Energy Unit Overview and Objectives

by Edward Choi & Brendan Palad

Define Power:

Rate of doing work; rate of energy conversion

Solve Problems involving Power, Work and Time:
An electric motor lifts an elevator that weighs 2.30 X 105 N a distance of 4.00 m in 17.0 s, calculate the power of the motor in watts, and hence the power in kilowatts.
In watts –
Given information:
  • F = 2.30 X 105 N
  • d = 4.00 m
  • t = 17.0 s
Unknown Variable(s):
  • P (power)
Basic Equation(s):
  • P = W / t
Where P is power, W is work done (Fd), and t is time taken
P = W / t = (Fd) / (t) = ((2.30 X 105 N)(4.00 m)) / (17.0 s) = 54117.64706 (Nm) / s = 5.41 X 104 J / s = 5.41 X 104 W
In kilowatts –
((5.41 X 104 W) / (1)) X ((1 kW) / (1000 W)) = 54.1 kW
Note: A reason why converting Power into kilowatts is common is because a watt (one joule / second) is a relatively small unit.

Define Efficiency:
Ratio of output work (or energy) to input work (or energy)

Calculate and Compare the Efficiencies of Common Devices:
  • Machines, whether powered by engines or humans, allow for easier work. A machine eases the load either by changing the magnitude or the direction of the force exerted to do the work.
  • The mechanical advantage, MA, is the ratio of resistance force to effort force (MA = Fr / Fe)
  • The ideal mechanical advantage, IMA, is the ratio of the displacements (IMA = de / dr). In all real machines, MA is less than IMA.
Origins of IMA = de / dr :
An ideal machine transfers all the energy, so the output work must equal the input work:
Wo = Wi,
And since Wo = the product of the resistance force and machine displacement (Frdr)
And recalling that Wi = the product of the effort force and displacement caused by the machine controller,
Then Frdr must equal Fede
This equation can be re = written (manipulated) to bring about a ratio equation. Simply divide dr and Fe on both sides of the equation. Variables with cancel to give the ration equation:
(Fr / Fe) = (de / dr)
As this is an ideal situation, the right side indicates IMA as equivalent to MA (left side). Again, this is only characteristic of an ideal machine. This is how de / dr is derived.

Here are two ways Efficiency can be expressed / calculated (it is a percent value).
  1. In terms of work: efficiency = (Wo / Wi) X 100%
  2. In terms of mechanical advantage and ideal mechanical advantage: efficiency = (MA / IMA) X 100%
The second expression is derived from the first expression. Recall that Wo = (Fr / Fe) and that Wi = (dr / de) so Wo / Wi can be communicated as MA / IMA.
Further Notes on Machines:
  • IMA of most machines is a fixed value maintained through the machine’s design
  • An efficient machine has an MA almost equal to its IMA. This is true because in fraction arithmetic, the closer the value of a numerator is to a larger denominator, the resulting quotient will approach 1, but still < 1. And because we know this quotient will be multiplied by 100% to yield an efficiency value, a quotient of 1 would give 100% efficiency (1 x 100% = 100%).
  • A less efficient machine will have a smaller MA, because the division of MA / IMA will result in a much smaller quotient.
  • A conclusive statement can be made: Lower efficiency means that a grater effort force is needed to exert the same resistance force.
Compound Machines–
  • All compound machines are combinations (sometimes complex) of up to 6 simple machines. The orientation or links of these component machines are situation so that the resistance force of one machine becomes the effort force of the subsequent machine. Essentially, the six simple machines serve as monomers for the construction and arrangement of compound machines (polymers).
Six Simple Machines
Wheel- and-axle
Inclined plane

Example Problem with the bicycle wheel:
A student uses the bicycle wheel with gear radius 3.00 cm and wheel radius 34.6 cm. When a force of 154 N is exerted on the chain, the wheel rim moves 13.0 cm. Due to friction; its efficiency is 94.0%.
Given information:
  • Effort force, Fe = 154 N
  • Gear radius = 3.00 cm
  • Wheel radius = 34.6 cm
  • Efficiency = 94.0%
  • Resistance displacement, dr = 13.0 cm
(a) What is the IMA of the wheel and gear?
IMA = de / dr = (gear radius / wheel radius) = (3.00 cm / 34.6 cm) = 0.0867052023 = 0.0867
(b) What is the MA of the wheel and gear?
Since efficiency is = (MA / IMA) X 100%
MA = eff X IMA / 100% = ((94.0%) (0.0867052023)) / (100%) = 0.0815028902 = 0.0815
(c) What force does the scale attached to the wheel read? (Find resistance force, Fr)
MA = (Fr / Fe), then Fr = (MA)(Fe)
So Fr = (0.0815028902) (154 N) = 12.55144509 N = 12.6 N
(d) How far did the student pull the chain? (Find effort displacement, de)
IMA = (de / dr)
So de = (IMA)(dr) = (0.0867052023) (13.0 cm) = 1.12716763 cm = 1.13 cm

Define Work in terms of force and displacement:
Work is the product of force and displacement in the direction of the force.

Solve Problems Involving Work, Force and Displacement:

Example 1

A Student lifts a box of books that weights 184 N. The box is lifted 0.700 m. How much work does the student do?

Given information:
  • mg = 184 N
  • d = 0.700 m

Unknown Variable(s):
  • W (work)

Basic Equation(s):
  • W = Fd

The student exerted enough force to lift the box, that is, enough to balance the weight of the box. It is important to remember that work is done on an object only if it moves in the direction of the force. Thus,

W = Fd = (184 N)(0.700 m) = 128.8 N(m) = 129 J

Example 2

A sailor pulls a boat along a dock using a rope at an angle of 50.0°. How much work does the sailor do if he exerts a force of 254 N on the rope and pulls the boat 29.0 m?

Given information:
  • θ = 50.0°
  • F = 254 N
  • d = 29.0 m

Unknown Variable(s):
  • W (work)

Basic Equation(s):
  • W = Fd cos θ

The Work done by the sailor is:
W = Fd cos θ = (254 N) (29.0 m) (cos 50.0°) = 4734.773533 J = 4.73 X 103 J

Define Energy:
Energy is a non – material property capable of causing changes in matter.

Define Gravitational Potential Energy:
Change of energy of object when moved in a gravitational field.

Solve Problems involving gravitational potential energy, mass, acceleration due to gravity and height above a reference point:

A 3.00-kg textbook is lifted from the floor to a shelf 3.10 m above the floor.

Given information:
  • m = 3.00 kg
  • Information to calculate change in heights

Unknown Variable(s):
  • Potential Energy, Ep

Basic Equation(s):
  • Ep = mgh

Gravitational Potential Energy relative to the floor

h = 3.10 m – 0.00 m = 3.10 m

Ep = mgh = (3.00 kg)(9.8 m / s2)(3.10 m) = 91.14 J = 91.1 J

Gravitational Potential Energy relative to the head of a 1.64 m tall person

h = 3.10 m – 1.64 m = 1.46 m

Ep = mgh = (3.00 kg)(9.8 m / s2)(1.46 m) = 42.924 J = 42.9 J

Define Kinetic Energy:

Kinetic energy is the energy of motion (the movement of an object, particle, or set of particles). Any object in motion uses kinetic energy.
Example) a thrown tennis ball, people walking, etc…

If the friction is zero, the work done on the mass equals the change in kinetic energy of the mass. If there is a friction force, some or all of the work is done against friction. All or part of the work done will become heat energy.

Objects that are not in motion possess potential energy, which is converted to kinetic energy when some force (like gravity) acts upon the object.
Example) Elastic potential energy is stored in a stretched rubber band and when the rubber is released, the stored energy is converted to kinetic energy.

Solve problems involving kinetic energy, mass, and velocity
A rocket of mass 1.5x104kg accelerates at 220m/s2 for 29s from aninitial speed of 5200m/s.

(a) How fast will be rocket be travelling after the 29s?
Solution: By using a=(v-u)/(t) when time is 29s, acceleration is 220m/s2, v(final velocity) is unknown, and u(initial velocity) is 5200m/s, we can rearrange this equation to find out the final velocity. Thus, a new equation gives us v-u=at, which is v=at + u. When numbers are plugged in, (220x29)+(5200) = 5200+6380, we get 11580m/s.

(b) How much Kinetic Energy has the rocket gained?
Solution: Calculate the kinetic energy of the rocket both before and after the acceleration and work out the difference. By using an equation, Ek =1/2mv2, we can find out the initial kinetic energy. Initial kinetic energy is 0.5x(1.5x104) x (5200) 2, which is 2.028x1011J. Then the final kinetic energy is 0.5x(1.5x104) x (11580) 2, which is 1.006x1012J. Thus, Kinetic Energy gained = Final Ek - Initial Ek, 1.006x1012J - 2.028x1011J = 8.032x1011J.

Define temperature, thermal energy, and specific heat capacity:

Temperature is a measurement of the average kinetic energy of the molecules in an object, measured with either a thermometer or a calorimeter. It’s a means of determining the internal energy contained within the system.
*Note that the temperature is different from the heat although these two concepts are linked. Temperature is a measure of the internal energy of the system while heat is a measure of how energy is transferred from one system to another.

Thermal Energy is the energy that is associated with the random motion of molecules and atoms. We know that when the thermal energy of an object increases, the molecular and atomic motion in the object increases. Thermal energy is kinetic energy at the molecular or atomic level. Thermal energy of an object can increase by 1)changing mechanical energy to thermal energy 2)transfer of heat from one object to another

Specific heat capacity(of a solid or liquid) is defined as the heat required to raise unit mass of substance by one degree of temperature. The equation is Q=mcΔT where:
Q = heat
m = mass
c = specific heat
ΔT = change in temperature

Examples of specific heat capacities:
  • water has a specific heat capacity of 4184 J/kg degree Celsius
  • copper has a specific heat capacity of 390 J/kg degree Celsius

Solve problems involving mass, specific heat capacity, and change in temperature

An insulated container (negligible specific heat capacity) contains 475g of water at a temperature of 15 degree Celsius. If 325g of hot water at a temperature of 80 degree Celsius is added, what is the temperature of the water in the container when it is completely mixed?

Heat gained by cold water + Heat lost by hot water = 0
(McΔTcCc + MhΔThCc = 0)

(0.475)(Tf -15)(4.18 x 103) + (0.325)(Tf – 80)(4.18 x 103) = 0

(1.99 x 103)(Tf – 2.98 x 104J + (1.36 x 103) Tf – 1.09 x 105J = 0

(3.34 x 103J/degree Celsius) Tf = 1.38 x 105J

Tf = 41.4 degree Celsius

Relate Energy Transformation to Work Done:
The relationship between energy transformation and work done is best represented with the work – energy theorem. The work – energy theorem states that the change in the kinetic energy of an object is equal to the net work done on it. This is simply expressed as W = Δ Ek

State The Law of Conservation of Energy:
The Law of conservation of Energy states that within a closed, isolated system, energy can change form, but the total amount of energy is constant. Energy can be neither created nor destroyed.

Solve Problems using the Law of Conservation of Energy Including Changes in Gravitational Potential Energy, Kinetic Energy, and Thermal Energy:
A large chunk of ice with mass 14.0 kg falls from a roof 7.00 m above the ground.
Given information:
  • m = 14.0 kg
  • g = 9.80 m / s2
  • h = 7.00 m
  • Eki = 0
  • Epf = 0
Unknown Variable(s):
  • We will be asked to find Ekf and vf
Basic Equation(s):
  • Eki + Epi = Ekf + Epf
Find the kinetic energy of the ice when it reaches the ground–
Eki + Epi = Ekf + Epf
0 + mgh = Ekf – 0
Therefore, Ekf = mgh = (14.0 kg)(9.80 m / s2)(7.00 m) = 960.4 J = 9.60 X 102
Find the velocity of the ice when it reaches the ground–
Recall that Ekf = (1/2)m(vf)2
(vf)2 = (2 (Ekf) )/ (m) = (2(960.4 J) )/ (14.0 kg) = 137.2 m2 / s2

The square root of 137.2 m2 / s2 gives vf = 11.71324037 m / s = 11.7 m / s