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PRACTICE PROBLEMS WORK (CHAPTER 11)
OLD EDITION NEW EDITION
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I l-8 Chapter I I
11.24. Visugltse The system loses l(X)O J of potential en€rgy. In the process of losing this energy" lt do€s500 J of work on fhe envhonment, which means lV*, = -500 J. Since the thermal efrbrgy incfoases l(X) J, we haveAE* = IQQ J. This is shown in the energy bar chart.
al
Ww
(J)
Kt + Ut. * W""t = Kf + Ur + AE6,
Energy (I)+l K +
lfi)ol : *--Ysq. =,--51. * -ur * -4&-
' -:
+ +fE
500
0
-500
-1000
Energy
+
11.25. Visuallze:
600,f00
zffi
0
Note that the conservation ofenergy equationKt+Ut+W*=Kt+Ur+LE6
requires that W* be equal to +400 J.
t1,26. Solve Please refer to Figure Exl 1.26. The ene{gy eonservation euuatioa yieldsl | , +Ut+Wen=Kr+II , +AE; +4J+1J+Iy€,t =! I+2J+lJ-rW* =-1 J
Thus, the work done to the envirounent is -l J. In other words, I J of energy is transferred from the sy$emenvironment. This is shown in the energy bar chart.
Energy (I)
Rt + Ui +W"o= Kf + Ur + A.E64
3a
I
0
- l
a
w. =.w.w.kw:&
Work l1-9
fi.n, Visualize: The tension of 20 N in the cable is an extemal force that does work on the block I7*, =
(20 I{X2.0 m) = 40 J, increasing the gravitational potential energy of the block. We placed the origin of our coordinatesystern on the initial resting position of the block, so we have 4 = 0 J and U, = mglr = (1.02 kSXS.t m/s2 X2.0 m) =
20 J. Also, K, = 0 J, and AE*= 0 J. The energy bar chart shows the energy fransfen and transformations.Energy (J)
Ki
Af
w
+U.r+l%*t=Kr+Ut+M,n
Solve: The conservation ofenergy equation is
Ki+IJ,+W* = Kr+(Jr+A.8, + 0 J+0J+40 J = )*r i +20 J+0 JL
=+ vr = JQ\ J)QY L0' k8 = 6.26 mls
11.28. Model: Model the elevator as a particle, and apply the conservation of energy equation.Solve: The tension in the cable does work on the elevator to lift it. Because the cable is pulled by the motor, wesay that the motor does the work of lifting the elevator.(a)Theenergyconsewation equationis K, +U1+Wot = K, +Ut + AE,,.Using6=0J,Kr=0J,and/E*=0Jgives
Wu, = ((J1 -U) = mg(h -/i) = (1000 kg)(9.8 m/s2Xt00 m) = 9.8x 105 J
(b) The power required to give the elevator this much energy in a time of 50 s is
p =wu*, - 9 '8 x 105 J = 1.96 x loa w
Lt 50s
Assess: Since I horsepower (hp) is 7a6 W, the power of the motor is 26 hp. This is a reasonable amount of powerto lift a mass of 1000 kg to a height of 100 m in 50 s.
11.29, Model: Model the steel block as a particle subject to the force of kinetic friction and use the energyconservation equation.Visualize:
l1'nownvlt = Yo" -- lJ mts. t0=Y0=0mfo=0s]r=0s t t=3'0s
Ft = 0'6
Findxt:.t'o =tr'x .rg, /0" rs
vor
Before
Solve: (a) The work done on the block is l7"o - d", 'Ay' wheredisplacement using kinematic equations and the force using Newton'sthex-direction is
After
Ai is the displacement. We will find thesecond law of motion. The displacement in
v
' At-rr =x0 +vo.*( l , - ro)+la.( t , - t6)2 =0m+(1.0m/s)(3.0s-0 s)+0m=3.0mz
Thus Ai = 3.01'm.
ll-10 Chapter 1I
The equations for Newton's second law along the.r and y components are
(4" ), = n - w =0 N + n = w = mg = (lO kg)(9.S m/s2 ) = 98.0 N
(4",), = F - n= 0 N =+ F = f* = 1t*n = (0.6)(98'0 N) = 58.8 N
= W"* = i* .li = FAxcos0o = (58.8 NX3.0 mxl) = 176.4 J
(b) The power required to do this much work in 3.0 s is
p =Y -176'4 J = 58.8 w
t 3.0s
11.30. Solve: The power of the solar collector is the solar energy collected divided by time. The intensity of fusolar energy striking the earth is the power divided by area. We have
^ LE l50xl06 J . . - - - - - . .^^^ wp = ---=- = 41,667 W and intensity=1000-;
Lt 3600 s m'
+ Area of solar collector ' 4l'667 w
- = 41.7 m2100OW/m'
11.31. Solve: The night light consumes more energy than the hair dryer. The calculations are
1.2 kW x10 min = 1.2x103 x 10 x60 I =7.2x105 J
l0Wx24hours = 10x24x60x60J = 8.64x105 J
11.32. Solve: (a) A kilowatt hour is a kilowatt multiplied by 3600 seconds. It has the dimensions of energy.(b) One kilowatt hour is energy
1 kwh = (1000 J/sX3600 s) = 3'6 x 106 J
Thus
500 kwh =
11.33. Model: Model the sprinter as a particle, and use the constant-acceleration kinematic equations and tbedefinition of power in terms of velocity.Visualize:
xgttg xPt lvor= 0 vl*
Solve: (a) We can find the acceleration from the kinematic equations and the horizontal force from Newton'ssecond law, We have
t^ lx - xo + vo*( t , - t )+ )a,G, - t ) ' = 50 m = 0 m+0 m+ |a,QIs-0 s)2
- a, =2.04m1s2
t F, = t/ta, =(50 kgX2.0 -lr'i=
1g2 P
O) We obtain the sprinter's power output by using P = F -i, where i is the sprinter's velocity. At t = 2 s the power is
P = (F,)fvo, + ct,(t - t)l= (102 NXO m/s + (2.04 m/s' X2.0 s - 0 s)l = 416 W
The power at t = 4 s is 832 W, and at t = 6 s the power is 1248 W.
11.34. Model: Use the definition of work for a constant force F, W = F . A3, where Ai is the displacement.
Visualize: Please refer to Figure P11.34. The force F = tei + Sj) N on the particle is a constant.
Solve: (a) l7*o =WeslWu, = F.(A.!)AB +F'1631uo
= (6i + 0j) N . (3i) m + (6i + sjl N. t+ji m = 18 J + 32 J = 50 J
(sook'h{T#)=,.**,0n,
(b) wocD = lryAc + IVCD = F' (43)Ac + F' (Ai)cD
= (6i +6i) N' (4i) m + (6i + 8j) N (3i) m = 32 J + 18 r = s0 r
(c) w* = F.ral t - =toi+0i t r ' r3 l++j lm=tsJ+32 J=s9J
1.lr" to# i, "on"".uative
because the work done is independent of the path'
11.35. Moiletl The force is conservative' so it has a potential energy'
iiJiit*i--pi""t" ,"i"t to FiSure Pl1'35 for the graph of the force' -. -
Solve: The definition of potenti'I "";
it At j-W<.i
-.O 3 "AAlti' work is the area under the force-velsus-
displacement graph. Thus tU = U, -ij'= -1area under the force curve) Since U' = 0 at 't = 0 m' the potential
enetgy at position .r i, uOf = <"'* onii' t'"'ltt* "t-"
ttrn o to r) From 0 m to 3 m' the area increases lineady
from 0 N m to -60 N m, so u increas"s?o^bi to eOr' ett= 4 nu dre area is -?0 J' Thus U= 70 J at.r=4 m' and
U doesn't change after ttat since tne iorcil' ttt"n ""'o'
Between 3 m and 4 m' where F changes lineady' U must
have a quadratic dependence on r (i'e', the potential energy curve is a parabola)' This information is shown on *re
potential energY graPh below u(n
Work 11- l I
, (m)
PleaserefertoFigufePll.36.Wewillf indtheslopeinthefollowiug'regions:0cm<x< I cm, 1<.x<3 cm'
012345
(b) Mechanicat energy is E=K+ U Fmm the craptr' U=20J atx= | m'
Thekineticenergvis1(=+nltr=t(0'l0k'g)'(25nVs)'?=31 25 J Thus E= 5l '25 J'
t"f ilr" ttrtf *".gi-fit" at 5i 25 J is shown on the graph above'
(d) The tuming point occurs where ilt"tJ;tCy Ii* "'*'o
the loteniial enersv cuwe' Itye can se€ from the
sraDh that this is at apFoximately z'i t' fot u to? u"curate value' the potential enirgy function is U = 20x J' The
ifii.. "t*r* "
,fte point wtrere 20* = 5l '25' sNch is "r = 2 56 m'
113f. Model: use the relationship between force and potential energy and the work-kinetic energy theorem'
Visualize! F (N)
3<r<5 cm, 5 <r<7cm, and 7 <r< 8 cm'
ir]"", "
t"ib, it ,rt" nugutiu" .lop" oith"
--versus-x
graph' for example' for 0 m< r<2m
dU _ 4J __461N=F,=+,t00Ndx 0.01 m
Calculating the values of F, in this way, we can draw the force-versus-position graph as shown'
(b) Since W = j:l 4 dr = area of the Flersus-x graph between 4 and xn the work done by the force as the particle
ffiil3"T,L#I'?J#"fi"iu?"*",*',=r,+uuwecanseeftomthegraphthatu'=0Ian<l 4=2Iin
;;;;;;;;t "m to.r =-6 cm rhe finat speed is v' = l0 m/s' so
2 J + +(0'010 kg)(10 0 m/s)'? =0J+*(0010kg)v''+v' = 22'4 mls
t1-14 Chapter 11
/o =0myl = l0ma = 1.0 fir/s2a = 100O kB
Find
Solve: (a) The work done by gravity on the elevator is
W" = - LU, = m,lo - mTt t = -mg(yl - y) = -(1000 kg)(9.8 m/s'? )(10 m) = -9.8 x 104 J
(b) The work done by the tension in the cable on the elevator is
Wr = T(Ay)cosO' = I(vr - Yo) = ?(10 n)
To find rwe write Newbon's s€cond law for 0re elevator:
),f, = f - * = r", = T = w + tM, = n k + 4y) = (1000 kgx9.8 m/s' + 1.0 m/s'z)
=1.08x10a N+ wr =(1.08x105 N11t0 m; = 1.93;a 1gs I
(c) The work-kinetic energy theorem is
W* =Ws+Wr = LK = Kr - X, = X, - j^r'" a K, = W, + 1t1t, a !^u3
+ Kr = (-9.8 x 104 J) + (1.08 x 105 J) + +(loo0 kgxo m/s)'z = 1.o x 10a Jz
l - l(d) &=-nt = l .Oxloa J =;(1000 ks)v;
- \=4.47 mts
11.40. Model Model the rock as a particle, and apply the work-kinetic energy theorem.Vicualize: L7
f r=0m t l= lm
' to=otE/s , l=30n/8
Solve: (a) The work done by Bob on the rock is
w* = LK = !nv: -!-ri, = !^ri = l1o.so t glro.r s)2 =225 12' 2 " 2 2 '
(b) For a constant force, W*o = F*oA.r + F"o = Ws.r /Ar = 225 N,(c) Bob's power output is PB.b = 4.byd. and qdll b€ a maximum when the rock has maximum sp€ed, This is jusr
as he releases the rock with v** = v, = l$ pts. Thus, P* = {*v, = 675Q W = 6.75 kW.
11-16 Chapter 11
where we have used Ar = hlsin2tr = 146 rn Now we can compute
vr =112(9.8m/s')(50 m)+{} = 15'7 m/st /) xg
(b) We will use a tilted coordinate system, with the -r-axis parallel to the slope. Newton's second law for Sam is
(R-). wsir20"- f,- cos20' ngsin20'- (* cos20'
^mmln
_ (75 kgx9.8 m/sr)sin20" - (200 N)cos2o' = 0.8rt6 m/s,7sw
Now we can use constant-acceleration kinematics as foUows:
Knowan=l0kgl=30N
fu=0rn, l r=3mt vl = o rn/s
Find
We define the.r-axis along the floor and the y-axis perpendicular to the floor.Solve: We need to first determine ,. Newton's second law in the y-direction is n + ?sin30o = rv -
mg s n =
rg - rsin30' = (lo kg)(9.8 mi s') - (30 N)(sin30") = 83.0 N. using n and the model of kin€tic fiction' fx= ltrn=(0.2X83'0 N) = 16'60 N. Thus,
Lt/"a = n. Ai+ i.Ai = (i )(Ax)(cos180') + (rx&)(cos30')= (16.60 N)(3.0 m)(-1)+ (30.0 N)(3.0 m)(0.8660) = 28.14 J
The other forces i and fr make an angle of 90" with Ai and do zero work, we can now use the work-kinetic
energy theorem to find the final velocity as follows:I
t f f*= K,- K,= y, -ot =lnvl -, ,
= ^ffr j i
=Jz.:r, ' , l l1l i ldfu =z.gt mrs
Ass€ss; A spe€d of 2.37 m/s or 5.3 mph is reasonable for the present problem.
11.44. Model: Assume an ideal spring that obeys Hooke's law. Model the box as a particle and use th€ modelqf kinetic ftiction.Visualize:
v! =.trt,+2a(x, - x) =2ax, + vr= lEA = f@"846;/-1G4'6.) = 15.7 m/s
Asscss: We used a vertical yaxis for energy analysis, mtherdetermined by its vertical position.
11.43. Modell Model Paul and the mat as a panicle, assumeftiction, and apply the work-kinetic etrergy theorem.Visualize: y
,b .t"= 0 'bvl'
x2't2t=
Solv€: When the hodzontal surface is frictionless, conservation of energy meansr ^ I ^ | -_!Hx^-x^t , =!nv?,=K, e 6, = l1to0 N/mX0.20 m - 0 m) ' = 2.0 J
'22
than a tilted coordinate system, because % is
the mat to be massless, use the model of kinetic
Work 11-15
11.41, Model: Model the crate as a padicle,
Visualizer
and use the work-kinetic energy theorem.
v
x0:rb=0 r! = h/sin0 = 5.85 m
m =5kg pr=0 Fp*1=25N
Solve: (a) The work-kinetic energy theore^ i" 6y = lmv! - infi = lntl = Wt"ra. Ttuee forces act on the box'
so W.,,, = W*" + Itl" + W",r,. The normal force is perpendicular to the motion, so lry. = 0 J. The other two forces do
lhe following amount of work:
w*n = fun -ai = 4*nArcos20" = 137.4 J wg* =It'Ai= w,Lx = (-ngsrn2o')A x = -98'0 J
Thus, L(*, = 39.4 1, l"ading to a speed at the top of the ramp equal to
(b) The.r-component of Newion's second law is
(F \ E"$cos2oo-vr'sin2oo _ 4*i cos20" - mgsin2o" = L34.1 nlsza,=a=::B:t '=J
m m
Constant-acceleration kinematics with 4 = h lsin2o" = 5.85 m gives the final speed
v,2 =vf ,+Za1xr-x)=2axrsvr=Jr*, = 16(134? r , tdx5.85 t") = 3.97 m/s
I 1.42. Model: Model Sam strapped with skis as a particle, and apply the law of conservation of energy.
Visualize: Aris for energy analysis y
-- 20"- \>
F-.pel""- 160p
I,I/ = F*ind ' Ar'= F*ina Ar cos 160'
Solver (a) The conservation of energy equation is
rr +%, + aEo\ =ro+%o+l l /* t
The snow is ftictionless, so AE = Q J, However, the wind is an extemal force doing work on Sam as he moves
down the hill. Thus,
W*, = W*,n =(& +%,)-(K. +%")
I t , \ i r t " \ / l __\ ^ t=l1nvi +na,
)- l ;^ , i , +.8y"
)=\ :nvi +o r
)-(o t +r l .svo\ =:mvi - mgvo
We compute the work done by the wind as follows:
W ."d = 4,"d . Ai = 4idArcos160o = (200 N)(1'16 m)cos160'= 11,4N I
)o=50m
Ar= xl -- /t/sin2o'= 146 m
rt=146m.yt=0,v|
* r, = ,lrrro *!*
Work 1r-l'l
Thatis,theboxislaunchedwith2oJofkineticenergyltwil l lose2'0Jofkineticenergyontheroughsurface'The work-kinetic energy theorem is
w*,=-h'Li=K"-4=0J -2 '01= 1'o t
+ /*(x, - x, )(cos 18 Oo) = -Pkmg(xz - x)= 1'O J
2.O J 2-O Ia (xz - x) =
r f f i=,ortoj 6,r f i^5 = 54'4 cm
Assessr Because the force of ftictiol does negative work on the box, energy is trausferred out of tlF box into the
environment. In response, the box slows down and comes to rest'
11.45. Model: Model the suitcase as a particle' use the model of kinetic friction' and use the work-kinetic
eDergy theorem.Visualize: Y
la7>
&=0mBelore After
The only force that does-work on the suitcase is the force of kinetic ftiction' i The forces fr ar.d fr are
Derpoldicutar to the displacement, and thercfore do not do work on the suitcase'
3olve: The work-kinetic energy theorem rs
r-r I ' I
w^^=M=!m', i - l^r i - i* L i=OJ-tnv'o + ( i X-rr - r0 )cosl80" = --mv;
* -ur*a, - ̂ t = -|."i' p^= ffi = ur, -ffiffi , -l
= o orz
Assess: Friction does negative wofk on the suitcase, ajd thus transfers energy out of the suitcase. In response, the
suitcase slows down and comes to rest'
11.46. Model: Identiff the truck and the loose gravel as the system' We need the gravel inside the system
because friction incre".", tt "
t"-p"rlirJ .iii" *"f ""a
,n" gravel. we will also use the model of kinetic friction
and the conservation of energy equation'
Vizualizet
n = 15,000 kg
Pk=o.qvo' = 35 n/s
tu=0myo=0myl =xl si'l 6o
rrno
0-r
We place the origin of our coordinate system at the base of the ramp itl such a wav that the -r-axis is along tle ramp
and the v-axis is vertical so that we "#:;i;;il;#J "o"tgy
rn" r*t-u"ay diagram of forces on the truck is
3.*# *" "onr"*ation
of energy equation is K' + urr + AEn = Ko + ur6 + w*, ' In the present case' w* = 0 J'
vr, = 0 rn/s, Ud = 0 J, vo' = 35 rnls The themal energy qeated by friction is
AE* = (/* )(-r, -.ro) = Grkn)(t! - ro) = Fk(ngcos6'0")(tr - to)
= (0.40)(15,000 kg)(9'8 m/s')(cos6'0")(r' - xo) = (58'478 J/m)(x' - xo)
n= t5kA
Work 1l-19
11.48. Model: Model the two blocks as patticles The two blocks make our system'
Visualize: (vi):=om/s (vf)3
We place the origin of our coordinate svstem at the locatioD of the 3'0 kg block'
Sotve; (a) rhe conr"t"o* o, *orv'|i'ul;; ;i;;4 ; oE'' = xi +ua+w-'' using oE*= 0 J and w- =0 J
we get
:m,p,;?, + l^,(,,tl
+ n,g{v; = f,m,o,t',
+ L ^,{u')!
+ n's(t')
Notitrg that (rr)z = (vr)r = !r and (v,), = (v,)r = 0 m/s' this becomes
|{^,* -)u? = -m2g(vr - Y)
2(2.0 kg)(9.8 m/s'Xl.50 m) = 3.43 m/s(2.0 kg + 3.0 kg)
(b) We will use the same energy consewation equation' However' dris time
^.Ed, = ff,XAt) = (pknxAr) = Fk(ltr3g)(Ax) = (0 15X3'O kg)(9'8 m/s'X1'50 m) = 6'615 J
The energy conservation equation is now
|^,t *|^*i +m,svr +6.6!s I =|-"<',1i+f,^,<v')l + n'gv' +0 J- mr(\ l ; + Zm'2" ' t ' ,2
!1^,+,,)ul+6.615 J = ',,s(y, -)r)+ vr = J(tfi;) ',t" '
-r, '-w;.,-,,^I \ ,nr+mr, ,
Assess: A reduced speed when ftiction is present compared to when there is no ftiction is reasonable'
11.49. Modetr Use the particte model' the definition of work * = i .N'
and the model of kinetic ftiction'
iiJJr^, w" pf""" O" "oo,ainut"
r'"t" on the incline so that its r-axis is along the incline'
Knolvnrr = 8.0 kgr= 120 N
*r= o.25xo=0mtl=50m
. r=0 m, y=0m
(v)2= (td3 = vf
yr-yt=Ay= -1.50 m
|. ' I,t.o *rxr.8 m/s'?xt.s0 m)-6.615 Jl = 3 02 m/sIs.0kgf
11.51. Modelrenergy theorem.Visualize:
After a
i ^'ln=50kg
Wind F .n6 = 4.0 N
yl = rlsin3o" =-0 05 m
Solve: (a) The work-kinetic energy theorem isr l
AK = !mr? -:nvl =W-,^*1ry,n
There is no kinetic friction atong rrer direction'of moion. static friction acts to prevent hT ilrllles from slipping
,'i;;;;;r';; tt" i"", tut this forci is perpendicular to the motion and does no work: I4,t" = 0 J. The angle between
i'o ancl Ai is I = 135', so
W .* = 4"d Ai = 4,*Avcos135'= (4 N)(100 m)cos135'= -282'8 J
Thus, her final speed is
, ,= l , i ** =2-t6mts
(lr) If the skates don't slip, she has no acceleration in the t-direction and so (F".r)' = 0 N' That is:
i - 4hd cos45" = 0 N -
I = 4hd cos45" = 2'83 N
Nowthereisanupperlimittothestaticfriction:I<(i).*=llsm8'Tonotsliprcquires
, , t / "= 283N = = o.oo:s'" - mg (50kgx9.8 m/s')
Thus, the minimum value of A is 0.0058.
.lt"o"t t "
rvo* aone by G wind on the ice skater is negative' because the wind slows the skater down'
11.52.Model:Modeltheicecub€asapart ic le, thespr ingasanideal thatobeysHooke's law,andthelawofconservation of energY.Visualize: t
Original endof spring
25N/m \f n = 0.050 kg
\ 6'rl =-0 lo
3Oo t =o
Solver (a) The normal force does no work and the slope is frictionless' so mechanical,:i:tg^::.:"1t:::*-Yi^l:Jorvc; ta' a tilted s-axis to measure distance alongLu\an t*o t"p-u," *es: a vertical y-axis to $eas:re pgtentil energy anj
il;i;. roih;;;; '*'" o.ieii.'.r'i"r' i! at.the poTt wtlere.ft:.'p'lc l:t::yf::*: llT"T.:"'l:5"i
*: :i[d?]::"'ffi.""1il, ;il;,;';; ;;;i; ;'"k* il "tu'ii"'potJnti'r "o"rs]' simprv u = ] &(s - so )' =
|ts2,Becauseenergyisconsewed,wecanrelatetheinit iatpoint-withthesprinscompressed-tothefinalpointwhere the ice cube is at maximum heighr we do not need io nna tlt" speed with which it leaves the spring we
haveK, * IJ
"z -tue = Kr+U"rrU"r
1 . 1 '2 | 2 1 '2- mi + ngY' + = ksf = = mv; + mgYt + - lc1:, z z 2 '
Work tl-21
use the particle model for the ice skater, the model of kineticy'static ftiction, and the wort-kinetic
yl= 100 m, rt
,
Before I ] Yo=0m.uo=qmls
1l-22 Chapter 11
It is important to noto that at the final point, when the ice cube is at y2, tlrc end of the spring is only at s6. The spring
does ao, stretch to J2, so U"2 is zrt {fts]. Three of the terms are zero, leaving
l . "mgy2=+ng\+;ts: + y, -yr = A) = hei9ht Eained = -:]- = 0'255 m = 25'5 cm
The distance traveled is As = A/sin 30" = 51.0 cm.
@) Using the energy equation and the model for kinetic ftiction:
K2 + u.2 + Il"z + aE6 = Krlurr*us*w",t AEr = /.Ar's = p*zAs
From the ftee-body diagram,
(F*), = o m = n - tgcos30o + n = rzgcos3O"
Now, having fouud AEi, = lr(mgcos30')As, the energy equation can be n'ritten
o t +.ngy2 + O J + tt*(tzgcos30")As = 0 J + ngy' + f t{'tf + 0 f
=*g1y"-yJ-lr*f + P,rrgcos3o"As = 0
Using Ay = (As)sin30o, the above equation simplifies to
mgArsin30" + lkrtgqqs3g"Ar = 1/<s; e Ar=
11.53. Modell Assume an ideal spring, so Hooke's law is obeyed, Treat the box as a particle and apply the
energy conservation law. Box, spring, and the ground make our system, and we also use the model of kineticftiction.Vlsualize: We place the origin of the coordinate systeur on the ground direcdy below the box's starting position.
ir,yl ,2,y2 4,y31't
'2 V3
Solve: (a) The energy conservation equation is
, i=0m t0=0rD/s,2-422.On Pn=0,25/o=5'0m,l=)2=}3=0m v3=0m/st=500N/m
' ' l=5.0kg
r Fitrdvt 12 ,t- 12
ks? = O.379 m=37.9 cm2rzg(sin3Oo + p* cos30o)
rr + ug1 +u"r +aEd, = (, +uro +u,, + w"*
L^r l+-gy,+0J+0J=1rzv3+ ngyo+oI+oJ + imvl
+o! = 0l + mgyo
+ ,, = .,lzgy, =.,12(9.8m/s'X5 m) = 9.90 m/s
(b) The wo* of friction creates thermal energy. The energy conservation equation for this part of the problem is
K2 + IJ e1 + IJ"2 + M6 = K, * I ) "1 i -u "1
|w"n lnv ' r+ot+ot+(f ' \ (xr-xr)- f ,ml +oJ+oJ+or
L-rl + pun1r, - r,l = )^'i
= |^t1
+ p,(ms')(x, - x) = L6vl
+ v, = llvl -2p*g(x, - x,1 =./19.90 m/sf - 2(0.25X9.8 m/s'?)(2.0 m) = 9.39 m/s
Vrork tl-23
(c) To find how much the spring is compressed, we apply tlre energy consewation once again:
K3 + U!, + U,r + AE6 = K, * lJ rz + lJ,z + w-t 0 J +0 J + 1&1x, - x zr2 + o r = Lnv: + OJ + O J +o J
Using r, = 9.39 n/s, & = 500 N/rn and m = 5.0 kg, the above "q,rutiln
rrurd, 1t - rr; =-4. = 93.9 6-.(d) The initial eterqy = mgyo = (5.0 kg)(9.8 mi s')(5.0 m) = 245 J. The energy transformed to tlennal energydu{ing each passage is
fk?z - x) = p*mg(.t, - 4) = (0.25)(5.0 kg)(9.8 m/s'X2.0 m) = 24.5 J
The number of passages is equal to 245 J /U5 J or lO.
11.54. Morle} Assume an ideal spring, so Hooke's law is obeyed. Treat the physics student as a particle andapply Ore law of conservation of energy. Our system is comprised of the spring, the student, and the ground. Wealso use the model ofkinetic ftiction.Visualize: We place the origin of the coordinate system on the ground direcdy below the end of the compressedspdng that is in contact with the student.
rb=0m v0=0tr/str-tu=0.50m t= 80,000 N/my0=yr= l0m ,.= 100 kg
trrt=0,15 Yz=0m/s
Flndrr tu =y2lsin 306
Solve (a) The energy consewation equatior isK, +U4 *Ua + AE, = Ko + Uro + U"q + W*r
!^ri * ̂ gy,* +k(.:r, -x")'z +0 J = !-ui *
^syo + !k(.r, -xo)':+0J
Since y, =yo= l0 m, r, =.r", vo=$I/s, /<= 80,000 N/m, m = !00 kg, and (.rr -q) =0.5 m,
!^r? =l tT2
-k(xt - xo)' +v, =.,/:(.t, -ro)=l4.l4rn/s
(b) The wodc of friction creates themal energy. Applying the conservation of energy equation once again;
K2 + u 22 + U"2 + Mt = Ko + uso + 40 + Iv*,
Lt.:nvl + mgy, +O I + /*As = 0 J + nr8?o + ;k(xt -
x)2 +0 J
With v, = 0 m/s and yz = (l'r)sin3f, the above equation is sirnplified to
rn8(Ar)sin 30" + An As = mgyo + ik(x,
- xo\'
From the free-body diagram for the physics student, we see that n = lrcos3o". Thus, the conservation of energyequation gives
Ar(mgsin30" + Atrgcos30) = pgyo .r- lt(r, - ro)'
Using m = l0O kg, /t = 80,m0 N/m, (4 - rb) = 0.50 m, yo = 10 m, and 4-= 9.15, *" U",t - .
mgyo+-R(4-xol-Ar=
,rg(sin3o'+ & cos30')--32.1m
Wotk rt-23
(c) To find how much the spring is compressed, we apply the energy conservatron once agaln:
,2 + u,2+w^t 0J+0J+l t ( , r r - xr \ '+oJ=!nv!+oJ+0J+oJ
Using r,, = 9.39 m/s, /c = 500 N/m and n = 5'0 kg' the above equation yields (x!-x)=Nt=93'9cm'
td) itre initiat etergy = mgyo= (5'0kgx9'8 m/s'?)(5'0 m) = M5 J' "fhe energy transformed to thermal energy
during each passage is
f*(x, - xr) = p*mg(x, -.r, ) = (0.25)(5.0 kgxg'8 m/s'zX2'0 m) = 24'5 J
The number of passages is equal to 2451124 5 J ot lO
11'54'Moi lehAssumeanidea|spr ing,soHooke.slawisobeyed..Treatthephysicsstudent-asapal t ic leandapplv the law of conservation or.n".'gf 6ut system is comprised of the spring' the student' and the ground we
iso use the model of kinetic friction.Visualize:Weplacetheoriginofthecoordinatesystemonthegrounddirectlybelowtheendofthecompressedspdng that is in contact with the student'
to=O* tO=OttV,
rr -ro =0.50 m t=80,000N/myo=yt= 10m z=100kg
lrr=0.15 vz=0m/s
Findv1 As = y2lsin 30"
Solver (a) The energy conservadon equation is
Kt +ust +(Jn + AE,l' = ro + Uso + U,o + W..r
| . I I ' 1111, - ;^yr +0J!rr ! +ngy,+lk(xr -x") '+oJ ==mv;+mgJ,* 2" ' " r "o ' - -
Sincey,=yo=l0m,. t l=r" ,vo=6p75,f t=80'000N/m'zr=100kg'ancl(r ' - ro)=0'5m'
! *r? = i*<r, -
xo)z + vt= l!1",
- ro) =,0.,o -^
(b) The work of friction creates themal energy' Apptying the conservation of energy equaion once agam:
K2 + IJ s2 + u"2 + AE,n = 1ro + 40 + U,0 + ly.-r
t , I
; mv; + m;y z + O J +/*A's = 0 J + ztSyo + iktr,
-'ro)' + 0 J
With v, = 0 m/s anct y, = 1As]sin30', the above equation is simplified to
m8(As)sin30' + P*lr Lt = ^gYo
+ ik(t,
- ')'
From the free-body diagram for the physics student' we see that z = wcos3oo' Thus' the conservation of energy
equation gives
Ar(mgsin3O" + pkmgcos30".) = 4gro 11&(r, - 'to)'?
using m = 100 kg' ft = 80,000 N/rn, (-t, - tJ = 0'50 m' )o = l0 m' and pr = 0 15' we get
1_mgyo +:k(-tr - xo )-
r .= z =-12- lmmg(sin30" + l| cos30")
tO'Yo .Il,Itto vt
ll-24 Chapter 11
Assessr y, = (As)sin3Oo = 16.05 m, which is greater than yo = l0 m' The higher value is due to tlrc tramfoma-
tion of the spring energy into gavitalional potential energy.
11.55. Modell Treat the block as a particle, use the model of kinetic ftiction, and apply the energy conservation
law. The block and the incline comprise our system.Vi*"|i"", We place the origin o'f the coordinate system directly below the block's starting position at the same
level as the hodzontal surface, On the horizontal surface the model of kinetic ftiction applies'
rO=0m rz-r t =Lyo=r, ] l =]2=0m
FindYl y3
Solve: (a) For the first incline, the conservation of energy equatif,n gives
Kr +%r +aEd = Ko*IJ"s*wur ] .u i *o l*o I = 0 ! +mLto+oJ = v1= " [ r*"
=Ath
(b) The work of fricrion creates thermal energy. Applying once again the conservation of energy equation, we have
K, + %3 + aE,' = K | + Il st + w*, ! rr! + ^gy,
+ f,(*, - *rl = ! ^vl
+ mgyt + w-l
Using v. = 0 pts, y, =0 m, tyd = 0 J,/k =A^g, n, =,[zgh, ard (xz -.rr) = t, we get
mgYr+ Pnmgl = !m(2gh) + Yr=h-IL*L
Assess: For p* = 0, y3 = ft which is predicted by the law of the conservation of energy'
Solve: For a conservative force the work doneindependent of the palh. We will show that WA-C-B
is given by
This means
on a particle as it moves from an initial to a final position is= WA-B for the spring force, Work done by a spring force F= -'/c
. , [w=JFdx=-Jkxdx
'P k, ^ ^, T k, . , T k, .w^ .. = - I kx dx = -;Gi -,i). w^ .c = - I kx b. = -;('i -'i). nd w,
-' = - I kx dx = -
r\xi' - -t Ir^t
11.56. Model: Assume an ideal spdng, so Hooke's law is obeyed.
Adding the last two:
wo-.-" = wo-" +wc -,=-f,{{-* '^*ri-xi)=w^-,
Work tt-25
11,57. Morlel: A "sprong" that obeys the force lavt F = -q(x- x''3' whete q is the sprong constant and t" i3
$li#,*t$"Tilflhe origin or tne coordinate system or the fteE end of the sprong, that is, r" = r; = 0 m'
;=2bs q =,l0,ooo N/m'
*xr
ilin,,fil ffiJl3Jl[5r}i"*; "*
a parabola. rhe rorce increases bv a factor or 8 every time x increases bv
a iactor of 2.
0
(d) Applying the fiergy conservation equation to the ball and- sprorg system:
K" +U, = 7r19,
| " ^ , ^ , ,qxi;m\+vI=et+j
= 10 m/s
11.58, Solvq (a) Because sin (cr) is dimensionless' Fo must have units of force in newtons'
(b) The Droduct cx is an argle o*""'] il'#'"iltJ thJ sine of-it' An argle has no rcal phFical units' If 'x has
H'tr; ; ; ililpt"d"ct ci is unitless' then c has to have units-of m-" ^
(c)Atxo=0m.theforc"i"R'i"tul=ii 'f"r' i i"iftti"it"Uf"tof*"e&:0monlvbecauseithasaninitialvelocity'(d) The force is a maxim* *n"n onfii = i' fus oc"utt *n"n
"* = r!2' or lor x* = tt]2c'
iej ffre gaptt is rne nnt quarter of a sine curve'
0 r .o- " ' *
(f) We can frnd the velocity vt at -xr =t- ftom tlre wort-kinetic energ;l theorem:
"2Wv; +-- rnt1a=L^,i -L^u1 =l^"i . f,^4
=$t +v'=
ll-26 Chapter I I
Thisisavariableforce'Astheparticlemovesftom{=0mtotr=T'-=ty'2c'thewor*doneonitis
.L .t l2c
w= J, 'F( ')&= 4Jo sin(cr)dr = -&cosorl;'" = -&(cos Z-'*"o)=*
Thus, lhe particte's speed at 't1 =:c,* = n/2c is v, =
1159, Visualize; We place the origin of the coordinate system at the base of the stairs on the first floor.
v
Third floor
S€€ond floor
Flrst floor0
Solv€; (a) We might estimate y2 -yr = 4.0 m=12ft- yt-yz' thus, yr -yr = 8'0m'
O) We might estimate the time to run up these two flights of stairs io-be 20 s'
i"j stti"tui" yo* lrru"t as m = 7ol<g =-150 Ib' Your power output while running up the stairs is
work done by you -
change in poiential enerEry -
m?(y,- yt')
time time time
_ (?0 ksxe.sgJ)(8.0 m) = now = (no wt'ffi) = o.rr nn
Assess: Your estimate may vary, depending on your mass and how fast you run'
the lawnmower as a particle and use the model ofkinetic ftiction'11.60. Model: ModelVisusliz€:
\
We placed the origin of our coordinate system on the lawnmower and drcw the ft€€-body diagram of forc€s'
Solve: The normal force i, which is related to the ftictional force, is not equal to li. This i8 due to the presence
of F. The rolling ftiction is /. = p,n, ot n = f,l p,. The lawnmower moves at constant velocity' so 4.. = 9' 11t"
two components of Newton's second lax' are
()4) = "
- w - r'skt37" = mo, = O N -
f, I IL, - nc- Fsin37' = 0 N =+ f, = p, ns + p,F s'i 37"
(I4) = r"ot:Z' -.4 = O N + Fcos37' - p.ns - p'Fsin37" = 0 N
- F - lt,mg -
(0'l5xl2 kgx9 8 n/s') - 29.4 N-' cos37" - trsin3T' (0.7986) - (0.15X0'6o18)
Thus, the power supplied by the gardener in pushing the lawnmower at a constant speed of l'2 m/s is P =
Fv = (2,9 NX1.2 m/s; = 29.9 1".
ll-26 Chapter II
This is a variable forpe' As the particle moves ftom4 = 0 m to4=:r * = fi/2c' the work dorc on it is
w = ),. F(x)dx = Folo sin(cr)dr = -&costcx;t[/'" = -4(cos i- *"o)=
*
Thus, the particle's spe.ed at 4= t(w = ltlzc is vr =
11,59. Visuatizc: We place the origin of the coordinate system at the base of tlle stairs on the first floor.
I mrd floot
Second floor
First floor0
Solve: (a) We might estimate y2 - yt = 4.O m-12ft= y3 -yz, thus, yi -j2r = 8.0m'
(b) We might estimate the time !o run up these two flights of stai$ to bo 20 s.
i"i E tl-ui" your -^s
as n = 70 kg = t50 lb. Your polver output while running up the stairs is
work done by you _ change in potential energy - ng(h - yl)
time time time
_ (70 kg)(e.s_g{s'?)(8.0 m) = 270 w = (270 w{ffi) = o.rr nn
A$s€ss: Your estimate may vary, depending on your mass and how fast you run
11.60. Model: Model the lawnmower as a particle and use the model of kinetic ftiction'
Visualize: Y
\
we placed the origin of our coordinate system oD the lawrunower and drew the ftee-body diagram of forces.
Solvq The normal force i, which is related to the ftictional force, is not equal to fi. This is due to fe nr3sence
of F. The rolling frictior is /, = p,n, ot n = f,lp,. The lawnmower moves at constant velocity' so 4Et = 0' The
two comoonents of Newton's second law are
(14,) = "
- w - rs;ttr3'l" = ma' = o N + f, / pt - ns - Fsin3?' = 0 N + f, = p, mg + p,F sn37"
(!4) = rcos:z' - /, = 0 N = Ircos3T' - l.ms - l.Fsin37" = 0 N
) F = -!43-- (0.15X12 kgx9.8 m/s'z) - 29.4 N
cos37" - Il sin 37" (0.7986) - (0' 15 X0.6018)
Thus, the power supplied by the gardener io pushing the lawnmower at a constant speed of 1'2 m/s is P =
Fv = (24.9 NXl.2 n ls) = 29.9 W.
Wo*
11.61. Solve: Power output during the push-off period is equal to the work done by the cat divided by the time
,f," "u, "ppli"a
,fr" f-"e. Since the foice on tlte floo; by the cat is_equal in magnitude to the force-on the cat by tlle
noor, Jort done by the cat can be found using the work-kinetic energy theorem dullg th.e nush-orr
;;;1, t"" = t"* = "AK. We do not need to explicitly calculate W*,, since we know that the cat's kiretic energy is
hansformed into its potential energy duriug the leap. That is'
AU, = mg1r, - r,1 = (5.0 kg)(9.8 m/s'?)(0.95 m) = z16'55 J
Thus, the average power output during the push-off period is
o= w"" =46.55J =z: :wr 0.20 s
11.52. solve: Using the convefsion 746 W = | hp, we have a power of 1492 J/s. This means w = Pr =
(1492 JAX1 lti = 53.112 x 105 J is the total work done by the electric motor in one hour. Furthermore,
W."* = -W, = us -U a = mg(yr - y) = mq(l0 m)
-- 4'd' - 5.1712\rc6 J = 5.481x loa ks = 5..18I x l0o kg
" l,lt"t = s4,800 li
"o" ' - e( toIn) - (9.8 m/s 'zXlo m) - - - l kc
11.63, Solve: (a) The change in the potential energy of l 0 kg of water in falling 25 m is AUr=-mgh=
-(1.0 kgx9.8 m/3'?)(25 m\ = -2A5 l.(b) The power required of the dam is
P=W =Y=SOx106 Watts+W=50x106 Jt ls
That is, 50 x 106 J of energy is required per second for the dam. Out of the 245 J of lost potential energy'
(245 D(0.S0) = 1g6 J is converted to electrical energy. Thus, the amount of water needed per second is
50 x 106 J0 kg/196 J) = 255,000 kg.
11.64. Solve: The force required to tow a water skier at a speed v is 4. = Av' Since power
p = Fv, the power requir€d to tow the water skier is P,o* = Fr*v = Av2 - We can find lhe constant A by noting that a
speed of v = 2.5 mph requires a power of 2 hp. Thus'
(2 hp) = A(2.s mph)' J A = 0.32Gh
Now, the power required to tow a water skier at 7.5 nph is
P,* = Av'| = 0'32;:3;'(7 s mPh)' = 18 hP"" (mph)-
Assessi Since p * v2, a three-fold increase in velocity leads to a nine-fold increase in power.
11.65.Solve:Bydef in i t ion, themarimumpoweroutputofahorseisP=lhp=746W'Atmaximumspeed'
wheo the horse is running at constant speed, i = 6' The propulsion force F**' provided by the horse pushing
asainst the qround, is balanced by the drag force of ail resistance: FL* = D' We learned in -Cllapter
5 that a
;;;ili;;;;;ii;; a-s it o = 1ev,, wh-ere A is the cross section area. since the power needed for force F to
push an object at velocity v is P= Fv, we have
P - t hp = 746 w = F;,.v = (+Av')v = +Av3
" =( !!\''' - ( +tz+o wl '1' = 15 m/s\ A / \(o.s m)(1.8 m) /
Assess: 15 m/s = 30 mph is a reasonable top spe€d for a well-trained horse'
11.56, Solve: The net forc€ on a car moving at a steady spetd is zero. The motion is opposed both by rolling
friction and by air resistance. Thus the n,oputrioit forc" prouia"a by the drive wheels mus t he F", = p,ng + f Avz 'where 14 is the rolling friction, m is the mass, A is the cross-section area, and v is the car's velocity. The power
required to move the car at spe€d v is I .
P=Fsy=hmgv+-Ay'
ll:28 Ch&pter 11
since the maximum power output is 2oo w lllrd 75% of the power reaches the drive wheels, P = (200 hpx0.75) =
t 50 hp. Thus,(1a6w\ I
trso hpi 'l! *
l= (0.02X1500 kgx9.8 m/s')v + ]{ i.6 m;11.4 m;u'' l \ lhp ' , | - +
+ O.56 vt + 294 v' 111900 = 0 + v = 55.5 m/s
The easiest way to solve Otis equation is through iterations by trial and error.Ass€ss3 A speed of 55.5 m/s -
I l0 mph is very reasonable.
11.67. Model: Use the model of static ftiction, kinematic equations, and the definition of power.
Solve: (a) The rated power of the Porsche is 217 hp = 161,882 W and the rl'eight of thecr is (1480 kgX9 8 rnls') =
14504 N. The weight of the car on the drive wheehls (l45MX?3) = 9670 N' Because the static ftiction of the tires
on road pushes the car forward, * -'"; ;it ; #::: j';']::"'' = *-
1480 kgp 16rr8?I = 16.? m/s(b) P = Fv.* =u* =F=-9670N
(c) using the kinematic equation, vmd =vo +4md(t!u" -lo) withvo=0m/sand'o=0s'weobtain
, . _ u.* = 16.7 m/s= = 2.56 "atu 6.53 m/s'
Assess: An accelemtion time of 2.56 s for the Porsche to reach a speed of =35 mph ftom rcst is reasouable'
11.68. (a) A student uses a string to pull her 2.0 kg physics book, startiry from lest, across a-2.o-m-long lab
beuch- The coefhcient of kinetic ftiction berween the book and the lab bench is 0.15. If the book's final speed is
4.0 m/s, what is the tension in the stdng?
vo=0rtVs Yl=4rds
(c) The tension does extemal work w"n. This work increases the book's kinetic energy and also caus€s an incfease
AE. itr th" tt *tt"l "nogy
of the book and the lab bench' Solving the equation gives I = 10.9 N'
11,69. (a) A 20 kg chicken crate slides down a 2.5-m-high, 40o ramp from the back of a nuck to the gound' The
coefficient of kin€tic ftiction betwe€n the crate and the ramp b€nch is 0.15. How fast are the chickens going at the
bottom of the ramp?(b) y
(b)
!o=25mvo=0m/s
Yr=0mn=2OkAtlt=0.15
trnd
0vr = 6.34 m/s.
m=2.Okg
(c)tt, Yt, v\
11.70. (a) If you expend 75 W of power to push a 30 kg sled on- a surface where the coefdcient of Hnetic friction
between the sled and the surta"e is p* =b26' iurt"t tp""a *il you be able to maintain?
(b)
n=3Okg
(b)
(c) F",n = (0 20)(30 kgxg'8 m/s'?) = 58'8 N + 75 w = (s8'9 N)v + v = J1a = 1'28 m/s
11.71. (a) A 1500 kg object is being accelerated upwad at 1'0 tr/s2 by a rope' How much power must the motor
,ufp[ u, m" in"*nt *tt"n the velocity is 2'0 m/s?
At a latertime 0)
aa = 1.0 m/sz I
eadier tilne(t=0)
(c) T = (1500 kg)(9.8 m/s'?) + 1500 kg(1'0 m/s'?) = 16'200 N
P =T,izmls\ = (162N N)(2'0 m/s) = 32'400 W = 32'4 kW
11.72. Moilel: Model the water skier as a particle, apply the law of conservation of mechanical enelgy, and use
the constant-acceleration kinematic equations'
Visualize:
I<nown
,r=2.0m.i2-rt = 5.0 m
Find
We placed the origin- of the.cooFdinate svstem at the base of the ""ttot"tt ffif
'"ro*s her to clear the shark tank.
ilit'*' w":il tiit uy linding the smalJest speed vr at the top of the rarnp
prom the vertical motion forjumping the shark tank'
|z = lt * v1"lt, - t,\+ lar(t, - trl '1
+ 0 m = (2.0 m)+0 m+;(-9 8 m/s'?x" -tr)' + (" -'r)= 0'639 s
*k= o.20
Motor
