This preview shows pages 1–4. Sign up to view the full content.
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: P427 P428 P430 Choose a coordinate system with East and North. 21‘: = m3 = 1.00 kg(10.0 m/sz) at 300°
(5.00 N); +i1 = (10.0 N)430.0°= (5.00 N)} +(8.66 N)i F1 = 8.66 N (East) First, consider the block moving along the horizontal. The only
force in the direction of movement is T. Thus, 213,, = ma T = (5 kg)a (1) Next consider the block that moves vertically. The forces on it are the tension T and its weight, 88.2 N.
FIG. ‘P4.2.8 We have EFy = ma
88.2 N — T = (9 kg)a (2) Note that both blocks must have the same magnitude of acceleration. Equations (1) and (2) can be
added to give 88.2 N = (14 kg)a. Then a=6.30 m/s2 andT=31.5N . m1 = 2.00 kg, m2 = 6.00 kg, 0 = 550°
(a) ZPx=ngsin€—T=mza
and iii T—m1g=m1a =ngsinQ—mlg_ 2
(b) T=m1(a+g>=
(c) Since 0, =0, vf =at= (3.57 m/sz)(2.00 s)=. FIG. P430 P432 First, consider the 3.00 kg rising mass. The forces on it are
the tension, T, and its weight, 29.4 N. With the upward
direction as positive, the second law becomes + u o
ZPy = my: T— 29.4 N = (3.00 kg)u (1) T Rlsmg Mass The forces on the falling 5.00 kg mass are its weight and T,
and its acceleration is the same as that of the rising mass. _. _.
Calling the positive direction down for this mass, we have (F8 )1 = 29.4 N (F3 )2 = 49 N P = ma : 49 N —T = 5.00 k a (2) FIG. P432.
y y 8 Equations (1) and (2.) can be solved simultaneously by adding them: T — 29.4 N + 49.0 N — T = (3.00 kg)a + (5.00 kg)a (b) This gives the acceleration as _ 19.6 N _ z
a _ 3.00 kg — '
(a) Then
T— 29.4 N = (3.00 kg)(2.45 111/52): 7.35 N. The tension is (c) Consider either mass. We have 1 1 2
yzvit+—2—at2 =0+E(2.45 m/sz)(1.00 s) =. P4.35 Forces acting on 2.00 kg block: T—m1g=m1a (1) Forces acting on 8.00 kg block:
PX —T=m2a (2.) 2kg
(a) Eliminate T and solve for a: a (m/SZVTTTZTITITT?  ++++++++4
15. 4.4.4444 ' ' ‘ 2 =Fx—m1g
m1+m2 a>0forFx >m1g=19.6N . (b) Eliminate a and solve for T: a m
T=——1
m1+m2 FIG. P4.35
T=0 for Fx g—ng=—78.4 N . (c) Px, N —100 —78.4 —50.0 o 50.0 100
ax, m/s.2 —12.5 —9.80 —6.96 —1.96 3.04 8.04 (Px +mZg) P4.44 (a) Following the inChapter Example about a block on a frictionless incline, we have a = gsin€ = (9.80 m/sz)sin30.0° 0.500 m
(b) The block slides distance x on the incline, with sin 300° = x x = 1.00 m: 012: = viz +2a(xf —xi)= 0+ 2(4.90 m/s2 )(1.00 m) fo = 2(1.00 m) 0f: 3.13 m/s aftertime ts=— m=0639s vf (c) Now in free fall yf —yl~ =vyit+%ayt2:
—2.00 = (—3.13 m/s) sin 300° t —%(9.80 m/sz)t2
(4.90 m/s2)t2 +(1.56 m/s)t — 2.00 m: 0 —1.56 m/si (1.56 m/s)2 —4(4.90 m/s2)(—2.00 m)
9.80 m/s2 t: Only one root is physical t = 0.499 5
xi = vxt = [(3.13 m/s)cos30.0°](0.499 s) = (d) total time = ts + t = 0.639 5+ 0.499 s = (e) The mass of the block makes no difference. P4.54 Since it has a larger mass, we expect the 8.00kg block to move down
the plane. The acceleration for both blocks should have the same
magnitude since they are joined together by a nonstretching string.
Define up the left hand plane as positive for the 3.50kg object and
down the right hand plane as positive for the 8.00kg object. 2 P1 = mlalz —mlgsin35.0°+T = mla
2132 = mzazz ngsin35.0°—T = mza FIG. P454 and —(3.50)(9.80) sin 35.0°+T = 3.50a
(8.00)(9.80) sin 35.0°—T = 8.000. Adding, we obtain +45.0 N — 19.7 N = (11.5 kg)a. (a) Thus the acceleration is a = 2.20 m/s2 . By substitution, —19.7 N + T = (3.50 kg)(2.20 m/sz) = 7.70 N. (b) The tension is . ...
View
Full
Document
This note was uploaded on 04/02/2008 for the course PHYS 6A taught by Professor Mahaashourabdalla during the Winter '07 term at UCLA.
 Winter '07
 MahaAshourAbdalla

Click to edit the document details