KINETIC AND POTENTIAL ENERGY CHANGES ANSWERS!!
Label the points at which:
__B____The skier is gaining speed
____C__ The point at which the skier is moving the fastest
____A____ The point at which the skier has the greatest
amount of potential energy
__B___ The point at which the potential energy is changing
to kinetic energy
___C__ The point at which the skier has the greatest amount
of kinetic energy
___D_ The point at which kinetic energy is changing to
potential energy
Explain why the skier will never return to the height of
point A.
Because PE D can never equal PE D, because you will
always lose some energy…..
Where and how is mechanical energy “lost?”
You lose the energy because of friction, and it turns
into heat.
A tennis ball undergoes several energy changes as it travels
towards a player, is struck, and rapidly speeds back over the net. Describe
these energy changes using the concepts of kinetic and potential energy. The ball has all
kinetic energy, as it strikes the racket, its energy turns into elastic
potential, and then back to kinetic energy again.
CONSERVATION OF ENERGY DIAGRAM SKILLS
A roller coaster with a mass of m moves along a smooth track
as diagrammed in the graph. The car leaves point A with no initial velocity and
travels to other points along the track. The zero energy level is taken as the
energy of point A.
1a) What is the car’s kinetic energy at point A? KE_{
A} = 0
b) What is the potential energy associated with the car at
point A? PE _{A} = mgh_{A}
c) What is the car’s kinetic energy at point B? KE_{B}= PE_{A} –
PE_{A} = ˝ mv^{2}
d) What is the cat’s potential energy at point B? PE_{B}
= mgh_{B}
2a) What is the speed of the car at point A? v_{A}=0
b) What is the speed of the car at point B? v_{B}=
sqrt(2KE_{B}/m) = sqrt (2(mgh_{A}-mgh_{B})/m)
3. Assume the mass of the car is 65 kg, and it starts at 30
m above the ground (each square is 5 m). Use the graph above to find the
kinetic energy, potential energy, and velocity for points C,D,E<F,G
Location |
KE_{A} |
PE_{A}=mgh |
KE _{location} |
PE _{location} |
V _{location} |
A |
0 |
65(9.8)(30) |
0 |
19 110 J |
0 |
B |
0 |
19 110 J |
19 110-6370=12740 J |
65(9.8)10=6370J |
19.79 |
C |
0 |
19 110 J |
19 110-9555=9555 J |
65(9.8)15=9555J |
17.1 |
D |
0 |
19 110 J |
19 110-6370=12740 J |
65(9.8)10=6370J |
19.79 |
E |
0 |
19 110 J |
19 110-3185=15925J |
65(9.8)5=3185J |
22.1 |
F |
0 |
19 110 J |
19 110-15925=3185 J |
65(9.8)25=15925J |
9.89 |
G |
0 |
19 110 J |
19 110-12740=6370 J |
65(9.8)20=12740J |
14 |
4. For each location, what do you notice about the sum of
the kinetic and potential energies? It remains constant….