energy skate park post-lab - gbwphysics.org 8 energy/unit 8 energy pdf/post-lab... · 1) start your...

Directions:

(a) Create a track for the skater that has at least one loop (you add track by clicking and dragging in pieces from the top left corner). DO NOT MAKE YOUR TRACK TOO BIG, OR THE PROGRAM WILL CRASH!!!

(b) Once you track is set up answer the following questions.(c) Once your track is created DO NOT PRESS THE “RESET”

BUTTON!!!! If your skater falls off the track, press the “Return Skater” button on the top right.

Energy Skate Park Post-Lab

NOTES...LIVE AND IN COLOR!!!

I would suggest you take notes like this:

(1) (2)

Point A:h = 10 mP.E. = mghP.E. = (75 kg)(10 m/s2)(10 m)P.E. = 7500 JStarts from restv = 0 m/sK.E. = 1/2mv2

K.E. = 1/2(75 kg)(0 m/s)2

K.E. = 0 JEtot = P.E. + K.E.Etot = 7500 J + 0 J = 7500 J

Point B:h = 3 mP.E. = mghP.E. = (75 kg)(10 m/s2)(3 m)P.E. = 2250 JEtot = 7500 J (C.O.E.)K.E. = ???Etot = P.E. + K.E.7500 J = 2250 J + K.E.K.E. = 7500 J ‐ 2250 J K.E. = 5250 J K.E. = 1/2mv2 5250 J = 1/2(75 kg)v2

v =

v = 11.8 m/s

2(5250 J) 75 kg√

Point C:h = 5 mP.E. = mghP.E. = (75 kg)(10 m/s2)(5 m)P.E. = = 3750 JEtot = 7500 J (C.O.E.)K.E. = ???Etot = P.E. + K.E.7500 J = 3750 J + K.E.K.E. = 7500 J ‐ 3750 J = 3750 JK.E. = 3750 JK.E. = 1/2mv2 3750 J = 1/2(75 kg)v2

v =

v = 10 m/s

2(3750 J) 75 kg√

C.O.E.(CONSERVATION OF ENERGY) C.O.E. Caclulations

A

B

10 m

3 m

5 m

E tot = 750

0 J

P.E. = 750

0 J

K.E. = 0 J

v = 0 m/ s

E tot = 750

0 J

P.E. = 225

0 J

K.E. = 525

0 J

v = 11

.8 m/ s

E tot = 750

0 J

P.E. = 375

0 J

K.E. = 375

0 J

v = 10

m/ s

Skater:

mass = 75

kg

A

CC

C B

C

1) Start your skater at the top of your track and let your skater do his thing. Draw a picture of your skate ramp below. Be sure to include where the ground is in relation to your track.

2) Click the “Show Pie Chart” box on the right.

B) Place a ‘B’ on your track where the kinetic energy is the greatest.

(There may be more than one place!!!!)

B) Place a ‘B’ on your track where the kinetic energy is the greatest. (There may be more than one place!!!!)

A) Place an ‘A’ on your track where the potential energy is the greatest. (There may be more than one place!!!)

A) Place an ‘A’ on your track where the potential energy is the greatest. (There may be more than one place!!!)

h

mgmg Our skater has mass, gravity, and a height above the ground to start, but he has no velocity!

What does this mean?

h

mg The skater has no kinetic or thermal energy. The total energy is all potential at the top.

mg

h

mg

Etot = PE + KE + TEEtot = PE + KE + TE

The skater has no kinetic or thermal energy. The total energy is all potential at the top.

0 0

mg

10m

Let's use real

values for emphasis

75kg75kg

PE = (75kg)(10m/s2)(10m)

PE = 7500 J

PE = (75kg)(10m/s2)(10m)

PE = 7500 J

10m

Let's use real

values for emphasis

75kg75kg PE = mghPE = mghPoint A:Point A:

PE = mghPE = mghPoint A:Point A:PE = (75kg)(10m/s2)(10m)

PE = 7500 J

PE = (75kg)(10m/s2)(10m)

PE = 7500 J

10m

Let's use real

values for emphasis

75kg

This is Etot

This is Etot

NOTE:Etot is the same for all points on the track (C.O.E.)

NOTE:Etot is the same for all points on the track (C.O.E.)

75kg

PEThe PE will be a maximum whenever the skater reaches

the same height again

PE = 7500 J

A AA APE = 7500 J

PEKE

KE

The KE will be at a maximum whenever the skater is at the lowest point on the track. The PE is at a minimum, and has been transformed to KE.

B B

A AA A

B B#2 on Lab

How much PE does the rider have at these points on the track?


B B

A AA A

B B

h=3m



B B

A AA A

B B

h=3m



B B

A AA A

B BPE = mghPE = mghPoint B:Point B:

PE = mghPE = mghPoint B:Point B:

h=3mPE = (75kg)(10m/s2)(3m)

PE = 2250 J



B B

A AA A

B BPE = (75kg)(10m/s2)(3m)

PE = 2250 J

How much KE does the rider have at these points on the track?


h=3mB B

A AA A

B B


Etot = 7500 J


Etot = 7500 JPE = 2250 JPE = 2250 J

h=3mB B

A AA A

B B

h=3mB B

A AA A

B B


Etot = 7500 J


Etot = 7500 JPE = 2250 JPE = 2250 J

h=3mB B

A AA A

B B


Etot = 7500 J


Etot = 7500 JPE = 2250 JPE = 2250 J

Etot = PE + KEEtot = PE + KE

Point B:Point B:

h=3m

7500J = 2250 + KE

KE = 5250J


7500J = 2250 + KE

KE = 5250JB B

A AA A

B B


Etot = 7500 J


Etot = 7500 JPE = 2250 JPE = 2250 J

Point B:Point B:

How fast is the skater going at these points?


B B

A AA A

B B

B B

A AA A

B B



B B

A AA A

B B



KE = 5250JKE = 1/2mv2KE = 5250JKE = 1/2mv2

Point B:Point B:

2KE/m = v2

√(2KE/m) = √(v2)2KE/m = v2

√(2KE/m) = √(v2)

√[2(5250)/75] = v11.8m/s = v

√[2(5250)/75] = v11.8m/s = v

KE = 5250JKE = 1/2mv2KE = 5250JKE = 1/2mv2

Point B:Point B:

B B

A AA A

B B



3) Click the “Energy vs. Time” button on the right ....... When are the kinetic and potential energy of the skater equal? Label these points as “C” your diagram above.

B B

A AA A

B B

PE KEHALFWAY

PE and KE are equal whenever the skater is located HALFWAY

to the ground.

#3 on LabB B

A AA A

B B

C C C CC C C C

CC

PE KEHALFWAY


to the ground.

C C CC C C

#3 on LabB B

A AA A

B B

CC

PE KEHALFWAY


to the ground.

C C C

B B

A AA A

B BEtot = 7500 JEtot = 7500 J

C C CPoint C:Point C:

CC

PE KEHALFWAY


to the ground.

C C C

B B

A AA A

B BEtot = 7500 JEtot = 7500 J

PE = 3750 JPE = 3750 J

KE = 3750 JKE = 3750 J

C C CPoint C:Point C:

CC

PE KEHALFWAY


to the ground.

C C C

B B

A AA A

B B

C C C

h=5m

CC

PE KEHALFWAY


to the ground.

C C C

B B

A AA A

B BPE = 3750 JPE = 3750 J

C C C

h=5mPE = mghPE = mgh

PE = (75kg)(10m/s2)(5m)PE = (75kg)(10m/s2)(5m)

Point C:Point C:

CC C C

B B

A AA A

B B

C C

h=5m

Guess what? We can figure out how fast the skater is going at this point too!


CC

C C C

B B

A AA A

B B

C C C

h=5m



CCC C C

B B

A AA A

B B

C C C

h=5m

KE = 3750 JKE = 1/2mv2KE = 1/2mv2



Point C:Point C:KE = 3750 J

CCC C C

B B

A AA A

B B

C C C

h=5m2KE/m = v2

√(2KE/m) = √(v2)2KE/m = v2

√(2KE/m) = √(v2)

√[2(3750)/75] = v√[2(3750)/75] = v

10 m/s = v10 m/s = v

KE = 3750 JKE = 1/2mv2KE = 1/2mv2



Point C:Point C:KE = 3750 J

I would suggest you take notes like this:

(1) (2)

Point A:h = 10 mP.E. = mghP.E. = (75 kg)(10 m/s2)(10 m)P.E. = 7500 JStarts from restv = 0 m/sK.E. = 1/2mv2

K.E. = 1/2(75 kg)(0 m/s)2

K.E. = 0 JEtot = P.E. + K.E.Etot = 7500 J + 0 J = 7500 J

Point B:h = 3 mP.E. = mghP.E. = (75 kg)(10 m/s2)(3 m)P.E. = 2250 JEtot = 7500 J (C.O.E.)K.E. = ???Etot = P.E. + K.E.7500 J = 2250 J + K.E.K.E. = 7500 J ‐ 2250 J K.E. = 5250 J K.E. = 1/2mv2 5250 J = 1/2(75 kg)v2

v =

v = 11.8 m/s

2(5250 J) 75 kg√

Point C:h = 5 mP.E. = mghP.E. = (75 kg)(10 m/s2)(5 m)P.E. = = 3750 JEtot = 7500 J (C.O.E.)K.E. = ???Etot = P.E. + K.E.7500 J = 3750 J + K.E.K.E. = 7500 J ‐ 3750 J = 3750 JK.E. = 3750 JK.E. = 1/2mv2 3750 J = 1/2(75 kg)v2

v =

v = 10 m/s

2(3750 J) 75 kg√

C.O.E.(CONSERVATION OF ENERGY) C.O.E. Caclulations

A

B

10 m

3 m

5 m

E tot = 750

0 J

P.E. = 750

0 J

K.E. = 0 J

v = 0 m/ s

E tot = 750

0 J

P.E. = 225

0 J

K.E. = 525

0 J

v = 11

.8 m/ s

E tot = 750

0 J

P.E. = 375

0 J

K.E. = 375

0 J

v = 10

m/ s

Skater:

mass = 75

kg

A

CC

C B

C

4) ..... Now change the location of your skater to space, the moon, and Jupiter...... Explain the changes in motion you observe when the skater goes to each new location.

Space:

Remember Newton's 1st

Law???

Space:


The skater is an object at rest with no unbalanced force acting on him until

you pressed the arrow keys.


Law???

Space:


Once the skater is moving on the track, he has no gravity to affect his velocity,

so he flies off forever.


Law???

Moon:


Moon:


Since the skater's acceleration due to gravity is less on the Moon, he speeds up and slows down at a lesser rate.

This makes him seem to move in slow motion.

Since the skater's acceleration due to gravity is less on the Moon, he speeds up and slows down at a lesser rate.

This makes him seem to move in slow motion.

Jupiter:


Jupiter:

Jupiter:

#4 on Lab


Jupiter's gravity is much greater than the Earth's so the skater speeds up and slows down at a greater rate (greater ACCELERATION)

Jupiter:

Jupiter's gravity is much greater than the Earth's so the skater speeds up and slows down at a greater rate (greater ACCELERATION)

CCCCCCCC

5) Change your skater....Using the pie graph, observe where the skater is on the track when potential and kinetic energies are equal. Label these points on your track as “D”.

#5 and #6 on LabB B

A AA A

B B

DD DD DD DD

Although the total energy is less with a smaller mass, the spots on the track that have equal PE and KE are the same.

g=9.81 m/s2 for all masses

#5 and #6 on LabB B

A AA A

B B

CCCCCCCC

Although the total energy is less with a smaller mass, the spots on the track that have equal PE and KE are the same.

g=9.81 m/s2 for all masses

#5 and #6 on LabB B

A AA A

B BPE = KEmgh = 1/2mv2

When:PE = KE

mgh = 1/2mv2

Mass has no affect

CCCCCCCCDD DD DDDD

h=5m

If the bulldog has a total energy of 2000 J atthe top of the track, how much PE and KE

does he have at this point?



20kg20kg

h=5m

PE = mgh





PE = mgh

20kg20kg

h=5m

PE = mgh

PE = (20kg)(10m/s2)(5m)PE = 1000 J





PE = mgh

PE = (20kg)(10m/s2)(5m)PE = 1000 J

20kg20kg

h=5m

PE = 1000 J

Etot = 2000 J





PE = 1000 J

Etot = 2000 J

20kg20kg

h=5m





PE = 1000 J

Etot = 2000 JPE = 1000 J

Etot = 2000 J

20kg20kg

h=5m





PE = 1000 J

Etot = 2000 JPE = 1000 J

Etot = 2000 J

KE = 1000 JKE = 1000 J

2000 J = 1000 J + KE


2000 J = 1000 J + KE

20kg20kg

Guess what? We can figure out how fast the bulldog is going at this point too!


h=5m

20kg20kg



KE = 1000 JKE = 1000 JKE = 1/2mv2KE = 1/2mv2

h=5m

20kg20kg

2KE/m = v2

√(2KE/m) = √(v2)2KE/m = v2

√(2KE/m) = √(v2)

√[2(1000)/20] = v√[2(1000)/20] = v

10 m/s = v10 m/s = v

KE = 1000 JKE = 1000 JKE = 1/2mv2KE = 1/2mv2



h=5m

20kg20kg

7) Does changing the mass of the skater change the total energy in the skaterramp system? Explain the relationship.

By lowering the mass of the skater, the total energy of the skater is less than before.

Total Energy = PEtopTotal Energy = mgh

Total Energy = PEtopTotal Energy = mgh

8) Finally, click the ‘Track Friction’ button on the right. Move the slider to the right. What happens to the skater? As the skater moves through the track, what type(s) of energy does he possess now?

8) Finally, click the ‘Track Friction’ button on the right. Move the slider to the right. What happens to the skater? As the skater moves through the track, what type(s) of energy does he possess now?

The kinetic energy has

been converted entirely to

thermal energyHello real life!

What have we eliminated?

9) Move the skater around until you find where the skater has only thermal energy. What TWO things have to happen for the skater to have only thermal energy?

9) Move the skater around until you find where the skater has only thermal energy. What TWO things have to happen for the skater to have only thermal energy?

PE = mgh

KE = 1/2mv2

PE = mgh

KE = 1/2mv2

h = 0 mv = 0 m/sv = 0 m/sh = 0 m

With no height (h) or velocity (v), all the energy has been converted to thermal energy

With no height (h) or velocity (v), all the energy has been converted to thermal energy

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energy skate park post-lab - gbwphysics.org 8 energy/unit 8 energy pdf/post-lab... · 1) start your...

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