homework 5 solutions

# homework 5 solutions - Chapter 5 2 We apply Newton’s...

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Unformatted text preview: Chapter 5 2. We apply Newton’s second law (Eq. 5—1 or, equivalently, Eq. 5-2). The net force applied on the chopping block is Fm = F] + [3'2 , where the vector addition is done using unit—vector notation. The acceleration of the block is given by 5 = + /m. (a) In the ﬁrst case A A E +152 = [(3.0N)i + (4.0N)J:|+ [(—3.0N)i + (~4.0N)]] = 0 so a=0. (b) In the second case, the acceleration 5 equals m 2.0kg (c) In this ﬁnal situation, 21' is 131 +132 _((3.0N)i+ (4.0N)3) +((3.0N)i+(~4.01\1)j):(3 Urn/\$2); m — 2.0 kg 31. The free-body diagram is shown below. E, is the normal force of the plane on the block and mgr is the force of gravity on the block. We take the +x direction to be up the incline, and the +y direction to be in the direction of the normal force exerted by the incline on the block. The x component of Newton’s second law is then mg sin 6 = —ma; thus, the acceleration is a = — g sin 6. Placing the origin at the bottom of the plane, the kinematic equations (Table 2—1) for motion along the x axis that we will use are v2 = v3 +2ax and v 2 v0 + at . The block momentarily stops at its highest point, where v = 0; according to the second equation, this occurs at time t =r—v0 / a . 1 v2 1 3.50m/ 2 x=~__o=__ ( 2 fl) =1.18 m. 5(9.8 m/s )sm 32.0° (b) The time it takes for the block to get there is = i 2 v0 3.50m/s _ =— 20.674 . a —gsm(9 —(9.8m/s2)sin 32.00 S .(c) That the retum-speed is identical to the initial speed is to be expected since there are no d1s31pative forces in this problem. In order to prove this, one approach is to set x = 0 and solve x 2 Vet + %at2 for the total time (up and back down) 1‘. The result is t _ 2120 2v0 2(3.50 m/s) ______5_' =_ 2 _ 21.353. a —gsrn6 ~(9.8m/s )31n32.0° The velocity when it returns is therefore v =v0 +at = v0 —gtsin 923.50 m/s — (9.8 m/s2)(1.35 s)sin32°=—3.50 m/s. The negative sign indicates the direction is down the plane. 81. The mass of the pilot is m = 735/9.8 = 75 kg. Denoting the upward force exerted by the spaceship (his seat, presumably) onthe pilot as F and choosing upward as the +y direction, then Newton’s second law leads .to F — mgmoon = ma :> F = (75 kg)(1.6 m/s2 + 1.0 m/s2)= 195 N. 5. The net force applied on the chopping block is Fla : 17; + 132 + F; , where the vector addition is done using unit-vector notation. The acceleration of the block is given by c7=(17“1 +13"2 +133)/m. (a) The forces exerted by the three astronauts can be expressed in unit-vector notation as follows: A = (32 N)(cos 30°? + sin 300]) = (27.7 N)i +(16 Nﬁ 2 = (55 N)(cos 0°? + sin 00]) = (55 N)i F; = (41 N)(cos(—60°)i + sin(—60°)3) = (20.5 N)i — (35.5 N)j. "111 .1111 The resultant acceleration of the asteroid of mass m = 120 kg is therefore (27.7i +163)N+(55i)N + (20.5? —35.53)N 120kg a: =(0.86m/s2)i — (0.16m/s2)j. (b) The magnitude of the acceleration vector is la] =,/aj + aj = ,/(0.86 m/sz)2 + (—0.16 m/SZ)2 = 0.88 m/s2 . (c) The vector Ei makes an angle Bwith the +x axis, where 2 t9=tan_l i =tan‘1 =~ll°. a 0.86m/s2 15. (a) — (c) In all three cases the scale is not accelerating, which means that the two cords exert forces of equal magnitude on it. The scale reads the magnitude of e1ther of these forces. In each case the tension force of the cord attached to the-salaml must bethe same in magnitude as the weight of the salami because the salaml 1S not accelerating. Thus the scale reading is mg, where m is the mass of the salam1. Its value 1s (11.0 kg) (9.8 m/s2) = 108 N. 25. (a) The acceleration is (1:5: 20N =0.022m/sz. m 900kg (b) The distance traveled in 1 day (= 86400 s) is s = éaﬁ :5 (0.0222 m/sz) (86400s)2 = 8.3 x 107 m . (c) The speed it will be traveling is given by v = at = (0.0222 m/s2)(86400 s) = 1.9 ><103 m/s. ...
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• Spring '08
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