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A wooden pallet carrying a load of 600 kg rests on a wooden floor.

  1. A forklift driver decides to push it without lifting it. What force must be applied to just get the pallet moving?
  2. After a bit of time, the pallet begins to slide. How fast is the pallet moving after sliding under the same force you calculated in part a. for half a second?
  3. If the forklift stops pushing, how far does the pallet slide before coming to a stop?
in Grade 12 Physical Sciences by anonymous | 52 views

1 Answer

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Four forces are acting on the pallet: the downward pull of Earth's gravity, the normal force of the floor pushing up, the forward push of the forklift, and the backward resistance of friction. Weight and normal are equal throughout this example since the floor is level. Friction changes from static to kinetic — static friction initially since the pallet isn't moving initially, then kinetic friction once the pallet gets going. The push also changes from nothing to the value needed to get the pallet moving, then back to nothing after 0.5 seconds of motion.

 

    • To get the pallet started, the driver must push it with a force equal to the maximum static friction.

    P = fs = µsN = µsmg
    P = (0.28)(600 kg)(9.8 m/s2)
    P = 1646 N
    • Once the pallet starts moving, the coefficient of friction drops from its static value to its kinetic value.

      fk = µkN = µkmg
      fk = (0.17)(600 kg)(9.8 m/s2)
      fk = 1000 N

      But the forklift is still pushing with 1650 N of force. Thus we have a nonzero net force.

      F = P − fk
      F = 1646 N − 1000 N
      F = 646 N

      A net force causes acceleration.

      a = ∑F/m
      a = (646 N)/(600 kg)
      a = 1.08 m/s2

      Acceleration goes with a change in velocity.

      v = v0 + at
      v = (1.08 m/s2)(0.5 s)
      v = 0.54 m/s
    • Once the forklift stops pushing, kinetic friction becomes the net force. This net force will cause an acceleration opposite the direction of motion. When one vector is opposite another, one of the two needs to be negative. The convenient thing to do for this problem is to let friction be the negative one.

      a = ∑F/m = fk/m
      a = (−1000 N)/(600 kg)
      a = −1.67 m/s2

      Pick the appropriate equation of motion

      v2 = v02 + 2as

      Eliminate the zero term (final velocity), solve for distance, substitute, and calculate. Watch how the negative signs disappear. This has to happen. An object moving forward should be displaced forward.

      s = −v022as = −(0.54 m/s)22(−1.67 m/s2)∆s = 0.087 m 
by Diamond (49,543 points)

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