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Unformatted text preview: Name: }< E Y . Instructor: Louisiana State University Physics 2101, Exam 2, September 25, 2007. v Please be sure to write your name and class instructor above.
 The test consists of 3 questions (multiple choice, no partial credit), and 3 problems. ' For the problems: Show your reasoning and your work. Note that in many of the
problems, you can do parts (b), (c) and (d} even ifyou get stuck on (a) or (b).  You may use scientiﬁc or graphing calculators, but you must derive and explain your
answer fuily on paper so we can grade your work. ~ Feel free to detach, use, and keep the formula sheet pages. No other reference material
is allowed during the exam. Good Luck! Question 1 (11 points) A stationary block attached to a spring of spring constant K
is displaced Ax by an applied force as shown in the ﬁgure.
Circle the carrect answer below. (b) (5 points) The spring force, when the block is displaced A? from the relaxed position We; is;
KW) E #35: ., k A? Wit” 993 ‘5‘”) [#4 m t‘MW
°‘ 5455) 4pm %t.m (1. mg
e em (b) (6 points) Work done by the agglied force to displace the block from the relaxed
position tom? , where the block is stationary at the ﬁnal position, is; a. mgzlx W 7: AK ”’1 E W : «— :”g‘é a)
b. —mgAx W 5—“ W31" WT: L
G) Maw)” W 1‘ “til/ﬂ
d. my e Wt Problem 1 (22 points) In the following ﬁgure mass of the block 15 10kg and it slides 4121 down the ramp, starting from rest. A:
(1;; (a)(11 points) What' IS the coefﬁcientgf ﬁ'iction, if the kinetic energy ofthe biockw at the bottomofthc4mrampis SOJ‘? M; AK+AU+ AE—‘ﬁu 54¢;sz 5465;;
AK: K; )s{”= Kg 96}
AIL}:  M? ﬂan/V16 ,K =w+atﬂcLJ
)1 K: Xian? W9 @‘9 woﬁgymﬁ (b) (11 points) In addition, if an external force is pushing the block down and the
kinetic energy of the 131001: at the botiom of the ramp is 1001, ﬁnd the work done by
tha external force. (Note: Friction is not zero.)
... :: W “I" N ‘f" W :2 } ...»
wrgg W W ? 1%. F 100 9w¥, 50}
T W
:13 K == 50}
m (“0 Question 2 (12 points) The following Figure shows a plot of potential energy U versus 1: of a. 0. 200 kg particle that can travel only along an x axis under the inﬂuence
ofa conservative force, where U551, Dar—7 SJ Uc=15J, and UD=201 The particle is released at the point x=2. 5111 where 1.1— U3, with kinetic energy 2.5 5 I? E :: 14?» u 2 5+7 lo], ‘1 i
i
5.
i
i
i
i
1
E 9H!!!
(a) (4 pts) The particle stops momentarily at, apgro imately; i 1:32.5111 and x=7 67m;
@1‘ —2. ﬂmand x=7 5m; « lash/1316511 4kg «ﬁrm? {>19ng (iii) 1r] .0m and x=8.01n; (iv) the particle never stops.
(b) (4 pts) The particle has maximum kinetic energy at:
x=6.5m
E r K 'i' u
(ii) x=5m; W (iii) x=2.51n; K: KW w Mm ﬂ: NW (iv) the particie has constant kinetic energy, so it has the same
energy at all points. (d) (4 pts) The particle is never found at the following point}:
(i) x=2.5m
(ii) x=5m
(iii) x=7m
Kix=8nr1 —> beyoM *l'IU‘Ml/L? 000M
“’7 Kj'vtf’F‘LC‘ 6% WM 196 regains bites?) Midi LC {wmbk Problem 2 (22 points) A breadhox of mass M on a frictionless inch'ne Surface of angle
9 is connected, by a cord that runs Over a pulley, to a ﬁght spring of spring constant K, as shown in the ﬁgure. The box is reieased from rest when the spring is unstretched. Assume that the pulley
is massless and ﬁ'ictionless. 4),._ O
O .— (a) {6 pts) Assume the breadbox has moved a distance "d” down the incline {when the spring is
not maximally stretched). Write an expression for the speed of the axis that point, in terms of 9,M,g,d,_andK. ‘55qu ”Q .,
*1; avg; + ﬁgériz WAsWQT—ﬁ (t) (5 pts) Draw a freebody diagram showing all a forces on the box. Haitian ~ __...._.__V_ if” (WW ’9] 5:.th ﬂ N 9). ts} Express the magnitude ofthe net force on the bkock M when it has géttteﬁtaﬁif')
(stopped in term ofE), M, g, L and K. ' " ' Question 3 (11 points) A 2.0 kg object is acted upon by a force in the x direction in a manner described by the graph shown. The object is initially at rest.
Circle the correct answer below. 2468 F (N) £2,
ieéﬂ’cegm (a) (3 pts) The impulse of the force on the object is (W M O 5—1": 5': 8 ) (i) 128mg. (a) 95m. iiiZ—r— “iii—‘5 4' 2mg 4» 2&9: 32+3’Zi‘ié
(iii) 5
@ 23.3.5 {34:80 “5 (b) (4 pts) The momentum aequired by the object in the first four seconds {t=Os to F45) is 0
(a) 4m. big; 334: Ff%
.23: fgdiéaﬂzmm ;
(iv) 5414.3
(e) (4 pts} What is the object doing between t=ch and t=8s? is meving in positive x direction and speeding up. 3F m oéTiW‘é‘ =?’ #Mdlo “
13)}? is moving in positive x direction and slowing down. 0; M 032‘. M M
(iii) It is moving in negative 3: direction and speeding up. We
(iv) It is moving in negative 3: direction and slowing down. ' M ”Ohmi erd m with spring constant of 128 Mm, is ﬁxed on a horizental part
lock with mass m; = 0.30 kg is placed against the spring,
compressing it by x 2 0.440 m. The block is not attached to the spring, so after the spring is released the
block slides down the track and hits a box of mass m, = 0.50 kg that is at rest on the lower portian of the
track. The track is frictionless. Just aﬁer the collisionﬂthe box m; moves femard with a speed of 2.0 mfs. Him: If is 3201 .tmwn Ifﬂw caiﬁsion is elastic!
.1 2. Problem 3 (22 points} A spring,
of a track that is 2.00 111 higher than the rest. A b . n: . p ._ .'~...9.‘~~.=". (a) (8 p13) galcuiate the speed by which the block as, leaves the spring. 05+ 3 {KiX);
r 1. a”:
ékxi‘: ital/y o (b) (’3 p13) Calculate the speed of the b'éock m, just before the m llision.
. d ° ‘ 2 H ‘3: ems emf tee—Wk .. ,
4; “.W «42,: gnaw sews
: th,g)(z)+Lz%wsro.isk/5 egaﬂwems (c) (7 pts) Calculate the spec: of the 131ch In; Just attes the collision. Formula Sheet for LSU Physics 2101,
Test 2, September 25, Fall ’07. wbzl: Vbi — 40c 23 Dot Product: 5  5 = nab; + ayby 4— azb, ..—_ [6 if)“ cos(¢) (423 is smaller angle between if and (3)
Cross Product: if X ii =2 (aybz — tabby)? + (azbz — £13,213);“ + (azby — await, l6 X 5‘ = I5} E sin(¢) Quadratic formula: for (12:2 + bx: + c = 0, 9.11,; = Equations of Constant Acceleration: linear equation missing
2: = 1:9 «1 at a: — :co
m—mo=v0t+§at3 '9
1): = mg + 26(113 — mo) 1:
1
3—mo=—(na+v)t a
2 1
a: — mo = ‘Ut —— in. ‘ 1:0 Vector Equations of Motion for Constant Acceleration: F = 1"", + ﬁat + in“, 13° = it, + {it Projectile Motion: I 2
1 a: sin 29
35:9th yzvayt—Egtz =_°__.i._;)
9 vmavmzconstant vy=voy—gt '
Force of Friction: Static: f, S jam 2: ”SF”, Kinetic: fie = 11'an Elastic (Spring) Force: Hooke’s Law F = —k:c (k = spring (force) constant) 1 1
Kinetic Energy (nonreiativistic): 'I‘ranslational: If = 5711122 Rotational: K = 51w: Work:
W = 13"  if (coast, force), W = f it." x 1" _.
1 F(m)dm (variable 1D force), W = f f F0")  dF (miabie 32) force) r.
Work  Kinetic Energy Theorem: W = AK = Kf — K;
Work done by weight (gravity close to the Earth surface): W = m g‘  C? as; as? 3?
Work done by spring force F = » k x: W = —ltf a: the = k _2_  f) Power:
dW W .— ..
—, P : F ~ "3.299 (const. force) Instantaneous: P = —, P = F  13' (const. force) A : =
verage P099 At dt Potentia} Energy Change: AU 2 “W (conservative force) . . wk)
Potentlal—Force Reiatzon: 17(3) = — dz
Gravitational (near Earth) Potential Energy: U(y) : mgy (at the height 3:)
1 Elastic (Spring) Fatential Energy: U = 58:32 (reiative to the relaxed spring) Cansemation of Energy: W = AK + AU + A3911 + AEint, where W is the extemai work done an the
system, and AER}, = fad. Center of Mass: M = 2 mi, 3mm 2 — Z mimg, ycom m ..... E VII—3‘95, 2.30m = — E mgz;
M M m M M .31 ,=1
1 g 1 if: 1 if; 1 f: .. I
me = — mif’i goon) = — 171113} Econ: = — mifii : — F'i
M i=1 M i=1 M i=1 M i=1
_. N
Deﬁnition of Linear Momentum: one particle: 35?: m6, system of particlazz P a Z 13} = M mom
{:1
d . — _, dP
Newton’s 2n Law for a System of Partlcles: Fnet = Macom = Ti?
Conservation of Linear Momentum of an Isolated System: z 55;; = $35}
.. f2 _. _.
Impulse  Linear Momentum Theorem (0118 dimension): A331 = 313 = F13 (ﬂdt = Favgggét
31
, m — m 2m 2m m — m
Elastic Collision (I Dun): 111; = 1—291,» + “ml—.92” vzf — __3._._..1,1£ + 3 11m m1+mg m1+m2 m1+mz m1 +5112 ...
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