This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 251 College Physics I
Exercise 4 Suggested solutions 1. (a) From the freebody Lliagral'u {Fig.3. 1) for the load being lifted. write
Newton's 211d law for the Vertical Llireetion. with up being positive. Fr Fr — mg : ma : 0.1601339
=I> FT 2 1.1[img : l.16(285kg}(9.8l11n/1~'2] = 3.24 X 1031\ (its) The net work done on the load is found from the net force. Him 2 Furieoso'; : (UJBUmgH
= 0.1609833kg)(9.80111_.f52)('22.0111) = 9.33 >< 103 J (t!) The work rlone by the cable on the load is: ll’Lﬂbsg = FTU’ eosU' = (1.1[img'1d = 1.10(285 kg‘}(9.80 mfsglﬂlﬁ m) = 7.13 x ll)"l . (d) The work done by gravity on the load is: “7.3 : mgdeosllo = —m.grf : —('285 kg)[9.8[}11'1fs2}[22.In} = —l3.l’l x 104.] {:3} Use the workenergy theor)r to ﬁnd the ﬁnal speech with an initial speed of U. ll'mac = ICE2 — 1031 = grueg — émi'?
LEI17m . 2(9.83 x 103.1] 2 _
‘ : I : — : 5
:> 12 m + t. (285 kg) + U 31 inf 2. Fig. 2 shows the motion of the three particles just after 3'! being given a velocity 1"
and the two end particles are about to eollide. v
V 1 M
m M m v v
m m
M M Fig. 2 (a) Conservation of momentum gives: MV = (AI + 2m)v
_ M V
_ M + 2771 =>v conservation Of mechanical CRETE" gives:
1 ‘2 1 2 1 2 2
_) / — _A I u _ ‘ l 2 [L +2(2"l)(v +U) When the two end particles are about to collide, the speed of each m is equal to m : ally/'2 “‘ 1"[1'2 = — = V + r") r
2m 2171. A1 + 2172 3. (a) Fig. 3 shows the motion of the wedge and the particle when the particle has moved
over a distance .5. ucos60° Fig. 3 Conservation of momentum in the horizontal direction gives: 0 = 2mV  m(ucos 60°  V)
=(hﬂV—:+V 2
11
V. : _
=’ 6
(b) By the conservation of mechanical energy.
1 1 1
mgs sin 60° 2 5771(1er08600 — V)2 + 5mm sin 60°)2 + 5(2111)lr"2
2
\/§ 1 u u 2 l uﬁ 1 u 2
’ 957567) +5 2 +5‘2’(§)
u2 3u2 (l2 11
3 =._ __ _.=._.2
=> \/_ gs 9 + 4 + 18 12¢: 12¢3ﬁ
2""=:V u ‘1. Use xmlsvrvutiun uf 011011;}: The: ImVtI of [11:3 bull n11 $110 llllt'UIllpI'!’SSWI wring taken as
the Zr'l'n 1: unﬂinn fur but]: gl’m‘itatinlml Hi (1/ : (I) null rlnstin' PE LI' 2 U). 'I‘akv up
Tu 1w pnsiI’iVel fur hull], Fig. ~l (a) Subsrript 1 rcprt 50111.»; the hall at fhv hmm'h point. and subscript ‘2 I‘I"1')I‘O>‘(‘I]l.5 the
Hill at Hm lut'ntium Wht'l't' i? jlls? bank [111‘ wring. 11? [110 1111w1111I1‘INSHI It'llfJIh. E1213: 1 7;, 1k 3 1 2 1]_ 2
Emil +1111)!“ + j .11 _ 3/1142 + Him/2 + T; .12 u 1 2 1 2
0+0+§kx1 =—mv2 +mgy2+0 (950N/m) (0.150m)2 — 2(0.30kg) (9.8m/52)(0.150m) v = 2 8.26m s
2 m /
[13} ﬁn'osn'rrzipt :3 represents 711913311 m in: liigllrfEi point.
£1 = +33
1 ._. I .1 I u l 2
:.> Emmi + mﬂyl + Eh'l : Em H3 + mi'lilm + Ekr'a
1 2
0+0+§kx1 =0+mgy3+0 _ kxlz _ (950N/m)(0.150m)2
3'3 _ 2mg _ 2(0.30kg)(9.8m/52) 2 3.64m . n. (H) Um" ("1’J11~1'1'\'z1!iu:11 of llH‘L‘lldniVﬂl nutI'm”. :Muming Ihvl‘v are no 1nmt‘ullSt‘rvmivu 1} 1111's. Subscript 1 represents the water at the top of the dam, and subscript 2 represents
the water as it strikes the turbine blades. The level of the turbine blades is the
zero location for PE (y = 0). we have v1 = U. y1 = 80 m. and y2 = 0. Solve for 19. E1 2 E2
1 ‘2 1 2
=> 511115 + mgyl 2 5172102 + mng
1 2
=> 17293]; = inn1’2 :5 1.9 : ‘l29y1 : \/2(9.81n/52)(81m) = 39.8 In (b) The energy of the water at the level of the turbine blades is all kinetic energy. and
so is given by émug. 58% of that. energy gets transferred to the turbine blades.
The rate of energy transfer to the turbine blades is the power developed by the
water: 1m?) (O.:38)(650kg/s)(39.8kg/s)2
2 2 P = 0.58 (— t = —— = 2.99 x 105W" 6. Consider the freebody diagram for the coaster at the bottom of the loop shown in
Fig. 5. The net force must be an upward centripetal force. N bottom
"lg Fig. 5
' mvgouom
2 F bottom = JrVbomoln _‘ mg : —R— 2
, mv
=> Mam = $5 + my Now consider the force diagram at the top of the loop shown in Fig. 6. Again, the net
force must be centripetal. and so must be downward. mg N”?
Fig. 6 _ 2
m stop za=tWW=R _ 2
m ‘L‘t DD R =;> Nmp = — mg Assume that the speed at the top is large enough that Afton ) U" and so own 5 Now apply the conservation of mechanical energy. Subscript 1 represents the. coaster
at the bottom of the loop1 and subscript 2 represents the coaster at the top of the
loop. we choose the level of the bottom of the loop to be the zero location for PE
(3; : 0). So we have yl : U and ya : 2R. Therefore. E1 = Es
1 2 1 2
2 5111111 + negy; = Eelv2 + mng :> lﬁmm : U30” + 491? The difference in apparent weights is the difference in the normal forces. .' 2 371112
Nbonom — Nu], = (w + my) — ( 1;") — mg) = 6mg Notice that the result does not depend on either 1! or R. U! ...
View
Full Document
 Fall '10
 drtong

Click to edit the document details