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Unformatted text preview: CHEF. 15] WORK. ENERGY. AND PUWER. Tl 6.2 6.4 iii A. block moves up a 3i}: incline under the action of certain forces. three of which are shown in
Fig. 6—2. F. is horizontal and of magnitude 40 N. F: is normal to the plane and of magnitude
2!} N. F3 is parallel to the plane and of magnitude Si} N. Determine the work done by each force
as the block {and point of application of each force) moves El] cm up the incline. Fig. «ti2 The component of F. along the direction of the displacement is
F. cos 30‘" = {40 N]I{D.EI5I5]I = 34.5 N Hence the work done by 151 is {34.6 Njitltil} m} = 23 .1. {Notice that the distance must be expressed in
meters.] Because it has no emnponent in the direction of the displacement. 51 does no work. The component of P3, in the direction of the displacement is 3t! N. Hence the work done by ii; is
{30 Nllilﬁl} in} = 24 I. A Elmg object slides 8!] cm along a horizontal tabletop. How much work is clone in overcoming
friction between the object and the table if the cccliicicnt of kinetic friction is 021'}? We ﬁrst Find the friction force. Since the normal force equals the weight of the object.
F} = HF” = (ﬁlm{0.3110 kg]{9.3l mj's1)= 0.533 N The work done overcoming friction is Frr cos I9. Because the friction force is opposite in direction to the
displacement. I9 = IEIJ". Therefore. Work = FIr cos ism = roses Nitaso m](— I} = —o.4r J The work is negative because the friction force slows the object: it decreases the object's kinetic energy. How much work is clone against gravityr in lifting a 3.0kg object through a vertical distance of
4!] am? An external force is needed to lift an object. tfthe object is Iilted at constant speed, the lifting force must
equal the weight of the object. The work done by the lifting force is what we refer to as work done against
gravity. Because the lifting force is mg. where m is the mass of the object. we have Work = [mglfhﬂcos I9} = {3.1) kg at 9.81leﬂﬁl} mlfl] = I2 I
In general, the work done against gravityI in lifting an object of mass in through a vertical distance ii is nigh. How much work is done on an object by the force that supports it as the object is lowered
through a vertical distance it? How much work does the gravitational force on it do in this
same process? 7'2 6.6 6.? 6.3 6.! WORK. ENERGY. AND POWER [CHAR 6 The supporting foroe is mg. where m is the mass of the object. It is directed upward while the displace
ment is downward. Hence the work it does is Fs cos it: {mgﬂhjicos ISTJ = —mgit The force of gravity acting on the object is also mg, but it is directed doumward in the same direction as the
displacement. The work done on the object by the forest of gravity is therefore Er cos 6 = [mgﬂiﬂicos 6°} 2 mgit A ladder 3.6 In long and weighing 200 H has its center cl'gravittr III] cm from the bottom. At its
top end is a SCIN weight. Compute the work required to raise the ladder from a horizontai
position on the ground to a vertical position. The work done {against gravity} consists of two parts, the work to raise the center of gravity 1.2K] m and
the work to raise the weight at the end through 3.6 m. Therefore Work done = (200 N}[1_2t’t m) + {50 mos n1} = 030 1LT IZionipute the work done against gravityr by a pump that discharges 60!} liters of fuel oil into a tank
26' nt above the pump's intake. lI'Ci'ne cubic centimeter of fuel oil has a mass of 0.32 g. One liter is
lﬂﬂﬂ ten13'. The mass lifted is 3 (e00 liters}(lttﬂﬂ %) (as: %) = 492000 1; = 492 kg The lifting work is then
Work = {mgjﬂr} = {492 leg at RBI mfsijfzﬂ m] = '96 it} a III—kg mass falls 400 cm. {a} How much work was done on it by the gravitational force?
{it} How much PEG did it iose'? Curravjtjtr pulls with a force mg on the object. and the disp]aeement is 4 m in the direction of the force.
The work done by gravityr is therefore testisDU mt = [2D its a 0st mom :0} = 73 .T The change in PE; of the object is High; — mgirf. where it, and ff! are the initial and ﬁnal heights of the
object above the referents: level. We then have lChange in PEG = High — mgh = rag{hf —fr.] = [1t] kg in: 9.3l NJf—dﬂ n1} = —?E J
The loss in PE; is ?E J. a. force of 1.51} N acts on a [LED—kg cart so as to accelerate it along an air track. The track and
force are horizontal and in line. How fast is the cart going after aooeleration from rest through
3t} cm, if friction is negligible? The work done by the fore: causes. and is equal to. the increase in HE ol' the cart. Therefore.
WEIfit. done = {KEjmi — {KEjmn or F3 cos ﬂ“ = grew} — 1']
Substituting gives
{1.50 human m}: 50.20 kgjqﬁ CHAP. t5] WORK, ENERGY. AND PDWER. T3 6.10 till ﬁsll 6J3 6.11 A. 0.50—ltg block slides across a tabletop with an initial velocity of 20 cmls and comes to rest in a
distance of 70 on. Find the average friction force that retarded its motion. The KB of the block is decreased because of the slowing action of the friction force. That is.
Change in RF; of block = worlt done on block by friction force il’l’l‘ii‘zlr — §mv§ = Flat eos ﬁ' Because the friction force on the block is opposite in direction to the displacement, oos I9 = —I. Using
or = 0. u; = 0.20 arts. and s = 0.?0' III. we ﬁnd a — i {use kgltllllﬂ me? = [otters m}{— 1}.
from which F' a 0.014 N. A car going I5 arts is bronght to rest in a distance of 2.0 m as it strikes a pile ofdirt. How large an
average force is exerted by seatbelts on a IitICi—ltg passenger as the ear is stopped? We assume the seatbelts stop the passenger in 2.0 tn. The force F they apply acts through a distance of
2.0 m and decreases the passenger‘s KE to zero. So Change in HE of passenger = work done by F
Iiiit?!) ksltlﬁ moi} = {Fires :rtjf—I]. where cos 0 = —I because the restraining force on the passenger is opposite in direction to the displacement.
Solving. we ﬁnd F = 5.1 Itbl. A projectile is shot upward from the earth with a speed of 20 mf's. How high is it when its speed is
3.0 mJI's'? Ignore air Friction. Because the projectiie‘s energy is conserved. we have
Change in KE+change in PEG = 0
smi — smut + (signs. — a.) = a We wish to ﬁnd Ii). — hf. After a little algebra. we obtain jfvi _ _W
lg _ 29ml mfsll hf—fr,= =ll'm In an Atwood machine [see Problem 3.23} the two masses are 300 g and T00 g. The system is
released from rest. How fast is the EDDg mass moving after it has fallen 120 cm? The T00g mass rises 120 not while the 300g mass falls l20 cm, so the net change in PE; is Change in asG = [are agitasl militias m} — {ass kgtissl mfszmlﬂ m) = —I.Is I which is a loss in PEG. Because energy is conserved, the RE of the maues must increase by LIE J.
Therefore. Change in KE. 2 LIE .l = “0.00 kgjfu} — erg} + H030 ItgHo}  Iii1} The system started from rest, so of = 0. We solve the above equation for a; and ﬁnd of = 1.15 thI's. Figure 6—3 shows a head sliding on a wire. lffriction forces are negligible and the head has a speed
of 200 cnn's at A, what will be its speed in) at point B? {bl At point C? We know lhe energy of lhe head is consented. so we can write
Change in KE+change in PEG = 0 “M 5.15 6.16 WORK. ENERGY. AND PGWER [Cl[AR 5 11ij —%mtl; + #thth — a} = a {it} Here. it, = 2.0 rnfs. h, = MI} in. and ”I = it. Using these values. while noticing that m cancels out. gives
air = 4A ants. {it} Here. 1'. = 2.0 this. in = 0.31} m. and II; = £1.51] In. Using these 1mines gives a; = 3. mls. Suppose the head in Fig. fir3 has a mass of 15 g and a speed of 2.0 rrn's at A. and it stops as it
reaches point C. The length of the wire from A to C is 25G em. How large an average friction
force opposed the motion of the head? When the head moves from A to C, it experiences a change in in total energy: it loses both KE and PEG.
This total energy change is equal to the work clone on the head by the friction force. Therefore. Change in PE; +change in KE = work done by friction force mettle  a} + latte}  at = or we a Notice thal cos I9 = —I, v: = IIII+ 1:". = 2.0 rrt.I's. It: FL. = —{l.3El n1, .3 = 1.51] rn. and m = 0.015 kg. Using
these values. we ﬁnd that Fr = 1103i} N. A [Elﬁnkg ear is coasting clown a 30° hill as shown in Fig. 5—4. At a time when the car‘s speed is
12 ms, the driver applies the brakes. 1What constant force F {parallel to the road] must result if
the car is to stop after traveling [Ell nt‘l‘ The change in total energy.r ol' the ear {HE + PEG} is equal to the work done on it by the braking force
F. This work is Fr cos IED“ because F retards the car's motion. We have lmtu} — 163+ mightr — a.) = tot—I] CHAR 6] WORK, ENERGY. AND PGWER 75 til? ﬁtlﬂ where m: I200 kg
'b'f=ﬂ
e; = I! rni's
it; —h, = [100 ml sin 30“
3 = tilt] in 1With these values. the equation yields F = 6.? k“. A ball at the end of a lﬁﬂncm long string swings as a pendulum as shown in Fig. 15.5. The hall’s
speed is 4th} chills as it passes through its lowest position. {a} To what height i: above this position
will it rise before stopping? {it} What angle does the pendulum then make to the vertical? {a} The pull of the string on the ball is always perpendicular to the hall's motion. and therefore dues no
work on the hall. Consequently, the hall’s total energy remains constant; it loses KE hut gains a like amount of PEG. That is.
Change in KE+change in PE; = [I
ﬁmtﬁ—kﬂﬁ+m‘gh =t] Sinoe a; = t} and e, = 4.00 nits. we ﬁnd it = 0.3 If: m as the height to which the ball rises.
{in} From Fig. 65, mks—rt: I _e.sle 1 .El} whieh gives rjl = 56.9“. Fiﬁ 6'5 Fig. 66 A Suﬁg hiecl: is shot up the incline in Fig. 66 with an initial speed of see cn‘tfs. How far up the
incline will it go if the coeﬂicient of friction between it and the incline is 13.150? We ﬁrst ﬁnd the friction force on the block as
Ft = #5:  trims “‘13 35°} As the block slides up the incline a distance D. it rises a distance D sin 25$". Because the change in
energy of the block equals the work done on it by the friction force, we have Change in KE+ change in PE; = FIB cos 180°
ﬂit}: — 11:2} + mgtﬂ sin 2511“} = ﬂed) T6 6.]?! 6.20 6.21 6.12 WORK. ENERGY. AND POWER [CHAR 6 We calculated Fr above. and we know u. = 1.00 nits and a; = 0. Notice that the mass of the [ﬂock cancels
out in this case {but only because F' is gitten in terms of it}. Substitution gives .0 = 0.365 1n. A 600ng train is being pulled up a [.0 percent grade {it rises 1.0 m for each horizontal lﬂﬂ in]
by a drawbar pull of 3.0 kN. The friction force opposing the motion of the train is 4.0 Hi. The
train*s initial speed is 12 mfs. Through what horizontal distance at will the train more before its
speed is reduced to 9.0 oil's? The change in total energy of the train is due to the work of the friction force and the drawbar pull:
Change in KE +change in PEG : dem, + Wirielion
imitf — uﬂ+mgft10l0ﬂ = {seen MittH1] + {was NHs‘jt—l}
from which s = HS in = 0.23 km. ﬁtn advertisement claims that a certain [EDDkg car can accelerate from rest to a speed of 25 role
in a time of 8.0 a. What average power must the motor produce to cause this acceleration? Ignore
friction losses. The work done in accelerating the car is given by
Work done = change in KE = smot — of} The time taken for this work is 3.0 s. Therefore.
work _ if tau eggs may"! = = W
Power tilne 3'0 5 41' k
Converting from watts to horsepower. we have
_ I he _
Power _ (46 900 w}('l‘46 W) _ 63 hp ﬁt [LEShp motor is used to lift a load at the rate of 5.0 cmfs. How great a load can it lift at this
constant spened'iII We assume the power output of the motor to be 0.25 hp = $6.5 W. [n 1.0 s, the load mg is lifted a
distance of 0.050 In. Therefore. Work done in Lil s = {weightlfheight change in 1.0 s} = [mgj{0.050 m] By deﬁnition. power = 1wor‘itg'titne. so that less w  {Lam mil
It] s Usingg = 9.81 misi. we find that m = 331 kg. The motor can lift a load ofabout 0.33 x I03 kg at this speed. Repeat Problem 6.20 if the data apply to a car going up a 20‘“ incline. 1Work must be done to lift the car as 1well as to accelerate it:
Work done = change in KE + change in PEG
= Emir}  tea13+ methi — as} where fir — Jr; = .t sin 20° and .s is the total distance the car travels in the it s under consideration. Knowing
or = 0. of = 15 mls. and t = 3.0 s, we have .t = om.t:l,l:1t; + of}! 2100 m
Then Work done = trons kg}[625 mzfszj +{Izoo tgjtsst matinee mom set) = are at 1:13 kJ CHAR 6] WORK. ENERGY. AND PGWER T? 6.16 6.31 6.32 W? k] _ _ 3
Polls —‘3T kWﬂ.l3x 1t] hp from which Power = [n unloading grain from the hold of a ship. an elevator lifts the grain through a distance of 12 in.
Grain is discharged at the top of the elevator at a rate of 2.1] kg each second. and the discharge
speed of each grain particle is 3.6 mi's. Find the minimumhorsepower motor that can elevate
grain in this way. The power output of the motor is change in KE+change in PEG _ imiﬂji  ”fl + my:
time taken ' r = ? Etso mzfsz} + {set mfsljiu ml] Power = The mass transported per second. min is 2.0 ltgis. Using this value gives the power as 0.24 kw. Supplementary Problems ﬁt. force of 3.0M acts through a distance of I2. m in the direction of the fame. Find the work done.
Ans. 351 A. dellkg object is lifted 1.5 In. {a} How much work is done against the Earth's gravity? {it} Repeal ifthe
object is lowered instead of lifted. dies. {at} 59 I; {it} —59 J A. uniform rectangular marble slab is 3.4 m long and 1.0 m wide. It has a mass ol‘ IE!) kg. If it is originally
lying on the ﬂat ground, how much work is needed to stand it on end? Ans. ELL} kJ How large a force is required to accelerate a lSlllJultg car from rest to a speed of 2!] mi's in a distance of
El} in? Arts. 3.3 lsN A. lZ‘ﬂ‘llkg car going 30 this applies its brakes and skids to rest. If the friction force between the sliding tires
and the pavement is all!) N. how far does the ear skid before coming to rest? Arts. 90 m a proton [m = LET x Io"n kg} that has a speed of as x to“ this passes through a metal lilrn or thickness
llﬂlt‘l mm and emerges with a speed of ll] Dt'. 1IIZI'El mils. How iarge an average foroe opposed its motion
through the slut? Ans. Is a; It‘l'9 N ﬁt. EDDkg cart is pushed slowly up an incline. How much work does the pushing force do in moving the cart
up to a platform 1.5 m above the starting point if friction is negligible? Ans. 2.9 U Repeat Problem 6.30 if lhe distance along the incline to the platform is "Lil In and a friction force of I50 H
opposes the motion. Ana. 4]] k] A St} Dimkg freight ear is pulled 1300 m up along a L2G percent grade at constant speed. {or} Find the work
done against gravity by the drawhar pull. {in} ]f the friction force retarding the motion is ISIIII N. ﬁnd the total work done. Arts. {a} 4.?!) MI: in} 5.90 M] A ISO—kg woman walks up a ﬂight of stairs that connects two ﬂoors 3.13 m apart. {:3} How much lifting work is
done on the woman? {is} How mueh lifting work is done by the woman? {a} By how mneh does the woman's PEG change“?r Ans. {a} LB k]; {b} LS k1; [is] 1.3 k5 ...
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 Winter '11
 Landen
 Physics

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