Unformatted text preview: DISCUSSION PROBLEM [7.1]: ... answer.
Both solutions correspond to positions where the kinetic
energy of the jumper is zero. Consider a scenario in DISCUSSION PROBLEM [7.1]: which the rope is a vertical spring and the jumper is a K = 0 : UE = 0 : UG = mgh = 78480J mass attached to the spring. l = 50 m h = 100m
∆l lowest point
yo 1
K = 0 : UE = k( ∆l) 2 : UG = mgyo
2
UG = 0 In part (a) of the previous problem we found two values
for the lowest point y o, i.e., y o = 25.89 m and
y o = 66.27 m. Clearly, the appropriate solution is the former. What does the latter solution correspond to? “highest” point
66.27 m
mid point of
oscillation
46.08m
lowest point
25.89 m When the spring is stretched and released, the mass will
oscillate upanddown . The solution y o = 25.89 m
corresponds to the lowest point of the oscillation; the
solution y o = 66.27 m corresponds to the highest point of
the oscillation! The midpoint is at y = 46.08 m. 1 DISCUSSION PROBLEM [7.2]:
Between 100 m < y < 50 m the jumper is in free fall
because the only net force acting on the jumper is his
weight downward. Therefore, his acceleration is g ( = 9.81m/s2 ). However, when y < 50 m an upwards
F = k∆y DISCUSSION PROBLEM [7.2]: mg force, F = k∆y , now acts on the
jumper where ∆y is the extension
of the rubber rope. So, the net
force acting downward on the Can you think of an alternative method for determining jumper is Fnet = mg − k∆y , so his acceleration is the position where the jumper’s speed is greatest? () a = g − k m ∆y ,
which decreases as he falls (and ∆y increases). When
k ∆y = g , his acceleration will be zero, i.e., he will
m () have reached his maximum speed. Then
mg 80 × 9.81
∆y =
=
= 3.92 m,
k
200
i.e., the jumper is 50 − 3.92 = 46.08 m above the ground.
(Note this is also the midpoint of his oscillation.) 2 DISCUSSION PROBLEM [7.3]: ... answer. DISCUSSION PROBLEM [7.3]: Esys = ∑ Ei = E mech + E therm + Echem.
i ∆Esys = ∆E mech + ∆E therm + ∆Echem = 0.
If she did no work (i.e., ∆Echem = 0), the work against Why does a cyclist need to keep pedaling when traveling friction, air drag, etc., ( ∆E therm) would have to come on a level road, even though she’s not changing her from somewhere ... it would come from a loss of speed (i.e., with no change in ∆K )? Isn’t the work mechanical energy (kinetic energy), since energy theorem ( W ⇒ ∆K ) violated? Explain. ∆E therm = − ∆E mech
and so she would slow down. To overcome these
“losses” and maintain constant speed, she has to do an
amount of work equal to the energy lost through friction,
air drag, etc., i.e., ∆Echem = ∆E thermal. 3 DISCUSSION PROBLEM [7.4]: ... answers. DISCUSSION PROBLEM [7.4]: The predominant forces are
gravitational ⇒ conservative
frictional ⇒ nonconservative.
Major changes of energy are:
Gravitational potential energy ⇒
kinetic energy + “friction”
Magnum XL200, Cedar Point Park, Ohio with track
• What forces are involved in a roller coaster?
• Are the forces conservative or nonconservative ?
• What changes of energy are involved? wheel bearings air (if slip)
sound and heat
“ lost”
This explains why a roller coastal can never reach the
same height as the “first” hill. 4 ...
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This note was uploaded on 07/13/2011 for the course PHY 2048 taught by Professor Guzman during the Spring '08 term at FAU.
 Spring '08
 Guzman
 Physics, Energy

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