«mn»2«/mn»«mo»§#8230;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«/mtable»«/math»
Then, enough information has been provided to determine height.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»E«/mi»«mi»p«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»=«/mo»«mn»89«/mn»«mo».«/mo»«mn»552«/mn»«mo»§#160;«/mo»«mn»2«/mn»«mo»§#8230;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»3«/mn»«mo».«/mo»«mn»7«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»g«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»9«/mn»«mo».«/mo»«mn»81«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«msup»«mi mathvariant=¨normal¨»s«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd»«mi»h«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»?«/mo»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»E«/mi»«mi»p«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»m«/mi»«mi»g«/mi»«mi»h«/mi»«/mtd»«/mtr»«/mtable»«/math»
To isolate h, you must divide each side by mg. To move mg to the other side, you must use the opposite operation. Division is opposite to multiplication.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨» «mtable columnspacing=¨0px¨ columnalign=¨right center left¨» «mtr» «mtd» «mfrac» «msub» «mi»E«/mi» «mi»p«/mi» «/msub» «mrow» «mi»m«/mi» «mi»g«/mi» «/mrow» «/mfrac» «/mtd» «mtd» «mo»=«/mo» «/mtd» «mtd» «mi»h«/mi» «/mtd» «/mtr» «mtr» «mtd» «mi»h«/mi» «/mtd» «mtd» «mo»=«/mo» «/mtd» «mtd» «mfrac» «mrow» «mn»89«/mn» «mo».«/mo» «mn»552«/mn» «mo»§#160;«/mo» «mn»2«/mn» «mo»§#8230;«/mo» «mi mathvariant=¨normal¨»J«/mi» «/mrow» «mrow» «mfenced» «mrow» «mn»3«/mn» «mo».«/mo» «mn»7«/mn» «mo»§#160;«/mo» «mi»kg«/mi» «/mrow» «/mfenced» «mstyle displaystyle=¨true¨» «mfenced» «mrow» «mn»9«/mn» «mo».«/mo» «mn»81«/mn» «mo»§#160;«/mo» «mi mathvariant=¨normal¨»m«/mi» «mo»/«/mo» «msup» «mi mathvariant=¨normal¨»s«/mi» «mn»2«/mn» «/msup» «/mrow» «/mfenced» «/mstyle» «/mrow» «/mfrac» «/mtd» «/mtr» «/mtable» «/math»
Step 4: Calculate the answer.
h = 2.467...m = 2.5 m (to two significant digits)
Do you want a bit more detailed explanation of how to use the efficiency formula in a calculation? Watch this video for more information on this calculation.
https://adlc.wistia.com/medias/qjvymvdj79
Digging Deeper
C10.3 bicycle
The bicycle is considered to be the most efficient machine that humans have ever made. In terms of the amount of energy that a person must use to travel a certain distance, cycling is considered to the most efficient self-powered way to travel.
Even more efficient than walking!
To learn more about the efficiency of a bicycle, click on the link. https://en.wikipedia.org/wiki/Bicycle_performance#Energy_efficiency
Learn More
To learn more about the efficiency of a bicycle, click on the link. https://en.wikipedia.org/wiki/Bicycle_performance#Energy_efficiency
Learn More
Read This
Please read pages 216 to 220 in your Science 10 textbook. Make sure you take notes on your readings to study from later. You should focus on how efficiency is calculated, and communicated. Remember, if you have any questions, or do not understand
something, ask your teacher!
Practice Questions
Complete the following practice questions to check your understanding of the concept you just learned. Make sure you write complete answers to the practice questions in your notes. After you have checked your answers, make corrections to your responses
(where necessary) to study from.
- If a light bulb is 6.34% efficient and it emits a total of 5.27 × 103 J of light energy, how much electrical energy does it use?
Step 1: List the variables.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mn»6«/mn»«mo».«/mo»«mn»34«/mn»«mo»%«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mo»?«/mo»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mn»5«/mn»«mo».«/mo»«mn»27«/mn»«mo»§#215;«/mo»«msup»«mn»10«/mn»«mn»3«/mn»«/msup»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«/mtable»«/math»Step 2: Identify the correct formula, and rearrange if necessary.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathcolor=¨#191919¨»percent«/mi»«mo mathcolor=¨#191919¨»§#160;«/mo»«mi mathcolor=¨#191919¨»efficiency«/mi»«mo mathcolor=¨#191919¨»=«/mo»«mfrac mathcolor=¨#191919¨»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«/mfrac»«mo mathcolor=¨#191919¨»§#215;«/mo»«mn mathcolor=¨#191919¨»100«/mn»«mo mathcolor=¨#191919¨»%«/mo»«/math»
Total input energy must first be calculated.
To isolate total input energy, you must multiple each side by total input energy. To move total input energy to the other side, you must use the opposite operation. Multiplication is opposite to division.
Then you would have the equation
(total input energy)(percent efficiency) = useful output energy × 100%
To isolate total input energy, you must divide each side by percent efficiency. To move percent efficiency to the other side, you must use the opposite operation. Division is opposite to multiplication.
Then you would have the equation
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»=«/mo»«mfrac»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/math»
Step 3: Substitute the values into the formula.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»5«/mn»«mo».«/mo»«mn»27«/mn»«mo»§#215;«/mo»«msup»«mn»10«/mn»«mn»3«/mn»«/msup»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mn»6«/mn»«mo».«/mo»«mn»34«/mn»«mo»%«/mo»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/mtd»«/mtr»«/mtable»«/math»
Step 4: Calculate the answer.
total input energy = 83 123.028...J = 8.31 × 104 (to three significant digits)
- An archer does work on a bow by exerting an average force of 173 N over a distance of 0.59 m to stretch the string and bend the bow. In the process, elastic potential energy is stored in the bow. When the archer releases the 0.062 kg arrow, its
initial speed is 51 m/s. What was the efficiency of transforming the archer’s work in the arrow’s kinetic energy? Hint: This question is very similar to the example in the video above.
Step 1: List the variables.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mo»?«/mo»«mo»%«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mo»?«/mo»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«mo»§#160;«/mo»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mo»§#160;«/mo»«mo»?«/mo»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»m«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mo».«/mo»«mn»062«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»v«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»51«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»F«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»173«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»N«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»d«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mn»0«/mn»«mo».«/mo»«mn»59«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mtd»«/mtr»«/mtable»«/math»Step 2: Identify the correct formula, and rearrange if necessary.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mi mathcolor=¨#191919¨»percent«/mi»«mo mathcolor=¨#191919¨»§#160;«/mo»«mi mathcolor=¨#191919¨»efficiency«/mi»«mo mathcolor=¨#191919¨»=«/mo»«mfrac mathcolor=¨#191919¨»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«/mfrac»«mo mathcolor=¨#191919¨»§#215;«/mo»«mn mathcolor=¨#191919¨»100«/mn»«mo mathcolor=¨#191919¨»%«/mo»«/math»
Total input energy and useful output energy must first be calculated.
Total input energy is the work done. Remember that work is a measurement of energy.
W = Fd
Useful output energy is the kinetic energy that the arrow has.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨» «msub» «mi»E«/mi» «mi»k«/mi» «/msub» «mo»=«/mo» «mfrac» «mn»1«/mn» «mn»2«/mn» «/mfrac» «mi»m«/mi» «msup» «mi»v«/mi» «mn»2«/mn» «/msup» «/math»
Step 3: Substitute the values into the formula.
«math xmlns=¨http://www.w3.org/1998/Math/MathML¨»«mtable columnspacing=¨0px¨ columnalign=¨right center left¨»«mtr»«mtd»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mi»m«/mi»«msup»«mi»v«/mi»«mn»2«/mn»«/msup»«/mtd»«/mtr»«mtr»«mtd»«msub»«mi»E«/mi»«mi»k«/mi»«/msub»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mn»1«/mn»«mn»2«/mn»«/mfrac»«mfenced»«mrow»«mn»0«/mn»«mo».«/mo»«mn»062«/mn»«mo»§#160;«/mo»«mi»kg«/mi»«/mrow»«/mfenced»«msup»«mfenced»«mrow»«mn»51«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«mo»/«/mo»«mi mathvariant=¨normal¨»s«/mi»«/mrow»«/mfenced»«mn»2«/mn»«/msup»«mo»=«/mo»«mn»80«/mn»«mo».«/mo»«mn»631«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«mtd»«/mtd»«mtd»«/mtd»«/mtr»«mtr»«mtd»«mi»W«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mi»F«/mi»«mi»d«/mi»«/mtd»«/mtr»«mtr»«mtd»«mi»W«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfenced»«mrow»«mn»173«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»N«/mi»«/mrow»«/mfenced»«mfenced»«mrow»«mn»0«/mn»«mo».«/mo»«mn»59«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»m«/mi»«/mrow»«/mfenced»«mo»=«/mo»«mn»102«/mn»«mo».«/mo»«mn»07«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mtd»«/mtr»«mtr»«mtd»«/mtd»«mtd»«/mtd»«mtd»«/mtd»«/mtr»«mtr»«mtd»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mi»useful«/mi»«mo»§#160;«/mo»«mi»output«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«mrow»«mi»total«/mi»«mo»§#160;«/mo»«mi»input«/mi»«mo»§#160;«/mo»«mi»energy«/mi»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/mtd»«/mtr»«mtr»«mtd»«mi»percent«/mi»«mo»§#160;«/mo»«mi»efficiency«/mi»«/mtd»«mtd»«mo»=«/mo»«/mtd»«mtd»«mfrac»«mrow»«mn»80«/mn»«mo».«/mo»«mn»631«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«mrow»«mn»102«/mn»«mo».«/mo»«mn»07«/mn»«mo»§#160;«/mo»«mi mathvariant=¨normal¨»J«/mi»«/mrow»«/mfrac»«mo»§#215;«/mo»«mn»100«/mn»«mo»%«/mo»«/mtd»«/mtr»«/mtable»«/math»
Step 4: Calculate the answer.
percent efficiency = 78.995 7...% = 79% (to two significant digits)