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Unformatted text preview: NAME (WWW/M 01m ﬁat/z LAB PARTNERS ELﬁJ! jg
Station Number “'0 3 Li? f/"i ééﬁ M . Work and Energy
Experiment 14 INTRODUCTION Work and energy are two of the most important concepts in Physics. In this experiment,
we will look at the relationship between the change in mechanical energy of a system and the
work done by a nonconservative force. The will be used to calculate the coefﬁcients of rolling and sliding friction. THEORY When we lift a box at home or we push a shopping cart in a store we do work. The work
done by a constant force is given by W = Fd cos6, where F is the magnitude of the applied force, d is the displacement of the object and (9 is the angle between F and d. Notice that work
can be positive or negative depending on the angle 6. If there is a component of the force in the
direction of the displacement the work done by that force will be positive. If the force or a
component of it acts in the opposite direction of the displacement then work done by that force
will be negative. What if the force acts perpendicular to the displacement? In most situations there is more than one force acting on an object. The total work done is
the sum of the work done by each individual force WW, = W] + W2 + W3 +  — . There is a connection between work and kinetic energy. According to the WorkEnergy theorem, the total . . . . . . l 1
work done on an ObjBCt is equal to the change in 1ts kinetic energy Wm, = AK = —mv/2v — —mv,2 . If the total work is positive, the speed of the object increases and when Wm, is negative its speed decreases. lf W = 0, then the object moves with constant speed. Iula/ When an object has a vertical displacement gravity does work on it. Here the amount of
work is related to the change in potential energy of the object bng = —AU . Using the deﬁnition
of work it can be shown that the change in potential energy is simply given by A = mg Ay
where Ay is the vertical displacement of the object. This relationship shows that an object gains potential energy as its height increases. If on the other hand the object falls down, it will lose
potential energy, AU will be negative. The mechanical energy of a system is the sum of its kinetic and potential energies,
E = K +U . If only conservative forces act on the system E is constant, in other words,
AK = —AU . If there are one or more nonconservative forces acting on the system AK and
—— AU are not equal, their difference will be equal to the work done by the nonconservative force. I) , , f ERA?“ E, yﬂé‘ {3‘}! M ‘9‘“, WT U N“ ( {pi/{3"
m m 93 it , M “A ‘ ” N t l ‘} .té ,, “432
Wm! {25? 1 : age“) EXPERIMENT NO. 14 . Measure the mass of the cart and the wooden block. Mm: Alan/Avg Mblock = (514$ 30) . Level the track by adjusting the screws on the track’s feet. . Attach the string to the cart and place it at the starting position of the track. Put the weight
hanger at the other end of the string and add 10 grams to it. This amount should be
enough to accelerate the cart. . Release the cart and click Start to record the position and velocity of the cart as it moves
along the track. Make a graph of the cart’s velocity. On the graph, click on the label of the
horizontal axis and select x(m). Your graph must show velocity as a function of position.
Don’t print the graph yet. . Replace the cart with the wooden block. You will need to add a larger amount of mass
(about 40g) to the weight hanger. Record a data set as the block slides along the track. . Print a v vs. x graph that includes both datasets. Make sure you clearly label the cart and
block data to avoid confusions (ask your TA if you do not know how). Complete the table below. Use the Smart T00] i: . on the graph’s top menu to obtain the desired values of position and velocity. Horizontal Track
Hanging mass xi
Can am: $51 ° l m Block avert): {192 .\ m
Tablel ‘ 7. As we saw in Experiment 10, there is frictional force acting on both the can and the
wooden block. For each case, calculate the change in kinetic energy, the change in potential energy and the work 99,13913y— @1991 9.13th system. Complete‘Table 2.
Show all your calculations in the space provided. Cart— Horizontal Track '2 2
NC: “‘7; Um xthB CV; “ V1) a
W: 1 1.05m Mow) t (64“?91 ( «’5 ”/s‘Y’J
up: .2 52:7 :1: WM * 60“}
:4 a O ‘2— I M in wt t m“) Block — Horizontal Trackz k M: ’L CYYV‘l NC) (Vf V?!" j x 7 K W} 21
a: 1%t.*04§\463+.\7)%3~405‘>USEY‘W’ t 5’" W
N4: (6 o%ﬁ\g§)(61§ " . \gd D Horizontal Track
Total AK Total AU Casel: W: i
Rolling_Cart  0‘23 ‘3 0 lg 3
Case2: ‘ ,. Sliding Block '0 03$ 3 ”  O‘Vf}
Table 2 Wfriction »pogj
w. 05% 8. Remove the feet on the initial side of the track. Carefully calculate the track’s angle of
inclination. Show your work. “‘ if 7 a Cart  Inclined track 2 X EV»: éigmit M\)( ‘3; tr Val ) ,, x ”5 ﬂ ﬂ:
h¥:%l£$%k~%mwﬁ(@@§mkﬁ”a(19““)j
b¥r£1w¥4¥itnlswxmﬂ§> 937W has: (mag » fag‘ghgﬁﬁ'
«A, ”3: ‘ ’ , .A , r a ‘
L mmﬂ 4: El Weigvm my MM 04: wot—61 9%)6 m3 Smf't‘ . g; 03% ,J>U\‘: m0??? ”n:.m3n+4ng3 mu» garnet
“raw“ wwn wwMMWWW_J Block Inclined track
by [M‘EMW‘E(V "viii“; g $2.,
:2, f 3 ,3
Mai pomp Maugfﬁ L "2% W32 ( ' mm 3
WC arm  WM
\Eﬁiv 00 MT NM," ﬂ .3 “Aging mﬁ ‘ a ‘
AW qu£ ﬁiéﬁ’ég 3U M313 (iﬁéréjgéqgé? 3W3; my? w . 09 5 wk:
\szowj 3 Inclined Track Total AK Total AU Casel:
R011in_ Cart
CaseZ:
Slidi11g_Bloc Table 4 k man 10 Calculate the coefﬁcient of friction for each case Remember that frictional force 18 given
by f= yN. Show your work ﬁpnio Wm 41 at? CC‘H’C . ”fl" 10053)}! 7 x7811 950,43, SJ
(Magyar/1 rm coatoﬂmvx) WW...” * \oiéijtﬁk, : H a L U§E§>W , u “:“M‘r (,myazj ")Cq gxmg“ ,1 arm” (Moll .z—a—J n9 Eganlauc (MM A n C“ OWE»)
Mamm‘n ﬂgleZﬂ??N \W‘ MOCHC a c/l, wﬂ(~10(;”3) Horizontal Inclined
Track Track cm W .omoi W .1101
Block “S: (2‘0“) H5: '38'0 QUESTIONS 1. Consider the case of the wooden block moving along the horizontal track. USe DataStudio to
plot AK — AU and W as a function of displacement. Include all three curves in the same graph. Can the AK be larger than — AU? Explain. N15” WCl Ct than: ~ [311 . no out git/111g 1:... 1101111116 21. My 11 FWW
li “% (kﬁﬁﬁwC wg11¥ 11111215Mtggr11 1m&?mﬁ
M! «111113 1211111111? "1111112115 11L gnaw {112% [M 1 , shot/H v14 {Mai meta 11} 11‘ arzaifd C2?) R?
91%: 11/1 2 Consider the case of the {9253312wa moving alon “gﬂhcmclinﬁddtxack a) How “much 15 the
total work done on the system. b) How much 15 the change 111 mechanical energy? c) Where did
that mechanical energy go? &\T0hﬁlv101k V 11 , i ’ ,1 ‘3'“ {1
initizg*.ééC9£C)L(1lg ’Q ‘ (i2 1” SEW; Mat/2111mm {1161539 """ naval“ 1:111 ...,
1112.011511—«1 03“”? § 65 Me 111011111 at f: t1 53%}? WE C 11%; 1173712” 0411116 W ~1 111.11 Work—Energy.ds 03/02/2009 02:51 PM
Graph 1 0.70  ' ~ Run 2— Block 0.60 0.20 " Work—Energy.ds 03/02/2009 03:21 PM 7 ‘ f 1 ‘
x Run5Can
f X Run #3
0.40 T L ,L  & Run #5
x 0.30 ~+~ w. m +
'3: 7 Run 3 Block
(I)
g 0
3'0 20 ‘ j ~1~
I g
0 t
0.10 l + T i
0.00 0.05 010 0.15 0.20 0.25 0.30 0.35 x(m) " WorkE nergy.ds 03/02/2009 04:17 PM & £
W as a function of
displacement 0.00 0 L 0.10 0 20 30.0 0.40
X( m ) ~.u’ NAME @221‘2/6‘11 11111114562222 PHYS l 101 Homework Experiment 14 1. DeﬁrieJ Work and state the Work Energy theorem. M1 J J 2% J J J, __ J
WW 220262 22,201 12221 222/, ”2 22:2 2 2.22 ”W128 LAX/1L 5,2522'11/“102/11, (264% C 2'22'22 “W2 222 22'
OV/X’Cll (5712216112J 21 22/25” 2,” 2,29 ”5' 2 r, 2 ”“22 “”5; 22/” M , . :2 7 ‘ “ ' ”I
I . . ‘”
'1” 23312225,)”; (1? “2/5 “ ”i2,
2. Give an example of a conservative force and a non con{servative force. AWQQXNWW 111 (12 {32’12CW0‘221‘ C DYHZ 12, ”1222222;
Amman/21212 0:“ 112m (omen/212 a: 2221/2: 2g .4; m 3 The ﬁgure below shows a system of two masses connected by a string As m falls a distance h, M accelerates up the plane The coefﬁcient of fnction between M and the
inclined plane is [I Show that the change of kinetic energy is given by AK = %(m + M )[v; — v1.2 ]. Do not assume the system starts at rest (km / (WA/(j  Z lib/2.51”:
~~2o1—'/22 m 12,2, 2. 1J2
Md? 2J {2222 1JJ J,JJ'2 , {132.21 1J2,” 1211" 2 2.22/2 2 12, 2‘21, ,2 What IS the work done by gravity on the system? 122 22.2 ’ 222 2 212212 01212222. 10221 11222221
$10“ 22', 22 22222 3 / What IS the onrk done) by friction on the system? J J _
11111:)” :21 .2 ! *2 ‘22, 2' 222221 :,  21 7‘2 ’2’ 2 ...
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 Spring '10
 LOWELLWOOD
 Physics

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