Energy Unit Overview and Objectivesby Jennifer Hizon + Nicole Bartsch + Jennifer Kim

Power

Power is the rate of doing work or the rate of using energy, which are numerically the same. It is the quantity, the quantity work has to do with a force causing a displacement. Work has nothing to do with the amount of time that the force acts to cause the displacement. Power is the rate at which work is down. It is the work/time ratio. Power is measured in watts (W). One watt is one joule of energy transferred in one second. It is a scalar unit.

equation:power = △work/△time

This can be written as:

P = △work/△time

= (Fd)/△time

= F △d/△t
= Fv

Problems Involving Power

Problem 1: Power
A forklift lifts a crate of mass 100 kg at a constant velocity to a height of 8m over a time of 4s. The forklift then holds the crate in place for 20s. Calculate how much power the forklift exerts in lifting the crate?

Knowns:
mass of crate: 100kg
height that crate is raised: h = 8m
time to raise crate: 4s
time that crate is held in place = 20s
*required to calculate the power exerted

Determine which formula to approach the problem:

P = F △x/△t
*to calculate power. The force required to raise the crate is equal to the weight of the crate.

Calculate calculate the power required to raise the crate:
P = F △x/△t
= m x g △x/△t
= (100kg) (9.8 ms^-2) 8m/4s
= 1960 W

Problem 2: Work
James Joule lifts an 855N James Watt up 1.45m in 2.34s. How much Power does Mr. Joule exert?

Knowns:
F = 855N
d = 1.45m
t = 2.34s
W = Fd

Solution:
W = 855N x 1.45m = 1239.75J
P = W/ △t
P = 1239.75J/ 2.34s = 529.81 J/s
P = 530 W

Problem 3: Time
Find the power of the man who pushes the box 8m with a force of 15N in 6 seconds

W = Fd
W = 15N x 8m
W = 120 joule

Power = Work/time
Power = 120N/ 6s
Power = 20 watts

The power of the man is 20 watts. In other words he does 20 joules of work in 6 seconds.

Efficiency

efficiency is the ratio of output work (or energy) to input work (or energy). The Law of conservation of energy says

Einitial = Efinal

Energy is conserved, but it doesn’t end up where we want. The potential energy is converted into heat at the bearings, wheels and kinetic energy. Efficiency also describes how much useful energy we get out of an energy conversion

Eff(%) = Workoutput/ Workinput x 100%

Eff(%) = △Energyoutput/ △Energyinput x 100%

Compare the efficiencies of common devices

Electric Motor - Efficiency %
Home Oil Furnace - 65
Home Coal Furnace - 55
Steam Boiler (power plant) - 89
Power Plant (thermal) - 36
Automobile engine - 25
Light Bulb Fluorescent - 20
Light Bulb incandescent - 5

Consider a car.

Suppose that the car’s gasoline has 1000J of chemical energy

700J of the chemical energy is converted into heat

Only 300J is converted in to Kinetic Energy

Gas approx 30%, Diesel approx 45% efficient

Example: A 1500W kettle heats 1.5kg of water from 18 degrees Celsius to 59 degrees Celsius in 180s (Cwater = 4180J/kg/〬K)
a.) How much electrical energy was used by the kettle?
P = E/t
E = P x t = 1500W x 180s = 270000 J
b.) How much heat was delivered to the water?
Eheat = mc△T
Eheat = (1.5kg)(41800J/kg/〬K)(59〬C-18〬C) = 260000J
c.) what is the kettle’s efficiency?
Eff = (Eoutput/Einput) x 100%
Eff = (260000J / 270000J) x 100% = 96%

Work in terms of force and displacement

For cases were the force is constant, we define work as the product of the force exerted on an object and the distance the object moves in the direction of the force. Work is a scalar quantity and has no direction. The SI unit is the joule. Work is done only on an object if the object moves. Work is only done when the force and displacement are in the same direction. It is a form of energy and is the transfer of energy by mechanical means.

Work is said to be done when a force produces a motion

Equation: W = Fd
W = the fork
F = the magnitude of force
d = the magnitude of displacement in the direction of the force

Problem 1: Work

Mr. Frank’s A motorcycle breaks down, and he pushes with 80N for 20m to the side of the road. How much work does Mr. Cox do?

Knowns:
F = 80N
d = 20m

Solution:
W = Fd
W = 80N x 20m = 1600nm = 2000J = 2kJ

Problem 2: Force

A 20N force accelerates a yo-yo from rest to 8.0m/s in 2.5s. Calculate the work done.

Knowns:
F = 20N, Vi = 0m/s, Vt = 8.0m/s, t = 2.5s

Solution:
d = 1/2 (Vi + Vt) t
d = 1/2 (0m/s + 8.0 m/s) (2.5s) = 10m
W = Fd
W = 20N x 10m = 200j

Problem 3: Displacement

What work is done to stop an eastward moving 8.0kg dust bunny with a 60N force in 2.0s?

Solution:
F△t = m△v, rearrange and solve for △V
△v = F△ t/m = -60N x 2.0s/ 8.0 kg = -15m/s
d = 1/2 (Vi + Vt) t
d = 1/2 (15m/s + 0m/s) 2.0s = 15m
W = Fd
W = 60N x 15m = 900J

Energy

Work done

An indirectly observed quantity

What enables things to move

Gravitational Potential Energy

Gravitational Potential Energy is energy that an object possesses due to the object's position on the gravitational field.

Problems Involving Gravitational Potential Energy

An apple has has a mass of 0.15 kg and is at a height of 4 m in a tree, what is this apple's gravitational potential energy?

gravitational potential energy = (m)(g)(h)=(0.15 kg)(9.8)(4 m)= 5.88 J The apple has a gravitational potential energy of 5.88J.

An cell phone sitting on a table has a mass of 0.30 kg and a gravitational potential engergy of 2.94 J, what is its height?

Height = GPE/(m * g)= (2.94 J) / ( 0.30 * 9.8 )= 1 m The height of the cell phone is 1m.

A vase has a height of 4.2 m and has a gravitational potential energy of J, what is its mass?

Mass = GPE /( g * h )= ( 40.34 ) / ( 9.8 * 4.2 )= 0.98 kg The mass of the vase is 0.98kg.

Kinetic Energy

The kinetic energy of an object is the energy it possesses because of its motion.

Problems Involving Kinetic Energy

A girl is riding on a bike. She has a mass of 53.5 kg and her bike has a mass of 13.3 kg. She is traveling at a velocity of 4m/s, what is this kinetic energy?

kinetic energy = (0.5)(m)(v)2= ( 0.5 )( 53.5kg + 13.3kg )(4m/s)2= 534.4 J The kinetic energy of the girl on the bike is 534.4 J.

An car is traveling at a velocity of 10 m/s and a kinetic energy 58,750 J what is its mass?

mass = (KE) / ( 0.5 * v2)= ( 58,750 J ) / ( 0.5 * 102 )= 1175kg The car's mass is 1175 kg.

A rolling tennis ball has a mass of 0.06 kg and a kinetic energy of 0.0019 J what is its velocity?

velocity = [(KE) / ( 0.5) (m) ] <- square rooted= square root of ( 0.0019 J )/( 0.5 )( 0.06 m )=0.25m/s The tennis ball is rolling at a velocity of 0.25m/s.

Temperature How does energy transformation relate to work done?

ΔE = W

W represents the work done

Change in energy, so basically the energy transformation is the work done

Law of Conservation of Energy

In the law of conservation of energy, total amount of energy in a system remains the same.

Energy cannot be created nor destroyed.

Energy can change their form

Ex) chemical energy to kinetic energy

Problems involving the law of conservation of energy

(including changes in gravitational potential energy, kinetic energy, and thermal energy)

If we change the question above into a skateboarder going down a cement hill with frictional forces present, then the kinetic energy of the skateboarder at the bottom of the hill will be smaller than the potential energy at the top of the hill. It is due to the friction force generating some heat energy loss in the skateboarder skating down the hill.

Temperature

It is hotness, measured on some definite scale

Temperature of gas=the average kinetic energy of the particles

Thermal Energy

Heat results from the thermal energy

Heat is basically the transfer of thermal energy

Specific Heat Capacity

It is amount of energy that is used and added in order to raise the temperature of a unit mass to one temperature unit.

Problems involving mass, Specific heat capacity and change in temperature.

An aluminum block with a mass of 17.7g is warmed to 53.2°C and plunged into an insulated beaker containing 32.5g of water intitally at 24.5°C. The aluminum and the water are allowed to come to thermal equalibrium. Assuming that no heat is lost, what is the final temerature of the water and aluminum? The specific heat capacity of aluminum is 0.903J/gC and teh specific heat capacity of water is 4.18J/gC.

The cooling of the aluminum must be the same as the warming of the water. The new temperature of both will be x.

The 17.7g of aluminum will have to cool by (32.2-x)° with a factor of 0.903J/gC. The 32.5g of water will have to warm by (x-24.5)° with a factor of 4.18J/gC.

Multiplying these together gives 17.7(32.2-x)0.903 J and 32.5(x-24.5)4.18 J somce g,C cancel. Take 17.7(32.2-x)0.903 = 32.5(x-24.5)4.18 and solve for x. The value of x is how much the aluminum cools by and how much the water warms by.

Energy Unit Overview and Objectivesby Jennifer Hizon + Nicole Bartsch + Jennifer Kim## Power

equation:power = △work/△time= Fv

Problems Involving PowerProblem 1: PowerA forklift lifts a crate of mass 100 kg at a constant velocity to a height of 8m over a time of 4s. The forklift then holds the crate in place for 20s. Calculate how much power the forklift exerts in lifting the crate?

Knowns:

mass of crate: 100kg

height that crate is raised: h = 8m

time to raise crate: 4s

time that crate is held in place = 20s

*required to calculate the power exerted

Determine which formula to approach the problem:

P = F △x/△t

*to calculate power. The force required to raise the crate is equal to the weight of the crate.

Calculate calculate the power required to raise the crate:

P = F △x/△t

= m x g △x/△t

= (100kg) (9.8 ms^-2) 8m/4s

= 1960 W

Problem 2: WorkJames Joule lifts an 855N James Watt up 1.45m in 2.34s. How much Power does Mr. Joule exert?

Knowns:

F = 855N

d = 1.45m

t = 2.34s

W = Fd

Solution:

W = 855N x 1.45m = 1239.75J

P = W/ △t

P = 1239.75J/ 2.34s = 529.81 J/s

P = 530 W

Problem 3: TimeFind the power of the man who pushes the box 8m with a force of 15N in 6 seconds

W = Fd

W = 15N x 8m

W = 120 joule

Power = Work/time

Power = 120N/ 6s

Power = 20 watts

The power of the man is 20 watts. In other words he does 20 joules of work in 6 seconds.

## Efficiency

efficiencyis the ratio of output work (or energy) to input work (or energy). The Law of conservation of energy saysCompare the efficiencies of common devicesElectric Motor - Efficiency %

Home Oil Furnace - 65

Home Coal Furnace - 55

Steam Boiler (power plant) - 89

Power Plant (thermal) - 36

Automobile engine - 25

Light Bulb Fluorescent - 20

Light Bulb incandescent - 5

Consider a car.

Example: A 1500W kettle heats 1.5kg of water from 18 degrees Celsius to 59 degrees Celsius in 180s (Cwater = 4180J/kg/〬K)

a.) How much electrical energy was used by the kettle?

P = E/t

E = P x t = 1500W x 180s = 270000 J

b.) How much heat was delivered to the water?

Eheat = mc△T

Eheat = (1.5kg)(41800J/kg/〬K)(59〬C-18〬C) = 260000J

c.) what is the kettle’s efficiency?

Eff = (Eoutput/Einput) x 100%

Eff = (260000J / 270000J) x 100% = 96%

Work in terms of force and displacementEquation:W = FdW = the fork

F = the magnitude of force

d = the magnitude of displacement in the direction of the force

Problem 1: WorkMr. Frank’s A motorcycle breaks down, and he pushes with 80N for 20m to the side of the road. How much work does Mr. Cox do?

Knowns:

F = 80N

d = 20m

Solution:

W = Fd

W = 80N x 20m = 1600nm = 2000J = 2kJ

Problem 2: ForceA 20N force accelerates a yo-yo from rest to 8.0m/s in 2.5s. Calculate the work done.

Knowns:

F = 20N, Vi = 0m/s, Vt = 8.0m/s, t = 2.5s

Solution:

d = 1/2 (Vi + Vt) t

d = 1/2 (0m/s + 8.0 m/s) (2.5s) = 10m

W = Fd

W = 20N x 10m = 200j

Problem 3: DisplacementWhat work is done to stop an eastward moving 8.0kg dust bunny with a 60N force in 2.0s?

Solution:

F△t = m△v, rearrange and solve for △V

△v = F△ t/m = -60N x 2.0s/ 8.0 kg = -15m/s

d = 1/2 (Vi + Vt) t

d = 1/2 (15m/s + 0m/s) 2.0s = 15m

W = Fd

W = 60N x 15m = 900J

Energy## Gravitational Potential Energy

## Problems Involving Gravitational Potential Energy

gravitational potential energy = (m)(g)(h)=(0.15 kg)(9.8)(4 m)= 5.88 JThe apple has a gravitational potential energy of 5.88J.Height = GPE/(m * g)= (2.94 J) / ( 0.30 * 9.8 )= 1 mThe height of the cell phone is 1m.Mass = GPE /( g * h )= ( 40.34 ) / ( 9.8 * 4.2 )= 0.98 kgThe mass of the vase is 0.98kg.Kinetic Energy## Problems Involving Kinetic Energy

kinetic energy = (0.5)(m)(v)2= ( 0.5 )( 53.5kg + 13.3kg )(4m/s)2= 534.4 JThe kinetic energy of the girl on the bike is 534.4 J.mass = (KE) / ( 0.5 * v2)= ( 58,750 J ) / ( 0.5 * 102 )= 1175kgThe car's mass is 1175 kg.velocity = [(KE) / ( 0.5) (m) ] <- square rooted= square root of ( 0.0019 J )/( 0.5 )( 0.06 m )=0.25m/sThe tennis ball is rolling at a velocity of 0.25m/s.TemperatureHow does energy transformation relate to work done?E=W## Law of Conservation of Energy

## Problems involving the law of conservation of energy

(including changes in gravitational potential energy, kinetic energy, and thermal energy)## Temperature

## Thermal Energy

## Specific Heat Capacity

## Problems involving mass, Specific heat capacity and change in temperature.

The cooling of the aluminum must be the same as the warming of the water.

The new temperature of both will be x.

The 17.7g of aluminum will have to cool by (32.2-x)° with a factor of 0.903J/gC.

The 32.5g of water will have to warm by (x-24.5)° with a factor of 4.18J/gC.

Multiplying these together gives 17.7(32.2-x)0.903 J and 32.5(x-24.5)4.18 J somce g,C cancel.

Take 17.7(32.2-x)0.903 = 32.5(x-24.5)4.18 and solve for x.

The value of x is how much the aluminum cools by and how much the water warms by.