Name
Date
Pd
Energy Storage and Transfer Model Worksheet 5:
Energy Transfer and Power
1. A student eats a tasty school lunch containing 700 Calories. (One food Calorie = 4186 joules.) Due
to basal metabolism, the student radiates about 100 joules per second into the environment.
a. How long would the student have to sit on a couch to radiate away all of the energy from lunch?
700Cal 
4186J
1s
1hr

 29,300s 
 8.1hrs
1Cal 100J
3600s
b. If all of the energy from the student’s lunch did something useful, like lifting pianos weighing
5000 newtons to the top of a 10-meter tall apartment building, how many pianos could be lifted
with the energy from lunch? (Ignore the energy radiated by the student.) Complete the energy
bar graph below to aid your solution.
System/Flow
student
pianos
Earth
Ek Eg
Ee
Eth
Energy (J)
Ek Eg Eel Ech
Energy (J)
Note: Ech category added
0
0
Position B
Position A
Energy Conservation Equation: Ech = Eg
To lift one piano: Eg  mg h  5000N 10m  50,000J

700Cal 

4186J 1piano

 58 pianos with 30,000J leftover.
1Cal
50,000J
2. Jill pulls on a rope to lift a 12 kg pail out of a well, while the clumsy Jack watches. For a 10 meter
segment of the lift, she lifts the bucket straight up at constant speed. How much power is required
to complete this task in 5 seconds? Complete the energy bar graph as part of your solution.
System/Flow
Jill
pail
Earth
Ek Eg
Ee
Eth
Energy (J)
Energy (J)
Ek Eg Eel Ech
0
0
Position A
Position B
Energy Conservation Equation:
Ech = Eg
N
E g 12kg 10 kg 10m
Power 

 240W
time
5s
  
©Modeling Instruction - AMTA 2013
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U8 Energy - ws 5 v3.1
3. Hulky and Bulky are two workers being considered for a job at the UPS loading dock. Hulky
boasts that he can lift a 100 kg box 2.0 meters vertically, in 3.0 seconds. Bulky counters with his
claim of lifting a 200 kg box 5.0 meters vertically, in 20 seconds. Which worker is more
powerful?
Hulky
N
2.0m
E g 100kg 10 kg
P

 667W
t
3.0s
 
P
E g
t

Bulky
N
200kg 10 kg
5.0m
 
20s
 500W
Hulky is more powerful.
4. The trains on the Boss rollercoaster are raised from 10 m above ground at the loading platform to a
height of 60 m at the top of the first hill in 45 s. Assume that the train (including passengers) has a
mass of 2500 kg. Ignoring frictional losses, how powerful should the motor be to accomplish this
task? Complete the energy bar graphs below.
System/Flow
trains
Earth
Position A
motor
P
E g
t

 
©Modeling Instruction - AMTA 2013
Eth
W
Position B
W = Eg
N
2500kg 10 kg
50m
45s
Ee
0
0
Energy Conservation Equation:
Ek Eg
Energy (J)
Energy (J)
Ek Eg Eel Ech
 28.4kW
2
U8 Energy - ws 5 v3.1
5. a. An aerodynamic 1,000 kg car takes about 270 newtons of force to maintain a speed of 25 m/s.
How much horsepower is required from the engine to maintain this speed? (1 hp = 746 W)
J N m
m
 N   W  F v
The units of power are W  
s
s
s
1hp
P  270N 25 ms  6750W 
 9.0hp
746W


b. How much horsepower is required for the same car to accelerate from 0-25 m/s in 6 seconds?
P
Ek

t
1
2

m v f2  v i2
t

1
2
1000kg 25   0
m
s
2

6.0s
312.5kJ
 52.1kW
6.0s
6. Your electric utility company sends you a monthly bill informing you of the number of kilowatthours of energy you have used that month.
a. What is a kilowatt-hour (kilowatt x hour, or kWh)? Determine how many Joules equal one
kilowatt-hour.
P
b.

E
E P t
t

E  1kW h  1000 Js 3600s  3,600,000J or 3,600kJ
A frost free, 17 cu. ft. refrigerator-freezer uses energy at a rate of 500 watts when you hear the
compressor running. If the fridge runs for 200 hours per month, how many kilowatt-hours of
energy does the refrigerator use each month?



E  P  t  500 Js 200hr 3600 hrs  360,000kJ
c. In the Phoenix area, electricity rates range from 8.0 cents per kilowatt-hour (winter) to
11.5 cents per kWh (summer). How much does the energy cost each month to run the
refrigerator?
Cost = rate  time
$0.080
 0.5kW 200hr  $8.00
Winter
kWh
$0.115
 0.5kW 200hr  $11.50
Summer
kWh



©Modeling Instruction - AMTA 2013

3
U8 Energy - ws 5 v3.1