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Unformatted text preview: ENSC 2123 Exam 1 Spring 2008 (25 points) A rocket powered sled starts from rest and accelerates at: a = 30 + 2t m/sz, until
it reaches a speed of 200 m/s. At that point, it releases a parachute brake, and decelerates at: = 0.3v m/s2. a) Determine the time required to reach 200 m/s during the acceleration phase. ‘U 012%? ~5 {éohfzfolv’y ﬁéo+2£>0ié =[0/D’ .1 f 0U . L V 30é+f 54/: U/g/ % U230f+1£
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Clam/”0°45 69w 2 z —— 30 : W ' 657 b) Determine the distance required to reach 200 m/s during the acceleratiorzphase. S t: (go'év‘fljo‘l‘é 0) Determine the distance required to come to a stop in the during the deceéeration phase. = ‘ ‘V’ g i: Hal/7f d5 '2 27d
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S _O.'3/U’ 0‘2 013 Zoé Co /6 2. (25 points) Ifthe Club the golfer is using gives the ball an frfiial angle of (90 2 50 degrees,
determine the initial velocity v0 that will cause the ball to land in the hole at the base of the ﬂag pole. from (I) C); 2 .3; M0 w $0530! “7!" Q} 3 ,, (65506
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/ (25 points) The coefﬁcients of friction between the box A and the bed of the utility vehicle are us I 0.4 and pk = 0.36. The vehicle is initially traveling up the hill at a speed of 20 ft/s. The hill
angle is 6’: 15 degrees. If the driver carefully applies the brakes, determine the smallest stopping
distance for which the load will not slide on the bed. ,o
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=— ”I'MJO '60 I: M3 6059 N mycosé § éﬁ; '3 MQLX, i , \é ~35~m35m©=max g , “5% I’d/e 60/ {2 ”as/V 5' ﬂ: my€@[email protected] \h 5 ) 3 N0 1 n/ M v \Q ,/ . {9/ . w//:,nuggé w,,a§/n©:/WQK 3U ‘ f, u ecceix MS/ J 03/ ’/ d 6 51.7: /./0 Q Name /(67 4. (25 points) A boy rides a skateboard on the concrete surface of an empty drainage canal
described by the equation shown in the ﬁgure. He starts at y = 20 ft, and the magnitude of his
velocity is approximated by the equation given below the ﬁgure. If the boy weighs 140 lbs,
determine the normal force between the boy and the concrete surface when he reaches x = 15 ft.
You may ignore friction and you may ignore the size of the boyJand the skateboard (treat them as one particle). Fiﬁ/IWIHQ’V (5 720 161C N
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Engineering Science 2123
Elementary Dynamics General Deﬁnitions ds
v = —— d1 dv
a = w dz
ads : vdv Constant Acceleration 5 2 so +v0(t—t0)+%ac(t—t0)2 v 2 V0 +ac(t—t0) 112 : v3 + 26h (s — 50) Cartesian Coordinates r=xi+yj+zk
v:)'ci+yj+z’k
aziéi+j§j+2k Normal and Tangential Components Equation Sheet Cylindrical Coordinates r=rer+ze: rzfer +r6e9 +z‘e: Relative Motion (Translating Axes) r3 :rA +rB/A
VB :VA +VB/A 33 Z 3A + aB/A
Kinetics ofa Particle d (mv)
dt
2F 2 ma (constant mass) 2F: p [1+(y')2]“2 [r+(r')2]“
_ 511 udzy
y—f(X), y —dx, y —dx2
dr dzr : 6 I:— ”2
' g( )’ r de’ 0192 ...
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 Spring '08
 Russell

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