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

Info icon This preview shows pages 1–3. Sign up to view the full content.

Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern