ch04 - x M g k M g k 2 2 M g h k x 2 2 M g k x 2 M g h k...

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Unformatted text preview: x M g k M g k 2 2 M g h k x 2 2 M g k x 2 M g h k Solving for x M g h M g x ( ) 1 2 k x 2 so E 1 E 2 But Note: The datum for zero potential is the top of the uncompressed spring E 2 M g x ( ) 1 2 k x 2 Total mechanical energy at instant of maximum compression x E 1 M g h Total mechanical energy at initial state Apply the First Law of Thermodynamics: for the system consisting of the mass and the spring (t spring has gravitional potential energy and the spring elastic potential energy) k 400 N m h 5 m M 3 k g The given data is Solution Find: Maximum spring compression Given: Data on mass and spring A mass of 3 kg falls freely a distance of 5 m before contacting a spring attached to the ground. If the spring stiffness is 400 N/m, what is the maximum spring compression? Problem 4.2 x 3 k g 9.81 u m s 2 m 400 N u 3 kg 9.81 u m s 2 m 400 N u 2 2 3 u kg 9.81 u m s 2 5 u m m 400 N u x 0.934m Note that ignoring the loss of potential of the mass due to spring compression x gives x 2 M g h k x 0.858m Note that the deflection if the mass is dropped from immediately above the spring is x 2 M g k x 0.147m Problem 4.4 Problem 4.5 Problem 4.7 T amb 278K T amb 5 273 ( ) K T init 298K T init 25 273 ( ) K Given data for cooling where T init is the initial temperature. The available data from the coolling can now be used to ob a value for constant A T t ( ) T amb T init T amb e A t Integrating dT T T amb A dt Separating variables where M is the can mass, c T is the temperature, and T amb is the ambient temperature A k M c where dT dt A T T amb or M c dT dt k T T amb The First Law of Thermodynamics for the can (either warming or cooling) is Solution Find: How long it takes to warm up in a room Given: Data on cooling of a can of soda in a refrigerator Problem 4.8 W 1.5hr W 5.398 10 3 u s W 1 A ln T init T amb T end T amb s 1.284 10 4 ln 283 293 288 293 Hence the time is T end T amb T init T amb e A W with T end 288K T end 15 273 ( ) K T amb 293K T amb 20 273 ( ) K T init 283K T init 10 273 ( ) K Then, for the warming up process A 1.284 10 4 u s 1 A 1 W ln T init T amb T T amb 1 3 hr 1 hr 3600 s u ln 298 278 283 278 u Hence t W 10 hr when T 283K T 10 273 ( ) K Problem 4.10 Given: Data on velocity field and control volume geometry Find: Several surface integrals Solution k wdy j wdz A d 1 k dy j dz A d 1 j wdz A d 2 j dz A d 2 k b j az V k j z V 5 10 (a) dy zdz k dy j dz k j z dA V 5 10 5 10 1 (b) 5 5 5 10 1 1 2 1 1 1 1 y z dy zdz dA V A (c) zdz j dz k j z dA V 10 5 10 2 (d) zdz k j z dA V V 10 5 10 2 (e) k j k z j z zdz k j z dA V V A 25 3 . 33 25 3 100 10 5 10 1 2 1 3 1 2 2 Problem 4.11 Given: Data on velocity field and control volume geometry Find: Volume flow rate and momentum flux through shaded area Solution k dxdy j dxdz A d j by i ax V j y i x V (a)...
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ch04 - x M g k M g k 2 2 M g h k x 2 2 M g k x 2 M g h k...

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