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Unformatted text preview: Name: E Y Instructor: Louisiana State University Physics 2101, Exam 2, September 25, 2007. ° 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 reascming and your work. Note that in many of the
problems, you can do parts (b), (c) and (d) even if you get stuck on (a) or (b).  You may use scientiﬁc or graphing calculators, but you must derive and explain your
answer fully 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! Qua/m a (a) pays 3 (m We) 3 E TQZjL ML 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 correct dimer below. (b) (5 points) The spring force, when the block is displaced A? from the relaxed positionéad—is—statioaeﬁIanhe—ﬁaal—posiﬁot) a Kw) I; e k A? 3m? lJJ mam
: 3 Em) [Elm e em) (b) (6 points) Work done by the applied force to displace the block from the relaxed
position to A)? , where the block is stationary at the ﬁnal position, is; 3 mgAx “it{AK t: EW:'—Wzﬂﬁé
b. —mgAx W :: W3 i“ WT: F g
69 em)” W; rat‘s“ Problem 1 (22 points) In the following ﬁgure mass of the block is 10kg and it slides
4m down the ramp, starting from rest. 1::
mzio f
,8: m / R
03:0 /. " u? 40° (a)(11 points) What is the coefﬁcienth friction, if the kinetic energy of the block at the. bottomofthe4mramgis SOJ? M;AK1—bu+ =AK
H #1.
AK: Kgy/l = K19 r ‘96} ( , (b) (11 points) In addition, if an external force is pushing the block down and the
kinetic energy of the block at the bottom of the ramp is 100.1, ﬁnd the work doue by
the external force. (Note: Friction is not zero.) MrzSK W ¥ﬁwk+wf 2: JOB} ;@ M, 43$“; 5‘0 Question 2 (12 points) The following Figure shows a plot of potential energy U versus x of a 0.200 kg particle that can travel only along an x axis under the inﬂuence of a conservative force, where UA=5J, UB=7.5J, Uc=15J, and UD=20J. The particle is released at the point x=2.51n where U= U3, with kinetic energy 2.5 J. 1? EM: u 5 5+] 9' Us awn “ : l0} 0  4 . 2 i 5 i
L {(111) ‘ '
(a) (4 pts) The particle stops momentarily at, apgro i mately
i x22.5m and x=7.67m;
i =2.0m and x=7.5m; m Jar5W 45¢ +utrmlm? US
(iii) x=1.0m and x=8.0m;
(iv) the particle never stops.
(b) (4 pts) The particle has maximum kinetic energy at: x=6.5m
E 2': Ki" 71.
 = . W
“0" K: mm were a: M (iii) x=2.5m;
(iv) the particle 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
x=8m —;. myom “inn/rim? from ‘
“5 Krvtellic «my? uremia! bf tie3M) 4L6?) Midi
is two; b L6 Problem 2 (22 points) A breadbox of mass M on a frictionless incline surface of angle 9 is connected, by a cord that runs over a pulley, to a light spring of spring constant K, as shown
in the ﬁgure. The box is released from rest when the spring is unstrctched. Assume that the pulley
is massless and frictionless. 4)..” 0 a un— (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 x at that point, in terms of 9,M,g,d,andK. £3be :59 ,2— .‘
137/1131! +5145 am?” 4 ._. Wdﬂugzﬁ “(SM (1)) (6 pts) Write an expression, in terms of 6, M, g and K, for the distance L down the incline 4); 5 O A K: 0
that the mass M moves from its point of release to the point where the box stops momentarily. 4}“ :2 D £9), ts) Express the magnitude of the net force on the block M when it has '_,
< stopped in term of 9, M, g, L and K. ' " Question 3 (I 1 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. F (N) +2,
timesm (a) (3 pts) The impulse of the force on the object is M O iii: ‘1': 8 > (i) 128Ns.
(ii) 96Ns l” llzdL’SJ—é’ + 2XI£ + 25%): SZ+SZHE
(iii) 64Ns 7i 2 Z N . { #80 s
SONs (b) (4 pts) The momentum acquired by the object in the first four seconds (t=Os to
F45) is o (c) (4 pts) What is the object doing between t=63 and t=83? \®‘ t is moving in positive x direction and speeding up. 3F Vi; 56+er =? cunfdl‘o “—
ii Hat is moving in positive x direction and slowing down. 0;
(iii) It is moving in negative x direction and speeding up. M6
(iv) It is moving in negative x direction and slowing down. ' Moi—DOLL ﬁrm lamb“ Problem 3 (22 points) A spring, with spring constant of 120 Mm, is ﬁxed on a horizontal part
of a track that is 2.00 to higher than the rest. A block with mass m; = 0.30 kg is placed against the spring,
compressing it by x = 0.40 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 portion of the
track. The track is frictionless. Just after the collisiomthe box In; moves forward with a speed of 2.0 mfs. Hint: It is not known :fthe collision is elastic!
l 2. (a) (8 pts) galculate the speed by which the block in; leaves the spring. .1. 145me
2. k" ’3 2— % Rodouermfi‘on o’ﬂ
magemu‘agﬂww o (b) (7 pts) Calculate the speed of the block In; just before the _11ision.
'° “3 246:.ng 'ng): " Haw/s is
(c) (7 pts) Calculate the spec of the leer the collision.
279.; #279,; l) : 2sz
54
“1/973, 5* Wﬂiﬂwﬂizr muffleIsak
m ‘1}
4r : “(Nola 2 E? ’ ML” : IOJéMQt—gffz):+6o€3‘z
:4 r 9‘ —— W L' 0.3
M ’3 W ‘l
i r o  A42}: 6.9.3 1% GOﬂJEVT/wfim of ﬁrm M : Formula Sheet for LSU Physics 2101,
Test 2, September 25, Fall ’07. —b:l: Vb! — 40c Quadratic formula: for 03:2 + bx: + c = 0, 931,2 = 2a Dot Product: ('1'  3 = (tab: + (ruby + (1:53;!z = 6 cos(q5) («p is smaller angle between (1' and 5)
Cross Product: ii X 5 = (my!)z — azbyﬁ + (nzbz — ambzbi + (azby — aybzﬂE, l6 x 5‘ = It'il sin(¢) Equations of Constant Acceleration: linear equation missing
1: = on + at 93 — $0
1
m—moz'votiEatz v
1:2 = 1:: + 2a(a: — mo) 1:
1
m—mo=§(vo+v)t a.
l
m—mo=vt—§at3 9., Vector Equations of Motion for Constant Acceleration: 1?" = 1"}, + ﬁat + gift}, 13’ = 1'1}, + at Projectile Motion: . m=vo¢t y=voyt—§gt2 R=ﬂgﬁ
11¢ = on = constant '09 = you — gt Force of Friction: Static: f, S jaw = psFN, Kinetic: fk = kaN Elastic (Spring) Force: Hooke‘s Law F = —km (E: = spring (force) constant)
1 1
Kinetic Energy (nonrelativistic): Translational: K = 5va Rotational: K = 51:»: Work:
W t E  J(const. force), W = f mi Work  Kinetic Energy Theorem: W = AK = Kr — K‘ ! F(:r)da: (variable 1D force)1 W = f 1! E0")  d1?" (variable 3D force) 1,. Work done by weight (gravity close to the Earth surface): W = m 9‘  (I m! 93:; at?
Work done by spring force F =  k x: W = —k/ a: dz: = —k ~— 2‘4
Power:
W  _. dW _. _.
Average: Pang = P = F  vans (const. force) Instantaneous: P = —, P = F  1; (const. force) 3’ Potential Energy Change: AU = —W (conservative force) . . til((3)
PotentialForce Relation: = — do:
Gravitational (near Earth) Potential Energy: U(y) = mgy (at the height 3;) 1
Elastic (Spring) Potential Energy: U = 516332 (relative to the relaxed spring) Conservation of Energy: W = AK + AU + AEth + AEint, where W is the external work done on the system, and AB“, = fkd. N 1 N 1 N 1 N
Center of Mass: M = 2 mi, 93mm = — Z mime, ycom = ——~ 2 mm, 2mm = — Z mgz;
‘ M i=1 M 12—1 M 1 g 1 1i 1 g: 1 it, _.
Femn = — miﬁi ﬁcom = — miﬁi acom = — midi = — F12
M i=1 M i=1 M i=1 M i=1
_. N
Definition of Linear Momentum: one particle: 33’ = mﬁ‘, system of particles: P = 35‘; = M 6mm
i=1
d . o _, dP'
Newton’s 2“ Law for a System of Particles: Fact = M acom = 3
Conservation of Linear Momentum of an Isolated System: 215; = 215}
Impulse  Linear Momentum Theorem (one dimension): A131 = J12 = F12 (t)dt = Favg,1gAt
£1
m — m 2m 2m m  m
Elastic Collision (1 Dim): 111,: = 1—21)“ + —202,, 132; = —1—'v1,; + —2——1‘Ugi m1+m2 m1+m2 m1+m2 m1+mz . . 41:... ...
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