Exam II

Exam II will take place during the lab period on Thursday, March 20. As before, you will have the full three hours. The exam will consist of 6 to 8 problems and cover from Chapter 5 through Chapter 11.

Below are a few sample exam problems.

You are responsible for all material covered in lecture and homework assignments. DO NOT ASSUME that a subject not covered in the sample exam below will not show up on Tuesday's exam! During the exam, you will be provided with an equation sheet. You may download the Equation sheet. You will be allowed to use only the equations on the sheet, or others that you derive from those on the sheet. I will email that sheet to you. Use only that when trying the problems below. Resist the temptation to look at the answers, as that will better prepare you for the exam.

  1. A 3,000 kg flatbed railroad car on a horizontal track has a 2,000 kg crate sitting on it. If a 10,000 N force acts to the right on the car, what net force acts on the crate, assuming it doesn't slip on the flatbed?

    First, we find the acceleration of the car and crate from F = ma, a = 10,000N / (3,000+2,000) = 2 m/s2. So the the crate must have horizontal force F = ma = 2,000 x 2.0 = 4,000 N. (There is also the weight and normal force acting on the crate, but they cancel.)

  2. The uniform sign shown has a mass of 20 kg and width 1.0 m. (Assume the space between the sign and wall is negligible.) Find the tension in the wire and the force(s) acting on the wall support.

    Take the weight to be 200N acting at the center. In the FBD, the tension T acts along the wire. The force(s) at the wall are best drawn as a vertical (Fv) and a horizontal (Fh) component. Sums of torques about the wall support gives St = 0.5(200) - 1.0(Tsin37o), gives T = 167 N. Sums of forces will give the wall forces, SFy = Fv - 200 + 167sin37o, gives Fv = 100 N, and SFx = Fh - 167cos37o, gives Fh = 136N.

  3. A 50 kg skier starts from rest at the top of a frictionless 37o slope and slides100 m down the incline to level ground.
    a) What was her speed at the bottom?
    b) Once on the level, a constant frictional force stops her in a distance of 200 m. What is the coefficient of friction between her skis and the snow.

    You can use F = ma, but I suggest conservation of energy.
    a) Utop = Kbottom --> mg(100sin37) = 1/2mv2, so v = 34 m/s.
    b) WNC = Ffriction(200m)cos180 = DKE + DPE = 0-1/2(50)(34)2 + 0-0, so Ffriction = 147 N. But Ffriction = mkN = mkmg, so mk = 0.3

  4. An elevator (with passengers) has a mass of 700 kg and the counterweight has mass 400 kg.
    a) How much work is done by the motor in raising the elevator (and thus lowering the counterweight) by 50 m?
    b) What minimum power must the motor generate in order to do this in 12 seconds?

    Wmotor = DK + DU = 0 - 0 + 700kg(9.8)(50-0) +400(9.8)(0-50) = 1.47 x 105 J
    Power= Work/time = 1.47 x 105 J / 12 = 12.3 kW.

  5. A 20 kg coyote jumps off a 30 m cliff with his new ACME Springy Shoes. At the bottom, the springs compress 20 cm in bringing him to rest.
    a) What is the spring constant for the shoes?
    b) What average force did the shoes exert on him over the 20 cm?

    a) Conservation of energy --> Ug at the top = Uspring at the bottom. mgy = 1/2kx2 --> k = 2.94 x 105 N/m. (x = 0.2 m, but y = 30.0 or 30.2 m, depending upon how your interpret the problem.)
    b) Work = Fave(.2 m) = either Ug at the top or Uspring at the bottom. --> Fave = 29,400 N ... a squashed coyote!

  6. Your 80 kg friend is running W to E at 5 m/s. You (70 kg) make a flying tackle with speed 10 m/s running S to N. You and your friend stick together.
    a) What is the velocity of you and your friend just after the tackle?
    b) Was KE conserved in this collision? If not, how much was lost or gained?

    a) In the x-direction 80kg(5m/s) + 70kg(0m/s) = (80+70)v'x. --> v'x = 2.7 m/s.
    In the y-direction 80kg(0m/s) +70kg(10m/s) = (80+70)v'y --> v'y = 4.7 m/s
    b) KE is lost. Find Kf - Ki = 1/2(150)(2.72 + 4.72) - [1/2(80)52 + 1/2(70)102]

  7. A 200 gm cue ball traveling at 3 m/s makes a head-on elastic collision with a 150 gm billiard ball initially at rest. What is the final velocity of each?

    Conservation of p --> 200(3) + 150(0) = 200v'1 + 150v'2 and velocity in = velocity out --> 3 - 0 = v'2 - v'1. Substitute to find v'1 = .43 m/s and v'2 = 3.43 m/s

  8. A stone grinding wheel is a disc of radius 30.0 cm and mass 50 kg. It has an inner pulley wheel with radius 15.0 cm. A rope of length 10.0 m is wrapped around the pulley. With what force (T) must the rope be pulled to get the wheel from rest up to an operational speed of 10 rev/s?

    We need to know what force (T) will give the torque required to accelerate the wheel up to speed. First, we need to find the angular acceleration. q = s/r = 10.0 m / 0.15 = 67 radians and w = 2p 10 rev/s = 63 rad/s. Then use w2 =wo2 + 2a q --> a = 30 rad/s2. The torque required to do this is St = (0.15)T = Ia = 1/2(50)(0.3)2(30), gives T = 450 N.

  9. A bowling ball of mass 5 kg and radius 12.0 cm rolls from rest without slipping down the inner surface of a semi-circular bowl of radius 1.0 m. If the ball starts from rest where the bowl surface is vertical as shown, a) what is its speed at the bottom of the bowl and b) what force must the bowl exert on ball at the bottom?

    a) We use conservation of energy, remembering that we have rotational K involved. If we take the top as zero potential energy, then 0 = 1/2(5)v2 + 1/2[2/5(5)(0.12)2]w2 + 5(9.8)(-0.88). Note the vertical distance the ball's c.o.m. moves through is 1.0 - 0.12 = 0.88! We also need to substitute w = v/(0.12m) and we get v = 3.51 m/s.
    b) At the bottom, SFy = N - mg = mv2/r, where again r = 1.0 - 0.12 = 0.88 m, so N = 119 N. Note that there is no tangential force as the tangential acceleration is zero at the bottom. That would not be true elsewhere. For extra practice, see if you can figure out direction of the tangential frictional forces on either side of the bottom by thinking about what is required for the ball to roll without slipping along the bowl.

    ANS - On the initial left side, there would be a tangential frictional force in the opposite direction the ball is moving (upward), while on the right side (after passing the bottom point) it would be in the direction the ball is moving (also upward)!