2005 SPRING PHYSICS 220A MIDTERM EXAMINATION 1

I.  Short Problem (2 points for each)
Please fill the blanks with your answers, you don not need to give the processes of solutions.

  1. Complete the motion diagram as shown.
  2. The particle’s position at t = 3 s is 11 m, as shown in its velocity-versus-time graph.
  3. The y-component of vector r = (100 m, 45˚ below +x-axis) is –70.7 m.
  4. A wooden block is held against a wall by your hand.  The forces exerting on the block are weight, your push force, normal force from the wall, and the static friction between the block and the wall.
  5. 4000-kg truck is parked on a 15˚ slope. The friction force on the truck is 1.0 x 104 N.

II  Solve the following problems. (5 points for each problem)
    You need to give a solution process for each problem.

  1. A car traveling 55 km/h slows down at a constant 0.50 m/s2. Calculate (a) the distance the car coasts before it stops, (b) the time it takes to stop, and (c) the distance it travels during the first and fifth seconds.
    Solution.

    (a) Find the distance from
         vf2 =  vi2 + 2aΔx; 
        0 = [(55 km/h)(1000m/km)/(3600 s/h)]2 + 2(-0.50 m/s2)Δx,  Δx = 233.4 m.
    (b) Find the time.
        vf = vi + aΔt;
        0 = (55 km/h)(1000m/km)/(3600 s/h) + (-0.50 m/s2)Δt, Δt = 30.6 s.
    (c) The distance during the first second is
         Δx = viΔt + 1/2a(Δt)2 = (15.3 m/s)(1.0 s) + 1/2(-0.50 m/s2)(1.0 s)2 = 15 m.
         The distance during the fifth second is
         Δx = x5 - x1 = [(15.3 m/s)(5.0 s) + 1/2(-0.50 m/s2)(5.0 s)2] - 15 m = 60.1 m.
  2. A skier, whose mass is 75 kg, takes off down a 50-m-high, 10˚ slope on his skis. His skis have a thrust of 200 N. The skier’s speed at the bottom is 40 m/s. What is the coefficient of kinetic friction of his skis on snow?
    Solution. The motion diagram and free-body diagram are as shown.

    Find the ski's acceleration from kinematics
    vf2 =  vi2 + 2axΔx;
    (40 m/s)2 = 0 + 2ax(50 m/sin10°), ax= 2.8 m/s2.
    Use Newton's second law
    ΣFx = F + mgsin θ - f = max;
    ΣFy = n - mgcos θ = may = 0;
    we have f = μkn. combine above equations we have
    max = F + mgsin θ - μkmgcos θ)
    μk = [F + mgsin θ - max]/(mgcos θ)
        = [200 N + (75 kg)(9.8 m/s2)sin10° - (75 kg)(2.8 m/s2)]/[(75 kg)(9.8 m/s2)cos10°]
        = 0.163.
  3. A projectile is fired with a initial speed of 51.2 m/s at an angle of 44.5˚ above the horizontal on a long flat firing range. Determine (a) the maximum height reached by the projectile, (b) the total time in the air, and (c) the total horizontal distance covered (that is, the range).
    Solution.
    (a) At the highest point, the vertical velocity vy = 0. We find the maximum height h from
          vy2 = viy2 + 2 ayh;
          0 = [(51.2 m/s)sin44.5°]2 + 2(-    9.8 m/s2)h, h = 65.7 m.
    (b) Because the projectile returns to the same elevation, we have
         y = yi + viyt + 1/2ayt2;
        0 = 0 + (51.2 m/s)sin44.5°t + 1/2(9.8 m/s2)t2, t = 0, 7.32 s.
    (c) We find the horizontal distance from
        x = vixt = (51.2 m/s)cos44.5°(7.32 s) = 267 m.