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Unformatted text preview: Name SOHJ‘EQ‘D—é _ Problem 1. (20 points) A crate is at rest at point A when it is nudged down the incline. The distance from point A to point B is
7 m. The. distance t‘mm point B to point (.2 is also 7 m. If the coefﬁcient of kinetic friction (#3:) between the
crate and the incline is 0.30 from A to B and 0.22 from B to C, detem'tine its Speeds at points B and C. meﬂwool 4
f1
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A41. 3 ax: “$1.81 (9m $10”O.Sccs 510‘): c.5899 Lja
v51: uoz ﬂax (his) : 3035899060 = 3.
Bic C 0w = (1,8ICSin10°*0.99 (‘03 :05) = 'O.42!9’ 2
“In
536: => 1333;);8? L3 SDt/t Vca —. V32 + am (x~ x0) = ease: +a(o. 4;: mm =7 V5,“: 1.53 3%
“(18+th 9 »work energy A +0 B 2
u : AT:T‘_ITV ‘3 (maﬁa 30“)(=})—(o.3m3cog :10“)(;t):_;13muE % U3 :3=8:}' u=AT:Tc/n/° ' Mac "‘3 (#sin :51“; ?.s‘m10“)~(o¢3mgcca MN?)  4 \C. _ 2
Answer: SpeedatB 1;; “(O‘QQmC‘jCO‘a IC' —é~mL/C
A m :. .. 22],. 11’]
Answer: Speed atC _] .83 ‘5 3 UL 6 3’ *1: Name 80 {ASH 10$ Prublem 2. (30 points)
A 3kg block at point A is released from rest in the 60" position Shown and subsequently strikes the lkg cart at point B.
(a) If the coefﬁcient of restitution for the collision is c = 0.7. determine the velocities of the block and cart just after the collision.
(b) Determine the maximum displacement s of the can beyond point C. Neglect friction. (0*) _,n , V30 ﬁfﬂr ’j‘jx m9“ \‘5 N
block A Stunt): down
um" ‘ .
9'15(i'3_18CGS cc>°)=E—/T,Z¥%yitg (61,31)(i.3)(1—ms c0”) : gm:
uqz‘: L: "“i‘ Congo!" c cch' on 0(3 meme niu m u
3"? ma VA, + mg':J : rYlAVﬂz +maVs2
{3)(4.aoa) :('3)VAZ+(1)V32
E z 09 l} : UBZ‘VAL : VBZ'VAL
VAE'L/r'1_ O 4.,309 d— _
.. c  2
(0.71 M4409): ﬁzzVA; 3.15431—355m 3c. : 3(5oﬁel
Sch/e $ch VAZ 4— v n Solve L2,) 5,: aas m
‘ $01K S
m
Va; 2 .2. 4; ‘5 —»
V32“: 5.3% LE —.»
. m . “(g m
Answers: Velomty of block 0?  4 I») S Velomty of cart f) " 5
Answer: Maximum displacement 9' 3 m Name SCiL‘xTioﬁé Problem 3. (35 points)
A 2kg block of wood with an initial velocity of 15 me’s slides along a surface for 10 at before reaching a
cliff. The coefﬁcient of kinetic friction between the surface and the block is to; = 0.35. (a) Find the velocity v] of the block at the edge of the cliff. (5 points) (b) Calculate the distance d from the edge n’r'the cliff to where the block hits the ground. (10 points) (c) Calculate the velocity v; of the block when it hits the ground. (10 points) (d) A platform supported by a spring constrained to move only in the vertical direction is placed at the
distance d calculated in part (b) even with the ground. The block lands on the platform at the speed
calculated in part (c) and strikes and maintains contact with the platform. The acceleration of the
body after impact is a = g—cy, where c is a positive constant and y is the displacement measured
from the original platform position. If the maximum compression of the spring is observed to he
ym, determine an expression for the constant c in terms of ym and v0. Assume v0 is the vertical
component of velocity v; that you calculated in part (c). Assume that the mass of the platform is negligible. (10 points) “"
__ W‘IMC’J . . Y .
(at kBD .  L
es" Pasca Tatuo_1= ti épy:l’h \f
«nus:5 KNﬂO :* U1 —m3+;\;=0
Misti—Z at}; frtiw" Ngmrd = 90%”: W02 N
.  to  . . ‘ I Fﬁucr = 03506.:22): (5.36:?UN!
vi  ‘3503 De 9 h——.+
. d
(b = (+6016 X— X1 ' '
) Y O) abbzcoszes rgﬁagfrlhﬂj Ufl m
Vt: :3": :13 5—53 9% Yaw “ByteEeit 6‘0. «a0 f; Rafﬂeta )7 calculated in part (c) forvn
X;~IOO o::oo+o+€5)tﬂ)éz _ 41/0
7 Z {1:4,er sec ._
too:‘l£i d ) VXZL’OX+%?f*Cc) g l
a. 12.933 V‘xTUay : la.S'C'ES M —»
oi: bot 4S4 m T 5’ VHF“? HM (f—éc)
V?! 2 0‘3 (4.5lf—Ci)
Vy: MiaMag. 4r : ~443¢H 9% at: bﬁ‘(RSCE)(4.S‘:S‘): sets m
_._‘3
VA; MATC i‘— 4.40;;93‘ m 6 Name I I . a Problem 3. Continued (workspace) if 2%" M
I T  ow Y 9:0 \f m 5:1 mm: n. v:c V0 0
(Chaneln: diféCf‘fonS) w “(=0 "
 _I_ ~ _ _ I 2 7m
J J aV ) _ 9y 30,!
Va 0
"—l ' :.  .. I 1
3‘0 3?“ 3C)?!”
[ '2.
:iCYm L jymiévo—‘L
0" Ym"
. "~ m
Answer: Part(a) velocity Vloftheblock IQ56‘3’ ‘3
Answer: Part (b) distanced r‘ 4 m
Answer: Part(c)velociryv;0ftheblock IQ: ?—44.9~43 ID
\ 3 Sky.“ +4," 2 2
Answer: Pan ((1) expression for oomlzmt c ( 'J o J m Name SO‘U I 11%. Problem 4. ( 15 points total)
(a) At an instant, given the position, velocity and acceleration vectors shown below. accurately sketch
and label the normal and tangential components of the acceleration. (2 points) 3 I
1/ a l
\ (b) At the instant shown above, is the speed of the particle increasing, decreasing, or staying the
same? (2 points) Answer: .l I"! C (8013"! ('1 C! (c) The car is rounding a curve with a radius of curvature p at a constant speed. The resultantforce
acting on the car is in the direction of which arrow shown? (write A, B, C, D, E, or F). (2 points) AHSWer: (d) A force P is applied in two separate impulses during a 2—second period as recorded by the instrumentation, which measures the force as shown on the graph. Calculate the value of the total
linear impulse. (2 points) genteel — ézosxs) : 5. 95
n. . . 31(— CLreot C
ﬁlitzknéjld .1. 44mm 2 Answer: 5: 3L5 ' 3 Name 80 ILL“ ot’l ~$ (e) Suppose you are dcsigning a rollercoaster track that will take cars through a vertical ioop of 40—ft
radius. If you decide that for safety the dovmward force exerted on a passenger by his Seat at the
top of the loop should be at least onehalf the passenger’s weight, what is the minimum safe velocity of the cars at the top of the loop? (4 points) (oar +013 .243 if
00am: leap) N W19 éFﬁZW‘Qn N _.‘: =sz?“ I _. ,1
+m9 mom 1 E ﬁ/zgm/g—M'Qt LuanfﬂUzim [513( :L2
‘9 g vr/ZI=5X3;.;)(4C)1
Answer: lg: 44 3.: _ h a
V: 43.0511. 342. C (D At the instant depicted, aSSume that the particle, which moves on a curved path, has the velocity v
and acceleration :1 indicated. Accurater sketch and label the polar components of the velocity (3
points). ...
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This note was uploaded on 04/08/2010 for the course MECHENG 240 taught by Professor Perkins during the Fall '09 term at University of Michigan.
 Fall '09
 PERKINS

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