Solutions to Week 3 Assignment

 

  1. The vertical velocity is positive for t = 0s to t =10s. It is negative for t =14s to t = 20s. It is zero for t = 10s to t = 14s. So, the elevator reaches its highest location ata t =10s and remains there until t = 14s before it goes down. The elevator is at its highest location from t = 10s to t =14s.
  2. At t = 2.0s, vx is equal to the average speed, so vx = (6.0m - 4.0m)/(3.0s - 1.0s) = 1.0m/s.
  3. Use the definition of average acceleration
      aav = Dv/Dt
            = (vf - vi)/Dt
            = (0 - 28m/s in the direction of the car's travel)/4.0s
            = 7.0m/s2 in the direction opposite the car's velocity.
  4. Use the Newton's second law and solve for the mass.
       m = F/a = (0.375N)/(0.30m/s2) = 1.3kg.
  5. (a) Let south be the positive direction.
         vx - v0x = axDt
         ax = (vx - v0x)/Dt
             = (6.00m/s - 24m/s)/(9.00s)
             = - 2.00m/s2
         The acceleration is 2.00m/s2 north.
    (b) Dx = v0xt + 1/2at2 = (24.0m/s)(9.00s) + 1/2(-2.00m/s2)(9.00s)2 = 135m.
  6. (a) vx2 - v0x2 = 2axDx
        vx2 - 0 = 2(Fx/m)Dx
                  vx = +Ö(2FxDx/m)
                  vx = Ö2(6.4 x10-17N)(0.020m)/(9.109 x 10-31kg)  (speed is always positive)
                       = 1.7 x 106m/s.
    (b) x = 0 to 2.0cm:
         Dx1 = v0xt1 + 1/2axt12
                = 0 + 1/2axt12
           t12 = 2Dx1/ax
                = 2Dx1/(Fx/m)
           t1 = Ö2mDx1/Fx (t1 > 0, since the process begins at t = 0.)
         x =2.0cm to 47cm:
        Dx2 = vxt2
        t2 = Dx2/vx
           Find the total time
       t =  t1 +  t2
         = Ö2mDx1/Fx + Dx2/vx
         = Ö2(9.109 x 10-31kg)(0.020m)/(6.4 x10-17N) + (0.45m -0.20m)/(1.7 x 106m/s)
         = 290ns.
  7. (a) |Dy| = 1/2gt2
                = 1/2(9.8m/s2)(3.0s)2
                = 44m.
    (b) vy2 = -2gDy
         vy = Ö-2gDy
              = Ö-2(9.8m/s2)(-2.5m)
              = 7.0m/s
    (c) vy = gt = (9.8m/s2)(3.0s) = 29m/s.
  8. (a) Up is the positive direction.
         W'L = mL(g + ay)
         W'Lb = (mL + mb)(g + ay)
             Solve for mL. (Luke's mass) in the first equation, substitute for mL in the second, and solve for ay
    .    mL = W'L'/(g + ay)
         Substitute.
         W'Lb  = [W'L/(g + ay) +  mb](g + ay)
                  = W'L +  mb(g + ay)
         (W'Lb  - W'L  )/mb = g + ay
           ay  = (W'Lb  - W'L  )/mb - g
                = (1.200 x 103N - 0.960 x 103N)/20.0kg - 9.8mm/s2
                = 2.2m/s2
          So, a = 2.2m/s2 up.
    (b) W'L = WL(1 + ay/g)
          WL= W'L/(1 + ay/g)
               = 0.960 x 103N/(1 + 2.2m/s2/9.8m/s2)
               = 784N.