discussion_problems7

# discussion_problems7 - DISCUSSION PROBLEM[7.1 answer Both...

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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 up-and-down . 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 mid-point 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 mid-point 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 ⇒ non-conservative. Major changes of energy are: Gravitational potential energy ⇒ kinetic energy + “friction” Magnum XL-200, Cedar Point Park, Ohio with track • What forces are involved in a roller coaster? • Are the forces conservative or non-conservative ? • 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.

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