CHAPTER 8

  1. Find the acceleration of the block.
       v2 = v02 + 2aDx;
      (4.0 m/s)2  = 0 + 2a(2.0 m), a = 4.0 m/s2.
      F1 = m1a = (20 kg)(4.0 m/s2) = 80 N.
      F -  T1= F -  F1 = m2a;
     100 N - 80 N = m2(4.0 m/s2), m = 5 kg.
  2. The constraint of acceleration is
      a1= -a2 = a
    For 2-kg block 
    F - T - fk1 - fk2 = m1a, fk1 = μk1(m1 + m2)g,  fk2 = μk2m1g.
    For 1-kg block 
    T - fk2 = m1a.
    Combine two equation
    F - fk1 - 2fk2  = (m1 + m2)a;
    F - μk1(m1 + m2)g - 2μk2m2g = (m1 + m2)a;
    a = [F - μkg(m1 + 3m2)]/(m1 + m2)
       = [20 N - (0.30)((9.8 m/s2)(2 kg + 3 x 1 kg)]/(2 kg + 1 kg)
       = 1.77 m/s2.
  3. a.  For the physics book
          SFx = - m1gsinθ - T - fk = m1a
          SFy = n - m1gcosθ = 0.fk = μkm1gcosθ.
          - m1gsinθ - T - μkm1gcosθ = m1a
         For the coffee cup
          T - m2g = m2a.
         Combining the equations we have
         -(m1sinθ + μkm1cosθ + m2)g  = (m1 + m2)a
          a = -[(m1sinθ + μkm1cosθ + m2)g]/(m1 + m2)
             = -{[(1.0 kg)sin20° + (0.20)(1.0 kg)cos20° + (0.5 kg)](9.8 m/s2)}/(1.0 kg + 0.5 kg)
             = - 6.73 m/s2
        Find the distance the book slides up
         v2 = v02 + 2aDx;
         0 = (3.0 m/s)2 + 2(-6.73 m/s2)Dx, Dx = 0.69 m.
    b. At the highest point for the book, at the instant when the book stops
        SFx = - m1gsinθ - T + fs = 0
        and for the cup
       T - m2g = 0.
       Then we have
      - m1gsinθ - m2g + fs = 0, 
       fs = m1gsinθ + m2g = [(1.0kg)sin20° + (0.5 kg)](9.8 m/s2) = 8.25 N > (0.50)(1.0 kg)cos20°(9.8 m/s2) = 4.6 N
     The book will slide back down.
        
  4. a. From the diagram

    For hamster: n1  - m1g = 0. n1 = m1g.
    For wedge:   n2 -  m2g  - n1= 0, n2 = m1g + m2g = (1000 g)(9.8m/s2) = 9.8 N.
    b. As shown in the diagram

    For hamster: SFy = n1 - m1gcos40° = 0, n1 = m1gcos40°.
    For the wedge: SFy = n2 - m2g - n1cos40° = 0.
    n2 = m2g + n1cos40° = (m2 + m1cos40°cos40°)g
        = (0.8 kg + 0.2 kgcos240°)(9.8 m/s2)
        = 8.99 N.